R help - Jul 2010 - After writing data in MMF using SEXP structure, can i reference in R?

Hi all,


   After writing data in MMF(Memory Map File) using SEXP structure, can i
   reference in R?
   If input data is larger than 2GB, Can i reference MMF Data in R?


   my work environment :
   R version : 2.11.1
   OS : WinXP Pro sp3

   Thanks and best regards.

   Park, Young-Ju

   from Korea.

     ---------[ ???????? ???????? ???????? ]----------
     ???????? : R-help Digest, Vol 89, Issue 25
     ???????? : 2010???? 7???? 25???? ????????????, 19???? 00???? 07???? +0900
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     Today's Topics:
     1. Re: how to calculate the product of every two elements in two
     vectors (Dennis Murphy)
     2. Re: Question regarding panel data diagnostic
     (Setlhare Lekgatlhamang)
     3. Re: Question regarding panel data diagnostic
     (Setlhare Lekgatlhamang)
     4. Re: Question regarding panel data diagnostic
     (Setlhare Lekgatlhamang)
     5. Re: Trouble retrieving the second largest value from each row
     of a data.frame (David Winsemius)
     6. local polynomial with differnt kernal functions
     (assaedi76 assaedi76)
     7. Weights in mixed models (David R.)
     8. Re: Odp: Help me with prediction in linear model
     (Research student)
     9. Re: union data in column (Hadley Wickham)
     10. Re: UseR! 2010 - my impressions (Frank E Harrell Jr)
     11. Re: , Updating Table (Charles C. Berry)
     12. Re: , Updating Table (Duncan Murdoch)
     13. Re: glm - prediction of a factor with several levels (Ben Bolker)
     14. Doubt about a population competition function
     (Bruno Bastos Gon?alves)
     15. Doubt about a population competition function (Gmail)
     16. Book on R's Programming Language (Matt Stati)
     17. Re: how to calculate the product of every two elements in two
     vectors (Henrique Dallazuanna)
     18. Re: Book on R's Programming Language (Joshua Wiley)
     19. Re: how to calculate the product of every two elements in two
     vectors (Gabor Grothendieck)
     20. Re: Book on R's Programming Language (Joseph Magagnoli)
     21. Re: Constrain density to 0 at 0? (Greg Snow)
     22. matest function for multiple factors (shabnam k)
     23. Re: How to deal with more than 6GB dataset using R? (Greg Snow)
     24. Using R to fill ETM+ data gaps? (Abdi, Abdulhakim)
     25. How to generate a sequence of dates without hardcoding the
     year (Felipe Carrillo)
     26. Re: How to generate a sequence of dates without hardcoding
     the year (Henrique Dallazuanna)
     27. Re: Trouble retrieving the second largest value from each row
     of a data.frame ([4]mpward at illinois.edu)
     28. Re: (no subject) (Paul Smith)
     29. Re: How to generate a sequence of dates without hardcoding
     the year (jim holtman)
     30. Re: glm - prediction of a factor with several levels (zachmohr)
     31. Re: Trouble retrieving the second largest value from each row
     of a data.frame (David Winsemius)
     32. Re: Trouble retrieving the second largest value from each row
     of a data.frame (Joshua Wiley)
     33. Re: Trouble retrieving the second largest value from each row
     of a data.frame (David Winsemius)
     34. Re: Trouble retrieving the second largest value from each row
     of a data.frame (David Winsemius)
     35. c-statiscs 95% CI for cox regression model (paaventhan jeyaganth)
     36. Re: UseR! 2010 - my impressions (Dirk Eddelbuettel)
     37. Re: c-statiscs 95% CI for cox regression model
     (Frank E Harrell Jr)
     38. Equivalent to go-to statement (Michael Haenlein)
     39. Outlier Problem in Survreg Function (Vipul Agarwal)
     40. Re: Equivalent to go-to statement (Gabor Grothendieck)
     ----------------------------------------------------------------------
     Message: 1
     Date: Sat, 24 Jul 2010 03:02:47 -0700
     From: Dennis Murphy <[5]djmuser at gmail.com>
     To: aegea <[6]gcheer3 at gmail.com>
     Cc: [7]r-help at r-project.org
     Subject: Re: [R] how to calculate the product of every two elements in
     two vectors
     Message-ID:
     <AANLkTinUJjoCiG47ptPq5Eo_fXYZXuQbUzB=+[8]KLCO6QX at mail.gmail.com>
     Content-Type: text/plain
     as.vector(t(outer(A, B)))
     [1] 9 10 11 12 18 20 22 24 27 30 33 36
     HTH,
     Dennis
     On Fri, Jul 23, 2010 at 8:11 AM, aegea <[9]gcheer3 at gmail.com>
wrote:
     >
     > Thanks in advance!
     >
     > A=c(1, 2, 3)
     > B=c (9, 10, 11, 12)
     >
     > I want to get C=c(1*9, 1*10, 1*11, 1*12, ....., 3*9, 3*10, 3*11,
3*12)?
     > C is still a vector with 12 elements
     > Is there a way to do that?
     > --
     > View this message in context:
     >
     [10]http://r.789695.n4.nabble.com/how-to-calculate-the-product-of-every-tw
     o-elements-in-two-vectors-tp2300299p2300299.html
     > Sent from the R help mailing list archive at Nabble.com.
     >
     > ______________________________________________
     > [11]R-help at r-project.org mailing list
     > https://stat.ethz.ch/mailman/listinfo/r-help
     > PLEASE do read the posting guide
     > [12]http://www.R-project.org/posting-guide.html
     > and provide commented, minimal, self-contained, reproducible code.
     >
     [[alternative HTML version deleted]]
     ------------------------------
     Message: 2
     Date: Sat, 24 Jul 2010 12:54:11 +0200
     From: "Setlhare Lekgatlhamang" <[13]SetlhareL at bob.bw>
     To: "amatoallah ouchen" <[14]at.ouchen at gmail.com>,
     <[15]r-help at r-project.org>
     Subject: Re: [R] Question regarding panel data diagnostic
     Message-ID:
     <[16]25D1D72D6E19D144AB813C9C582E16CF03F7EA27 at BOB-EXCHANGE.bob.bw>
     Content-Type: text/plain; charset="iso-8859-1"
     My thought is this:
     It depends on what you have in the panel. Are your data cross-section data
     observed over ten years for, say, 3 countries (or regions within the same
     country)? If so, yes you can perform integration properties (what people
     usually call unit root test) and then test for cointegration. But if the
     data are quarterly or monthly, these techniques are not relevant.
     Hope this helps.
     Lexi
     -----Original Message-----
     From: [17]r-help-bounces at r-project.org
     [mailto:[18]r-help-bounces at r-project.org] On Behalf Of amatoallah ouchen
     Sent: Friday, July 23, 2010 12:18 AM
     To: [19]r-help at r-project.org
     Subject: [R] Question regarding panel data diagnostic
     Good day R-listers,
     I'm currently working on a panel data analysis (N=17, T=5), in order
     to check for the spurious regression problem, i have to test for
     stationarity but i've read somewhere that i needn't to test for it
as
     my T<10 , what do you think? if yes is there any other test i have
     to perform in such case (a kind of cointegration test for small T?)
     Any hint would be highly appreciated.
     Ama.
     *
     * For searches and help try:
     * [20]http://www.stata.com/help.cgi?search
     * [21]http://www.stata.com/support/statalist/faq
     * [22]http://www.ats.ucla.edu/stat/stata/
     ______________________________________________
     [23]R-help at r-project.org mailing list
     https://stat.ethz.ch/mailman/listinfo/r-help
     PLEASE do read the posting guide
     [24]http://www.R-project.org/posting-guide.html
     and provide commented, minimal, self-contained, reproducible code.
     DISCLAIMER:\ Sample Disclaimer added in a VBScript.\ ...{{dropped:3}}
     ------------------------------
     Message: 3
     Date: Sat, 24 Jul 2010 13:00:05 +0200
     From: "Setlhare Lekgatlhamang" <[25]SetlhareL at bob.bw>
     To: "Setlhare Lekgatlhamang" <[26]SetlhareL at bob.bw>,
"amatoallah ouchen"
     <[27]at.ouchen at gmail.com>, <[28]r-help at r-project.org>
     Subject: Re: [R] Question regarding panel data diagnostic
     Message-ID:
     <[29]25D1D72D6E19D144AB813C9C582E16CF03F7EA28 at BOB-EXCHANGE.bob.bw>
     Content-Type: text/plain; charset="us-ascii"
     Let me correct an omission in my response below. The last sentence
     should read "But if the data are 10 quarterly or monthly values, these
     techniques are not relevant".
     Cheers
     Lexi
     -----Original Message-----
     From: [30]r-help-bounces at r-project.org
     [mailto:[31]r-help-bounces at r-project.org]
     On Behalf Of Setlhare Lekgatlhamang
     Sent: Saturday, July 24, 2010 12:54 PM
     To: amatoallah ouchen; [32]r-help at r-project.org
     Subject: Re: [R] Question regarding panel data diagnostic
     My thought is this:
     It depends on what you have in the panel. Are your data cross-section
     data observed over ten years for, say, 3 countries (or regions within
     the same country)? If so, yes you can perform integration properties
     (what people usually call unit root test) and then test for
     cointegration. But if the data are quarterly or monthly, these
     techniques are not relevant.
     Hope this helps.
     Lexi
     -----Original Message-----
     From: [33]r-help-bounces at r-project.org
     [mailto:[34]r-help-bounces at r-project.org]
     On Behalf Of amatoallah ouchen
     Sent: Friday, July 23, 2010 12:18 AM
     To: [35]r-help at r-project.org
     Subject: [R] Question regarding panel data diagnostic
     Good day R-listers,
     I'm currently working on a panel data analysis (N=17, T=5), in order
     to check for the spurious regression problem, i have to test for
     stationarity but i've read somewhere that i needn't to test for it
as
     my T<10 , what do you think? if yes is there any other test i have
     to perform in such case (a kind of cointegration test for small T?)
     Any hint would be highly appreciated.
     Ama.
     *
     * For searches and help try:
     * [36]http://www.stata.com/help.cgi?search
     * [37]http://www.stata.com/support/statalist/faq
     * [38]http://www.ats.ucla.edu/stat/stata/
     ______________________________________________
     [39]R-help at r-project.org mailing list
     https://stat.ethz.ch/mailman/listinfo/r-help
     PLEASE do read the posting guide
     [40]http://www.R-project.org/posting-guide.html
     and provide commented, minimal, self-contained, reproducible code.
     DISCLAIMER:\ Sample Disclaimer added in a VBScript.\ ......{{dropped:14}}
     ------------------------------
     Message: 4
     Date: Sat, 24 Jul 2010 13:01:24 +0200
     From: "Setlhare Lekgatlhamang" <[41]SetlhareL at bob.bw>
     To: "amatoallah ouchen" <[42]at.ouchen at gmail.com>,
     <[43]r-help at r-project.org>
     Subject: Re: [R] Question regarding panel data diagnostic
     Message-ID:
     <[44]25D1D72D6E19D144AB813C9C582E16CF03F7EA29 at BOB-EXCHANGE.bob.bw>
     Content-Type: text/plain; charset="us-ascii"
     Let me correct an omission in my response below. The last sentence
     should read "But if the data are 10 quarterly or monthly values, these
     techniques are not relevant".
     Cheers
     Lexi
     -----Original Message-----
     From: [45]r-help-bounces at r-project.org
     [mailto:[46]r-help-bounces at r-project.org]
     On Behalf Of Setlhare Lekgatlhamang
     Sent: Saturday, July 24, 2010 12:54 PM
     To: amatoallah ouchen; [47]r-help at r-project.org
     Subject: Re: [R] Question regarding panel data diagnostic
     My thought is this:
     It depends on what you have in the panel. Are your data cross-section
     data observed over ten years for, say, 3 countries (or regions within
     the same country)? If so, yes you can perform integration properties
     (what people usually call unit root test) and then test for
     cointegration. But if the data are quarterly or monthly, these
     techniques are not relevant.
     Hope this helps.
     Lexi
     -----Original Message-----
     From: [48]r-help-bounces at r-project.org
     [mailto:[49]r-help-bounces at r-project.org]
     On Behalf Of amatoallah ouchen
     Sent: Friday, July 23, 2010 12:18 AM
     To: [50]r-help at r-project.org
     Subject: [R] Question regarding panel data diagnostic
     Good day R-listers,
     I'm currently working on a panel data analysis (N=17, T=5), in order
     to check for the spurious regression problem, i have to test for
     stationarity but i've read somewhere that i needn't to test for it
as
     my T<10 , what do you think? if yes is there any other test i have
     to perform in such case (a kind of cointegration test for small T?)
     Any hint would be highly appreciated.
     Ama.
     *
     * For searches and help try:
     * [51]http://www.stata.com/help.cgi?search
     * [52]http://www.stata.com/support/statalist/faq
     * [53]http://www.ats.ucla.edu/stat/stata/
     ______________________________________________
     [54]R-help at r-project.org mailing list
     https://stat.ethz.ch/mailman/listinfo/r-help
     PLEASE do read the posting guide
     [55]http://www.R-project.org/posting-guide.html
     and provide commented, minimal, self-contained, reproducible code.
     DISCLAIMER:\ Sample Disclaimer added in a VBScript.\ ......{{dropped:14}}
     ------------------------------
     Message: 5
     Date: Sat, 24 Jul 2010 08:40:05 -0400
     From: David Winsemius <[56]dwinsemius at comcast.net>
     To: <[57]mpward at illinois.edu>
     Cc: [58]r-help at r-project.org
     Subject: Re: [R] Trouble retrieving the second largest value from each
     row of a data.frame
     Message-ID: <[59]D09340C5-3B64-47FA-A168-8EA347F79747 at comcast.net>
     Content-Type: text/plain; charset=US-ASCII; format=flowed; delsp=yes
     On Jul 23, 2010, at 9:20 PM, <[60]mpward at illinois.edu> wrote:
     > I have a data frame with a couple million lines and want to retrieve
     > the largest and second largest values in each row, along with the
     > label of the column these values are in. For example
     >
     > row 1
     > strongest=-11072
     > secondstrongest=-11707
     > strongestantenna=value120
     > secondstrongantenna=value60
     >
     > Below is the code I am using and a truncated data.frame. Retrieving
     > the largest value was easy, but I have been getting errors every way
     > I have tried to retrieve the second largest value. I have not even
     > tried to retrieve the labels for the value yet.
     >
     > Any help would be appreciated
     > Mike
     Using Holtman's extract of your data with x as the name and the order
     function to generate an index to names of the dataframe:
     > t(apply(x, 1, sort, decreasing=TRUE)[1:3, ])
     [, 1] [, 2] [, 3]
     1 -11072 -11707 -12471
     2 -11176 -11799 -12838
     3 -11113 -11778 -12439
     4 -11071 -11561 -11653
     5 -11067 -11638 -12834
     6 -11068 -11698 -12430
     7 -11092 -11607 -11709
     8 -11061 -11426 -11665
     9 -11137 -11736 -12570
     10 -11146 -11779 -12537
     Putting it all together:
     matrix( paste( t(apply(x, 1, sort, decreasing=TRUE)[1:3, ]),
     names(x)[ t(apply(x, 1, order, decreasing=TRUE)
     [1:3, ]) ]),
     ncol=3)
     [, 1] [, 2] [, 3]
     [1, ] "-11072 value120" "-11707 value60" "-12471
value180"
     [2, ] "-11176 value120" "-11799 value180" "-12838
value0"
     [3, ] "-11113 value120" "-11778 value60" "-12439
value180"
     [4, ] "-11071 value120" "-11561 value240" "-11653
value60"
     [5, ] "-11067 value120" "-11638 value180" "-12834
value0"
     [6, ] "-11068 value0" "-11698 value60" "-12430
value120"
     [7, ] "-11092 value120" "-11607 value240" "-11709
value180"
     [8, ] "-11061 value120" "-11426 value240" "-11665
value60"
     [9, ] "-11137 value120" "-11736 value60" "-12570
value180"
     [10, ] "-11146 value300" "-11779 value0" "-12537
value180"
     --
     David.
     >
     >
     >> data<-data.frame(value0, value60, value120, value180, value240,
     value300)
     >> data
     > value0 value60 value120 value180 value240 value300
     > 1 -13007 -11707 -11072 -12471 -12838 -13357
     > 2 -12838 -13210 -11176 -11799 -13210 -13845
     > 3 -12880 -11778 -11113 -12439 -13089 -13880
     > 4 -12805 -11653 -11071 -12385 -11561 -13317
     > 5 -12834 -13527 -11067 -11638 -13527 -13873
     > 6 -11068 -11698 -12430 -12430 -12430 -12814
     > 7 -12807 -14068 -11092 -11709 -11607 -13025
     > 8 -12770 -11665 -11061 -12373 -11426 -12805
     > 9 -12988 -11736 -11137 -12570 -13467 -13739
     > 10 -11779 -12873 -12973 -12537 -12973 -11146
     >> #largest value in the row
     >> strongest<-apply(data, 1, max)
     >>
     >>
     >> #second largest value in the row
     >> n<-function(data)(1/(min(1/(data[1, ]-max(data[1, ]))))+
     >> (max(data[1, ])))
     >> secondstrongest<-apply(data, 1, n)
     > Error in data[1, ] : incorrect number of dimensions
     >>
     >
     > ______________________________________________
     > [61]R-help at r-project.org mailing list
     > https://stat.ethz.ch/mailman/listinfo/r-help
     > PLEASE do read the posting guide
     [62]http://www.R-project.org/posting-guide.html
     > and provide commented, minimal, self-contained, reproducible code.
     ------------------------------
     Message: 6
     Date: Sat, 24 Jul 2010 04:25:10 -0700 (PDT)
     From: assaedi76 assaedi76 <[63]assaedi76 at yahoo.com>
     To: [64]r-help at r-project.org
     Subject: [R] local polynomial with differnt kernal functions
     Message-ID: <[65]853644.1608.qm at web45210.mail.sp1.yahoo.com>
     Content-Type: text/plain
     Hi, R users
     ???br> I need to use the function (locpoly) to fit a local poynomial
     regression model, The defult for kernal function is " normal" ,
but???I need
     to use different kernal functions such as :Uniform, Triangular,
     Epanechnikov, ......
     Could someone help me define these functions to fit local polynomial
     regression model?.
     Email:[66]assaedi76 at yahoo.com
     ???br> ???br> Thanks alot
     [[alternative HTML version deleted]]
     ------------------------------
     Message: 7
     Date: Sat, 24 Jul 2010 04:14:44 -0700 (PDT)
     From: "David R." <[67]drbn at yahoo.com>
     To: [68]r-help at r-project.org
     Subject: [R] Weights in mixed models
     Message-ID: <[69]217686.73973.qm at web113215.mail.gq1.yahoo.com>
     Content-Type: text/plain; charset=iso-8859-1
     Hello everyone,
     I wonder if sample size?can be used as weight? in a weighted mixed model.
     Or
     should I use just the inverse of the variance?
     For example, the class'lmer' in the 'lme4' package?have an
option
     'weights'
     (as in the class 'lm' of 'stats'). In the help of lme4
there is an example
     using
     'herd size' as weight in a mixed model.
     ?
     ?So, if I have a measurement data (eg height) of 10 groups (sample size
     ranging
     from?30 to 3000 individuals?for each group) can?I??use this number (N,
     sample
     size) in the 'weights' option? Is this wrong?
     ?
     Finally, what to do if the results (coefficients) of weighing by
'inverse
     of the
     variance' or by 'sample size' are very different, even
opposite?
     Thank you very much in advance
     David
     ------------------------------
     Message: 8
     Date: Sat, 24 Jul 2010 02:48:11 -0700 (PDT)
     From: Research student <[70]vijayamahantesh_s at dell.com>
     To: [71]r-help at r-project.org
     Subject: Re: [R] Odp: Help me with prediction in linear model
     Message-ID: <[72]1279964891930-2300991.post at n4.nabble.com>
     Content-Type: text/plain; charset=us-ascii
     Thanks Murphy and pikal,
     I need another help, for fitting first fourier transformation , i used
     following thing .Please advise on this
     beer_monthl has 400+ records
     EXample:
     > head(beer_monthly)
     beer
     1 93.2
     2 96.0
     3 95.2
     4 77.1
     5 70.9
     6 64.8
     time<-seq(1956, 1995.2, length=length(beer_monthly))
     sin.t<-sin(2*pi*time)
     cos.t<-cos(2*pi*time)
     beer_fit_fourier=lm(beer_monthly[, 1]~poly(time, 2)+sin.t+cos.t) #this is
     not
     working
     beer_fit_fourier=lm(beer_monthly[, 1]~time+time2+sin.t+cos.t) #it is
     working
     #prediction is not workinng
     tpred_four <- data.frame(time = seq(1995, 1998, length = 20))
     predict(beer_fit_fourier, newdata = tpred_four)
     Is there any way to fit first fourier frequency ,
     Please assist.
     Thanks in advance
     --
     View this message in context:
     [73]http://r.789695.n4.nabble.com/Help-me-with-prediction-in-linear-model-
     tp2297313p2300991.html
     Sent from the R help mailing list archive at Nabble.com.
     ------------------------------
     Message: 9
     Date: Sat, 24 Jul 2010 07:53:23 -0500
     From: Hadley Wickham <[74]hadley at rice.edu>
     To: Jeff Newmiller <[75]jdnewmil at dcn.davis.ca.us>
     Cc: [76]r-help at r-project.org, Fahim Md <[77]fahim.md at gmail.com>
     Subject: Re: [R] union data in column
     Message-ID:
     <AANLkTi=eA4KHr2q7fija+qGbTYnHJPLrLLHgw25+Ki=[78]z at mail.gmail.com>
     Content-Type: text/plain; charset=ISO-8859-1
     On Sat, Jul 24, 2010 at 2:23 AM, Jeff Newmiller
     <[79]jdnewmil at dcn.davis.ca.us> wrote:
     > Fahim Md wrote:
     >>
     >> Is there any function/way to merge/unite the following data
     >>
     >> ?GENEID ? ? ?col1 ? ? ? ? ?col2 ? ? ? ? ? ? col3 ? ? ? ? ? ? ?
?col4
     >> ?G234064 ? ? ? ? 1 ? ? ? ? ? ? 0 ? ? ? ? ? ? ? ? ?0 ? ? ? ? ? ? ?
? ? 0
     >> ?G234064 ? ? ? ? 1 ? ? ? ? ? ? 0 ? ? ? ? ? ? ? ? ?0 ? ? ? ? ? ? ?
? ? 0
     >> ?G234064 ? ? ? ? 1 ? ? ? ? ? ? 0 ? ? ? ? ? ? ? ? ?0 ? ? ? ? ? ? ?
? ? 0
     >> ?G234064 ? ? ? ? 0 ? ? ? ? ? ? 1 ? ? ? ? ? ? ? ? ?0 ? ? ? ? ? ? ?
? ? 0
     >> ?G234065 ? ? ? ? 0 ? ? ? ? ? ? 1 ? ? ? ? ? ? ? ? ?0 ? ? ? ? ? ? ?
? ? 0
     >> ?G234065 ? ? ? ? 0 ? ? ? ? ? ? 1 ? ? ? ? ? ? ? ? ?0 ? ? ? ? ? ? ?
? ? 0
     >> ?G234065 ? ? ? ? 0 ? ? ? ? ? ? 1 ? ? ? ? ? ? ? ? ?0 ? ? ? ? ? ? ?
? ? 0
     >> ?G234065 ? ? ? ? 0 ? ? ? ? ? ? 0 ? ? ? ? ? ? ? ? ?1 ? ? ? ? ? ? ?
? ? 0
     >> ?G234065 ? ? ? ? 0 ? ? ? ? ? ? 0 ? ? ? ? ? ? ? ? ?1 ? ? ? ? ? ? ?
? ? 0
     >> ?G234065 ? ? ? ? 0 ? ? ? ? ? ? 0 ? ? ? ? ? ? ? ? ?0 ? ? ? ? ? ? ?
? ? 1
     >>
     >>
     >> into
     >> GENEID ? ? ?col1 ? ? ? ? ?col2 ? ? ? ? ? ? col3 ? ? ? ? ? ? ?
?col4
     >> ?G234064 ? ? ? ? 1 ? ? ? ? ? ? 1 ? ? ? ? ? ? ? ? ?0 ? ? ? ? ? ? ?
? ? 0
     >> // 1 appears in col1 and col2 above, rest are zero
     >> ?G234065 ? ? ? ? 0 ? ? ? ? ? ? 1 ? ? ? ? ? ? ? ? ?1 ? ? ? ? ? ? ?
? ? 1
     >> // 1 appears in col2 , 3 and 4 above.
     >>
     >>
     >> Thank
     >
     > Warning on terminology: there is a "merge" function in R
that lines up
     rows
     > from different tables to make a new set of longer rows (more columns).
     The
     > usual term for combining column values from multiple rows is
     "aggregation".
     >
     > In addition to the example offered by Jim Holtzman, here are some
other
     > options in no particular order:
     >
     > x <- read.table(textConnection(" GENEID col1 col2 col3 col4
     > G234064 1 0 0 0
     > G234064 1 0 0 0
     > G234064 1 0 0 0
     > G234064 0 1 0 0
     > G234065 0 1 0 0
     > G234065 0 1 0 0
     > G234065 0 1 0 0
     > G234065 0 0 1 0
     > G234065 0 0 1 0
     > G234065 0 0 0 1
     > "), header=TRUE, as.is=TRUE, row.names=NULL)
     > closeAllConnections()
     >
     > # syntactic repackaging of Jim's basic approach
     > library(plyr)
     > ddply( x, .(GENEID), function(df)
     > {with(as.integer(c(col1=any(col1), col2=any(col2), col3=any(col3),
     col4=any(col4))))}
     > )
     You can do this a little more succinctly with colwise:
     any_1 <- function(x) as.integer(any(x))
     ddply(x, "GENEID", numcolwise(any_1))
     Hadley
     --
     Assistant Professor / Dobelman Family Junior Chair
     Department of Statistics / Rice University
     [80]http://had.co.nz/
     ------------------------------
     Message: 10
     Date: Sat, 24 Jul 2010 08:55:01 -0500
     From: Frank E Harrell Jr <[81]f.harrell at Vanderbilt.Edu>
     To: Ravi Varadhan <[82]rvaradhan at jhmi.edu>
     Cc: "[83]r-help at r-project.org" <[84]r-help at
r-project.org>
     Subject: Re: [R] UseR! 2010 - my impressions
     Message-ID: <[85]4C4AF0B5.6070300 at vanderbilt.edu>
     Content-Type: text/plain; charset="ISO-8859-1"; format=flowed
     On 07/23/2010 06:50 PM, Ravi Varadhan wrote:
     > Dear UseRs!,
     >
     > Everything about UseR! 2010 was terrific! I really mean
"everything" -
     the tutorials, invited talks, kaleidoscope sessions, focus sessions,
     breakfast, snacks, lunch, conference dinner, shuttle services, and the
     participants. The organization was fabulous. NIST were gracious hosts, and
     provided top notch facilities. The rousing speech by Antonio Possolo, who
     is the chief of Statistical Engineering Division at NIST, set the tempo
     for the entire conference. Excellent invited lectures by Luke Tierney,
     Frank Harrell, Mark Handcock, Diethelm Wurtz, Uwe Ligges, and Fritz
     Leisch. All the sessions that I attended had many interesting ideas and
     useful contributions. During the whole time that I was there, I could not
     help but get the feeling that I am a part of something great.
     >
     > Before I end, let me add a few words about a special person. This
     conference would not have been as great as it was without the tireless
     efforts of Kate Mullen. The great thing about Kate is that she did so much
     without ever hogging the limelight. Thank you, Kate and thank you NIST!
     >
     > I cannot wait for UseR!2011!
     >
     > Best,
     > Ravi.
     >
     > ____________________________________________________________________
     >
     > Ravi Varadhan, Ph.D.
     > Assistant Professor,
     > Division of Geriatric Medicine and Gerontology
     > School of Medicine
     > Johns Hopkins University
     >
     > Ph. (410) 502-2619
     > email: [86]rvaradhan at jhmi.edu
     I want to echo what Ravi said. The talks were terrific (thanks to the
     program committee and the speakers) and Kate Mullen and her team did an
     extraordinary job in putting the conference together and running it. I
     am proud to have been a part of it. Thank you all!
     Frank
     --
     Frank E Harrell Jr Professor and Chairman School of Medicine
     Department of Biostatistics Vanderbilt University
     ------------------------------
     Message: 11
     Date: Sat, 24 Jul 2010 08:25:59 -0700
     From: "Charles C. Berry" <[87]cberry at tajo.ucsd.edu>
     To: Marcus Liu <[88]marcusliu667 at yahoo.com>
     Cc: [89]r-help at r-project.org
     Subject: Re: [R] , Updating Table
     Message-ID: <[90]Pine.LNX.4.64.1007240817250.21422 at tajo.ucsd.edu>
     Content-Type: text/plain; charset="x-unknown";
Format="flowed"
     On Fri, 23 Jul 2010, Marcus Liu wrote:
     > Hi everyone,
     >
     > Is there any command for updating table withing a loop??
     "Loops? We don't need no stinking loops!"
     (From 'The Good, the Bad, and the Rgly')
     tab <- table(data.raw, findInterval(seq(along=data.raw), ind+1 ) )
     tab %*% upper.tri(tab, diag=T)
     or
     tab2 <- tapply( factor(data.raw), findInterval(seq(along=data.raw),
ind+1
     ), table)
     Reduce( "+", tab2, accum=TRUE )
     HTH,
     Chuck
     p.s. See the posting guide re including a reproducible example with
     requests like yours.
     > For instance, at i, I have a table as ZZ = table(data.raw[1:ind[i]])
     > where "ind" = c(10, 20, 30, ...).?Then , ZZ will be as
follow
     >
     > "A" "B" "C"
     > ?3??? 10?? 2
     >
     > At (i + 1), ZZ = table(data.raw[(ind[i]+1):ind[i+1]])
     >
     > "A" "B" "D"
     > ?4 ?? 7??? 8
     >
     > Is there any command that can update the table ZZ for each time so
that
     in the above example, ZZ will be
     >
     > "A" "B" "C" "D"
     > ?7??? 17?? 2??? 8
     >
     > Thanks.
     >
     > liu
     >
     >
     >
     >
     > [[alternative HTML version deleted]]
     >
     >
     Charles C. Berry (858) 534-2098
     Dept of Family/Preventive Medicine
     E mailto:[91]cberry at tajo.ucsd.edu UC San Diego
     [92]http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego
     92093-0901
     ------------------------------
     Message: 12
     Date: Sat, 24 Jul 2010 11:51:05 -0400
     From: Duncan Murdoch <[93]murdoch.duncan at gmail.com>
     To: "Charles C. Berry" <[94]cberry at tajo.ucsd.edu>
     Cc: [95]r-help at r-project.org
     Subject: Re: [R] , Updating Table
     Message-ID: <[96]4C4B0BE9.7050409 at gmail.com>
     Content-Type: text/plain; charset=UTF-8; format=flowed
     On 24/07/2010 11:25 AM, Charles C. Berry wrote:
     > On Fri, 23 Jul 2010, Marcus Liu wrote:
     >
     >> Hi everyone,
     >>
     >> Is there any command for updating table withing a loop??
     >
     > "Loops? We don't need no stinking loops!"
     > (From 'The Good, the Bad, and the Rgly')
     Actually, that quote comes from the TreasR of the SieRa MadRe.
     Duncan Murdoch
     > tab <- table(data.raw, findInterval(seq(along=data.raw), ind+1 ) )
     > tab %*% upper.tri(tab, diag=T)
     >
     > or
     >
     > tab2 <- tapply( factor(data.raw), findInterval(seq(along=data.raw),
     ind+1 ), table)
     > Reduce( "+", tab2, accum=TRUE )
     >
     > HTH,
     >
     > Chuck
     >
     > p.s. See the posting guide re including a reproducible example with
     > requests like yours.
     >
     >> For instance, at i, I have a table as ZZ =
table(data.raw[1:ind[i]])
     >> where "ind" = c(10, 20, 30, ...).?Then , ZZ will be as
follow
     >>
     >> "A" "B" "C"
     >> ?3??? 10?? 2
     >>
     >> At (i + 1), ZZ = table(data.raw[(ind[i]+1):ind[i+1]])
     >>
     >> "A" "B" "D"
     >> ?4 ?? 7??? 8
     >>
     >> Is there any command that can update the table ZZ for each time so
that
     in the above example, ZZ will be
     >>
     >> "A" "B" "C" "D"
     >> ?7??? 17?? 2??? 8
     >>
     >> Thanks.
     >>
     >> liu
     >>
     >>
     >>
     >>
     >> [[alternative HTML version deleted]]
     >>
     >>
     >
     > Charles C. Berry (858) 534-2098
     > Dept of Family/Preventive Medicine
     > E mailto:[97]cberry at tajo.ucsd.edu UC San Diego
     > [98]http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego
     92093-0901
     >
     >
     >
     >
------------------------------------------------------------------------
     >
     > ______________________________________________
     > [99]R-help at r-project.org mailing list
     > https://stat.ethz.ch/mailman/listinfo/r-help
     > PLEASE do read the posting guide
     [100]http://www.R-project.org/posting-guide.html
     > and provide commented, minimal, self-contained, reproducible code.
     ------------------------------
     Message: 13
     Date: Sat, 24 Jul 2010 15:52:52 +0000 (UTC)
     From: Ben Bolker <[101]bbolker at gmail.com>
     To: [102]r-help at stat.math.ethz.ch
     Subject: Re: [R] glm - prediction of a factor with several levels
     Message-ID: <[103]loom.20100724T175114-259 at post.gmane.org>
     Content-Type: text/plain; charset=us-ascii
     blackscorpio <olivier.collignon <at> live.fr> writes:
     > I'm currently attempting to predict the occurence of an event
(factor)
     > having more than 2 levels with several continuous predictors. The
model
     > being ordinal, I was waiting the glm function to return several
     intercepts,
     > which is not the case when looking to my results (I only have one
     > intercept). I finally managed to perform an ordinal polytomous
logisitc
     > regression with the polr function, which gives several intercepts.
     > But does anyone know what was the model performed with glm and why
only
     one
     > intercept was given ?
     It's not sufficiently clear (to me at least) what you're trying to
     do. Please provide a minimal reproducible example ... As far as I know,
     polr is the right way to do ordinal regression; it's not clear how you
     were trying to use glm to do it.
     Ben Bolker
     ------------------------------
     Message: 14
     Date: Sat, 24 Jul 2010 12:13:30 -0300
     From: Bruno Bastos Gon?alves <[104]brubruzao at hotmail.com>
     To: <[105]r-help at r-project.org>
     Subject: [R] Doubt about a population competition function
     Message-ID: <[106]SNT111-DS23835E9F6F65737C7B0BEDBBA40 at phx.gbl>
     Content-Type: text/plain
     Hi,
     I'm doing a function that describe two populations in competition.
     that's the function that i wrote:
     exclusao<-function(n10, n20, k1, k2, alfa, beta, t){
     n1<-k1-(alfa*n20)
     n2<-k2-(beta*n10)
     if(t==0){plot(t, n10, type='b', xlim=range(c(1:t), c
     (1:t)), ylim=range(n10, n20), xlab='tempo',
     ylab='tamanho populacional')
     points(t, n20, type='b', col="red")
     points(t, n10, type="b", col="black")
     legend("topleft", c("Pop1", "Pop2"), cex=0.8,
col=c
     ("black", "red"), pch=21:21, lty=1:1);
     }
     if(t>0){
     for (i in 1:t){
     n1[i==1]<-n1
     n2[i==1]<-n2
     n1[i+1]<-k1-alfa*n2[i]
     n2[i+1]<-k2-beta*n1[i]
     if(n1[i]==0){n1[i:t]==0}
     if(n2[i]==0){n2[i:t]==0}
     }
     plot(c(1:t), n1[1:t], type='b', xlim=range(c(1:t), c
     (1:t)), ylim=range(n1[1:t], n2[1:t]), xlab='tempo',
     ylab='tamanho populacional')
     points(c(1:t), n2[1:t], type='b', col="red")
     legend("topleft", c("Pop1", "Pop2"), cex=0.8,
col=c
     ("black", "red"), pch=21:21, lty=1:1);
     }}
     Where n10: size population in time 0, n20: size population in time 0, k1:
     carrying capacity of the population 1, k2: carrying capacity of the
     population 2, alfa: competition coefficient of population 2 in population
     1, beta: competition coefficient of population 1 in population 2, t: time.
     and when some population becomes 0 (ZERO), i want that population still 0
     (ZERO) until the end of "t". i have tried to put "
     if(n1[i]==0){n1[i:t]==0} if(n2[i]==0){n2[i:t]==0}" after
     "n2[i+1]<-k2-beta*n1[i]" in the for function, but nothing
happens. What
     may i do ?
     Thanks
     Bruno
     [[alternative HTML version deleted]]
     ------------------------------
     Message: 15
     Date: Sat, 24 Jul 2010 12:15:24 -0300
     From: "Gmail" <[107]goncalves.b.b at gmail.com>
     To: "[108]r-help at r-project.org"@stat.math.ethz.ch
     Subject: [R] Doubt about a population competition function
     Message-ID: <822A5DC79A42471AB493F6C4A0817C8E at NotebookBruno>
     Content-Type: text/plain
     Hi,
     I'm doing a function that describe two populations in competition.
     that's the function that i wrote:
     exclusao<-function(n10, n20, k1, k2, alfa, beta, t){
     n1<-k1-(alfa*n20)
     n2<-k2-(beta*n10)
     if(t==0){plot(t, n10, type='b', xlim=range(c(1:t), c
     (1:t)), ylim=range(n10, n20), xlab='tempo',
     ylab='tamanho populacional')
     points(t, n20, type='b', col="red")
     points(t, n10, type="b", col="black")
     legend("topleft", c("Pop1", "Pop2"), cex=0.8,
col=c
     ("black", "red"), pch=21:21, lty=1:1);
     }
     if(t>0){
     for (i in 1:t){
     n1[i==1]<-n1
     n2[i==1]<-n2
     n1[i+1]<-k1-alfa*n2[i]
     n2[i+1]<-k2-beta*n1[i]
     if(n1[i]==0){n1[i:t]==0}
     if(n2[i]==0){n2[i:t]==0}
     }
     plot(c(1:t), n1[1:t], type='b', xlim=range(c(1:t), c
     (1:t)), ylim=range(n1[1:t], n2[1:t]), xlab='tempo',
     ylab='tamanho populacional')
     points(c(1:t), n2[1:t], type='b', col="red")
     legend("topleft", c("Pop1", "Pop2"), cex=0.8,
col=c
     ("black", "red"), pch=21:21, lty=1:1);
     }}
     Where n10: size population in time 0, n20: size population in time 0, k1:
     carrying capacity of the population 1, k2: carrying capacity of the
     population 2, alfa: competition coefficient of population 2 in population
     1, beta: competition coefficient of population 1 in population 2, t: time.
     and when some population becomes 0 (ZERO), i want that population still 0
     (ZERO) until the end of "t". i have tried to put "
     if(n1[i]==0){n1[i:t]==0} if(n2[i]==0){n2[i:t]==0}" after
     "n2[i+1]<-k2-beta*n1[i]" in the for function, but nothing
happens. What
     may i do ?
     Thanks
     Bruno
     [[alternative HTML version deleted]]
     ------------------------------
     Message: 16
     Date: Sat, 24 Jul 2010 09:39:10 -0700 (PDT)
     From: Matt Stati <[109]mattstati at yahoo.com>
     To: [110]r-help at r-project.org
     Subject: [R] Book on R's Programming Language
     Message-ID: <[111]289386.7557.qm at web43507.mail.sp1.yahoo.com>
     Content-Type: text/plain
     Can someone please recommend to me a book on the programming language that
     R is based on? I'm looking for a foundational book that teaches the
logic
     of the S language. It seems that knowing the underpinnings of the language
     can only make using R a bit easier.
     Any leads are greatly appreciated . . .
     Matt.
     [[alternative HTML version deleted]]
     ------------------------------
     Message: 17
     Date: Sat, 24 Jul 2010 13:39:53 -0300
     From: Henrique Dallazuanna <[112]wwwhsd at gmail.com>
     To: aegea <[113]gcheer3 at gmail.com>
     Cc: [114]r-help at r-project.org
     Subject: Re: [R] how to calculate the product of every two elements in
     two vectors
     Message-ID:
     <AANLkTi=DCF=Jv9gzLQK8p=[115]XrwenTLja3ZTrwKE9USm4z at
mail.gmail.com>
     Content-Type: text/plain
     Try this:
     c(as.matrix(B) %*% A)
     On Fri, Jul 23, 2010 at 12:11 PM, aegea <[116]gcheer3 at gmail.com>
wrote:
     >
     > Thanks in advance!
     >
     > A=c(1, 2, 3)
     > B=c (9, 10, 11, 12)
     >
     > I want to get C=c(1*9, 1*10, 1*11, 1*12, ....., 3*9, 3*10, 3*11,
3*12)?
     > C is still a vector with 12 elements
     > Is there a way to do that?
     > --
     > View this message in context:
     >
     [117]http://r.789695.n4.nabble.com/how-to-calculate-the-product-of-every-t
     wo-elements-in-two-vectors-tp2300299p2300299.html
     > Sent from the R help mailing list archive at Nabble.com.
     >
     > ______________________________________________
     > [118]R-help at r-project.org mailing list
     > https://stat.ethz.ch/mailman/listinfo/r-help
     > PLEASE do read the posting guide
     > [119]http://www.R-project.org/posting-guide.html
     > and provide commented, minimal, self-contained, reproducible code.
     >
     --
     Henrique Dallazuanna
     Curitiba-Paran???Brasil
     25???25' 40" S 49???16' 22" O
     [[alternative HTML version deleted]]
     ------------------------------
     Message: 18
     Date: Sat, 24 Jul 2010 10:13:52 -0700
     From: Joshua Wiley <[120]jwiley.psych at gmail.com>
     To: Matt Stati <[121]mattstati at yahoo.com>
     Cc: [122]r-help at r-project.org
     Subject: Re: [R] Book on R's Programming Language
     Message-ID:
     <AANLkTiki+9endGiR_T5pf9e8FofXbcyE5AL+-+[123]v0shNs at
mail.gmail.com>
     Content-Type: text/plain; charset=UTF-8
     Hi Matt,
     [124]http://www.r-project.org/doc/bib/R-books.html
     Lists a variety of books, and seems to include most (i.e., my
     searches through google, amazon, and barnes and noble, didn't really
     turn up others) books dedicated to R. I have always been under the
     impression that Programming with Data (the Green Book) is a classic.
     [125]http://cran.r-project.org/manuals.html
     has the official manuals
     Similar questions have been asked several times on this list so you
     can also search for previous threads (e.g.,
     [126]http://tolstoy.newcastle.edu.au/R/help/04/06/0063.html )
     Best regards,
     Josh
     On Sat, Jul 24, 2010 at 9:39 AM, Matt Stati <[127]mattstati at
yahoo.com>
     wrote:
     > Can someone please recommend to me a book on the programming language
     that R is based on? I'm looking for a foundational book that teaches
the
     logic of the S language. It seems that knowing the underpinnings of the
     language can only make using R a bit easier.
     >
     > Any leads are greatly appreciated . . .
     >
     > Matt.
     >
     >
     >
     >
     > ? ? ? ?[[alternative HTML version deleted]]
     >
     > ______________________________________________
     > [128]R-help at r-project.org mailing list
     > https://stat.ethz.ch/mailman/listinfo/r-help
     > PLEASE do read the posting guide
     [129]http://www.R-project.org/posting-guide.html
     > and provide commented, minimal, self-contained, reproducible code.
     >
     --
     Joshua Wiley
     Ph.D. Student, Health Psychology
     University of California, Los Angeles
     [130]http://www.joshuawiley.com/
     ------------------------------
     Message: 19
     Date: Sat, 24 Jul 2010 13:15:34 -0400
     From: Gabor Grothendieck <[131]ggrothendieck at gmail.com>
     To: aegea <[132]gcheer3 at gmail.com>
     Cc: [133]r-help at r-project.org
     Subject: Re: [R] how to calculate the product of every two elements in
     two vectors
     Message-ID:
     <AANLkTinRzLax+[134]dPbmg9YRPUcE2fQH_0Sqs2UJKFVJ3BS at
mail.gmail.com>
     Content-Type: text/plain; charset=ISO-8859-1
     On Fri, Jul 23, 2010 at 11:11 AM, aegea <[135]gcheer3 at gmail.com>
wrote:
     >
     > Thanks in advance!
     >
     > A=c(1, 2, 3)
     > B=c (9, 10, 11, 12)
     >
     > I want to get C=c(1*9, 1*10, 1*11, 1*12, ....., 3*9, 3*10, 3*11,
3*12)?
     > C is still a vector with 12 elements
     > Is there a way to do that?
     Here are yet a few more. The first one is the only one so far that
     uses a single function and the last two are slight variations of ones
     already posted.
     kronecker(A, B)
     c(tcrossprod(B, A))
     c(outer(B, A))
     c(B %o% A)
     Here is a speed comparison. The fastest are as.matrix, %outer% and
     %o% . They are so close that random fluctuations might easily change
     their order and since %o% involves the least keystrokes that one might
     be a good overall choice. Although not among the fastest the
     kronecker solution is the simplest since it only involves a single
     function call so it might be preferred on that count.
     > A <- B <- 1:400
     > out <- benchmark(
     + as.matrix = c(as.matrix(B) %*% A),
     + crossprod = c(tcrossprod(B, A)),
     + outer = c(outer(B, A)),
     + o = c(B %o% A),
     + kronecker = kronecker(A, B),
     + touter = as.vector(t(outer(A, B))))
     > out[order(out$relative), ]
     test replications elapsed relative user.self sys.self
     user.child sys.child
     1 as.matrix 100 0.92 1.000000 0.62 0.28
     NA NA
     3 outer 100 0.93 1.010870 0.59 0.35
     NA NA
     4 o 100 0.94 1.021739 0.66 0.28
     NA NA
     2 crossprod 100 1.11 1.206522 0.67 0.43
     NA NA
     5 kronecker 100 1.45 1.576087 1.25 0.21
     NA NA
     6 touter 100 1.84 2.000000 1.40 0.43
     NA NA
     ------------------------------
     Message: 20
     Date: Sat, 24 Jul 2010 11:55:08 -0500
     From: Joseph Magagnoli <[136]jcm331 at gmail.com>
     To: Matt Stati <[137]mattstati at yahoo.com>, rhelp
     <[138]r-help at r-project.org>
     Subject: Re: [R] Book on R's Programming Language
     Message-ID:
     <AANLkTi=+umQatto9rVeujEwjNyqOSiHOas+[139]eWg9Wcb80 at
mail.gmail.com>
     Content-Type: text/plain
     Matt,
     you might want to check out programming with data by John Chambers.
     [140]http://www.amazon.com/Programming-Data-Guide-S-Language/dp/0387985034
     /ref=sr_1_1?ie=UTF8&s=books&qid=1279990404&sr=8-1
     Best Joe
     On Sat, Jul 24, 2010 at 11:39 AM, Matt Stati <[141]mattstati at
yahoo.com>
     wrote:
     > Can someone please recommend to me a book on the programming language
     that
     > R is based on? I'm looking for a foundational book that teaches
the
     logic of
     > the S language. It seems that knowing the underpinnings of the
language
     can
     > only make using R a bit easier.
     >
     > Any leads are greatly appreciated . . .
     >
     > Matt.
     >
     >
     >
     >
     > [[alternative HTML version deleted]]
     >
     > ______________________________________________
     > [142]R-help at r-project.org mailing list
     > https://stat.ethz.ch/mailman/listinfo/r-help
     > PLEASE do read the posting guide
     >
    
[143]http://www.R-project.org/posting-guide.html<[144]http://www.r-project
     .org/posting-guide.html>
     > and provide commented, minimal, self-contained, reproducible code.
     >
     --
     Joseph C. Magagnoli
     Doctoral Student
     Department of Political Science
     University of North Texas
     1155 Union Circle #305340
     Denton, Texas 76203-5017
     Email: [145]jcm0250 at unt.edu
     [[alternative HTML version deleted]]
     ------------------------------
     Message: 21
     Date: Sat, 24 Jul 2010 12:55:18 -0600
     From: Greg Snow <[146]Greg.Snow at imail.org>
     To: "Farley, Robert" <[147]FarleyR at metro.net>,
"[148]r-help at r-project.org"
     <[149]r-help at r-project.org>
     Subject: Re: [R] Constrain density to 0 at 0?
     Message-ID:
     <[150]B37C0A15B8FB3C468B5BC7EBC7DA14CC633A53DF11 at
LP-EXMBVS10.CO.IHC.COM>
     Content-Type: text/plain; charset="us-ascii"
     Look at the logspline package. This is a different approach to density
     estimation from the kernel densities used by 'density', but does
allow you
     to set fixed boundaries.
     --
     Gregory (Greg) L. Snow Ph.D.
     Statistical Data Center
     Intermountain Healthcare
     [151]greg.snow at imail.org
     801.408.8111
     > -----Original Message-----
     > From: [152]r-help-bounces at r-project.org [mailto:r-help-bounces at
r-
     > project.org] On Behalf Of Farley, Robert
     > Sent: Monday, July 19, 2010 7:57 PM
     > To: [153]r-help at r-project.org
     > Subject: [R] Constrain density to 0 at 0?
     >
     > I'm plotting some trip length frequencies using the following
code:
     >
     > plot( density(zTestData$Distance, weights=zTestData$Actual),
     > xlim=c(0, 10),
     > main="Test TLFD",
     > xlab="Distance",
     > col=6 )
     > lines(density(zTestData$Distance, weights=zTestData$FlatWeight),
col=2)
     > lines(density(zTestData$Distance, weights=zTestData$BrdWeight ),
col=3)
     >
     > which works fine except the distances are all positive, but the
     > densities don't drop to 0 until around -2 or -3.
     >
     > Is there a way for me to "force" the density plot to 0 at 0?
     >
     >
     >
     > Thanks
     >
     >
     >
     > Robert Farley
     > Metro
     > 1 Gateway Plaza
     > Mail Stop 99-23-7
     > Los Angeles, CA 90012-2952
     > Voice: (213)922-2532
     > Fax: (213)922-2868
     > www.Metro.net
     >
     >
     >
     > [[alternative HTML version deleted]]
     >
     > ______________________________________________
     > [154]R-help at r-project.org mailing list
     > https://stat.ethz.ch/mailman/listinfo/r-help
     > PLEASE do read the posting guide
[155]http://www.R-project.org/posting-
     > guide.html
     > and provide commented, minimal, self-contained, reproducible code.
     ------------------------------
     Message: 22
     Date: Sun, 25 Jul 2010 01:16:57 +0530
     From: shabnam k <[156]shabnambioinfo at gmail.com>
     To: r-help <[157]r-help at r-project.org>
     Subject: [R] matest function for multiple factors
     Message-ID:
     <AANLkTi=mQ5bhntKQbfUapHpWNWpVLn+[158]Hzat_FD6jpZF6 at
mail.gmail.com>
     Content-Type: text/plain
     Hi,
     I am using maanova package for analysis. In my dataset, two fixed factors
     time and treatment and sample as random factor is there. Am able to get
     madata object and fitmaanova object. But, am unable to do f-test with two
     factors, but i have done f-test seperately for two factors.
     fit.full.mix <- fitmaanova(madata, formula = ~Sample+Time+Treatment,
     random = ~Sample)
     ftest.all = *matest*(madata, fit.full.mix, test.method=c(1, 1),
     shuffle.method="sample", *term="Time+Treatment"*,
n.perm= 100)
     Can u please suggest me, how to represent multiple factors in the
     above function simultaneously in term.
     [[alternative HTML version deleted]]
     ------------------------------
     Message: 23
     Date: Sat, 24 Jul 2010 13:55:59 -0600
     From: Greg Snow <[159]Greg.Snow at imail.org>
     To: "[160]babyfoxlove1 at sina.com" <[161]babyfoxlove1 at
sina.com>,
     "[162]r-help at r-project.org" <[163]r-help at
r-project.org>
     Subject: Re: [R] How to deal with more than 6GB dataset using R?
     Message-ID:
     <[164]B37C0A15B8FB3C468B5BC7EBC7DA14CC633A53DF25 at
LP-EXMBVS10.CO.IHC.COM>
     Content-Type: text/plain; charset="us-ascii"
     You may want to look at the biglm package as another way to regression
     models on very large data sets.
     --
     Gregory (Greg) L. Snow Ph.D.
     Statistical Data Center
     Intermountain Healthcare
     [165]greg.snow at imail.org
     801.408.8111
     > -----Original Message-----
     > From: [166]r-help-bounces at r-project.org [mailto:r-help-bounces at
r-
     > project.org] On Behalf Of [167]babyfoxlove1 at sina.com
     > Sent: Friday, July 23, 2010 10:10 AM
     > To: [168]r-help at r-project.org
     > Subject: [R] How to deal with more than 6GB dataset using R?
     >
     > &nbsp;Hi there,
     >
     > Sorry to bother those who are not interested in this problem.
     >
     > I'm dealing with a large data set, more than 6 GB file, and doing
     > regression test with those data. I was wondering are there any
     > efficient ways to read those data? Instead of just using read.table()?
     > BTW, I'm using a 64bit version desktop and a 64bit version R, and
the
     > memory for the desktop is enough for me to use.
     > Thanks.
     >
     >
     > --Gin
     >
     > [[alternative HTML version deleted]]
     >
     > ______________________________________________
     > [169]R-help at r-project.org mailing list
     > https://stat.ethz.ch/mailman/listinfo/r-help
     > PLEASE do read the posting guide
[170]http://www.R-project.org/posting-
     > guide.html
     > and provide commented, minimal, self-contained, reproducible code.
     ------------------------------
     Message: 24
     Date: Sat, 24 Jul 2010 17:03:44 -0400
     From: "Abdi, Abdulhakim" <[171]AbdiA at si.edu>
     To: "[172]r-help at r-project.org" <[173]r-help at
r-project.org>
     Subject: [R] Using R to fill ETM+ data gaps?
     Message-ID:
     <[174]97679C0A11332E48A01E0D463E8B3FF103AF0CE275 at
SI-MSEV02.US.SINET.SI.EDU
     >
     Content-Type: text/plain; charset="us-ascii"
     Hi,
     I was wondering if anyone knows of a method (or if it's possible) to
use R
     in interpolating Landsat ETM+ data gaps?
     Regards,
     Hakim Abdi
     _________________________________________
     Abdulhakim Abdi, M.Sc.
     Conservation GIS/Remote Sensing Lab
     Smithsonian Conservation Biology Institute
     1500 Remount Road
     Front Royal, VA 22630
     phone: +1 540 635 6578
     mobile: +1 747 224 7006
     fax: +1 540 635 6506 (Attn: ABDI/GIS Lab)
     email: [175]abdia at si.edu
     [176]http://nationalzoo.si.edu/SCBI/ConservationGIS/
     ------------------------------
     Message: 25
     Date: Sat, 24 Jul 2010 14:07:25 -0700 (PDT)
     From: Felipe Carrillo <[177]mazatlanmexico at yahoo.com>
     To: [178]r-help at stat.math.ethz.ch
     Subject: [R] How to generate a sequence of dates without hardcoding
     the year
     Message-ID: <[179]418059.32636.qm at web56602.mail.re3.yahoo.com>
     Content-Type: text/plain; charset=iso-8859-1
     Hi:
     I have a dataframe named 'spring' and I am trying to add a new
variable
     named
     'IdDate'
     This line of code works fine:
     spring$idDate <- seq(as.Date("2008-07-01"),
as.Date("2009-06-30"),
     by="week")
     But I don't want to hardcode the year because it will be used again the
     following year
     Is it possible to just generate dates with the month and day?
     I tried the code below:
     seq(as.Date("7-1", "%B%d"), as.Date("6-30",
"%B%d"), by="week")
     and got this?error message:
     Error in seq.int(0, to - from, by) : 'to' must be finite
     Thanks for any pointers
     ?
     Felipe D. Carrillo
     Supervisory Fishery Biologist
     Department of the Interior
     US Fish & Wildlife Service
     California, USA
     ------------------------------
     Message: 26
     Date: Sat, 24 Jul 2010 18:09:45 -0300
     From: Henrique Dallazuanna <[180]wwwhsd at gmail.com>
     To: Felipe Carrillo <[181]mazatlanmexico at yahoo.com>
     Cc: [182]r-help at stat.math.ethz.ch
     Subject: Re: [R] How to generate a sequence of dates without
     hardcoding the year
     Message-ID:
     <[183]AANLkTimd6DLagDiHKbFuF3hE5dES4d6DmC5T50Vn25N3 at
mail.gmail.com>
     Content-Type: text/plain
     Try this:
     format(seq(as.Date("2008-07-01"),
as.Date("2009-06-30"), by="week"),
     "%d/%m")
     On Sat, Jul 24, 2010 at 6:07 PM, Felipe Carrillo
     <[184]mazatlanmexico at yahoo.com>wrote:
     > Hi:
     > I have a dataframe named 'spring' and I am trying to add a new
variable
     > named
     > 'IdDate'
     > This line of code works fine:
     > spring$idDate <- seq(as.Date("2008-07-01"),
as.Date("2009-06-30"),
     by="week")
     >
     > But I don't want to hardcode the year because it will be used
again the
     > following year
     > Is it possible to just generate dates with the month and day?
     >
     > I tried the code below:
     > seq(as.Date("7-1", "%B%d"),
as.Date("6-30", "%B%d"), by="week")
     >
     > and got this error message:
     > Error in seq.int(0, to - from, by) : 'to' must be finite
     > Thanks for any pointers
     >
     >
     > Felipe D. Carrillo
     > Supervisory Fishery Biologist
     > Department of the Interior
     > US Fish & Wildlife Service
     > California, USA
     >
     >
     >
     >
     > ______________________________________________
     > [185]R-help at r-project.org mailing list
     > https://stat.ethz.ch/mailman/listinfo/r-help
     > PLEASE do read the posting guide
     > [186]http://www.R-project.org/posting-guide.html
     > and provide commented, minimal, self-contained, reproducible code.
     >
     --
     Henrique Dallazuanna
     Curitiba-Paran???Brasil
     25???25' 40" S 49???16' 22" O
     [[alternative HTML version deleted]]
     ------------------------------
     Message: 27
     Date: Sat, 24 Jul 2010 15:54:51 -0500 (CDT)
     From: <[187]mpward at illinois.edu>
     To: "Joshua Wiley" <[188]jwiley.psych at gmail.com>
     Cc: [189]r-help at r-project.org
     Subject: Re: [R] Trouble retrieving the second largest value from each
     row of a data.frame
     Message-ID: <[190]20100724155451.CHG28413 at expms6.cites.uiuc.edu>
     Content-Type: text/plain; charset=iso-8859-1
     THANKS, but I have one issue and one question.
     For some reason the "secondstrongest" value for row 3 and 6 are
incorrect
     (they are the strongest) the remaining 10 are correct??
     These data are being used to track radio-tagged birds, they are from
     automated radio telemetry receivers. I will applying the following formula
     diff <- ((strongest- secondstrongest)/100)
     bearingdiff <-30-(-0.0624*(diff**2))-(2.8346*diff)
     Then the bearing diff is added to strongestantenna (value0 = 0degrees) if
     the secondstrongestatenna is greater (eg value0 and value60), or if the
     secondstrongestantenna is smaller than the strongestantenna, then the
     bearingdiff is substracted from the strongestantenna. The only exception
     is that if value0 (0degrees) is strongest and value300(360degrees) is the
     secondstrongestantenna then the bearing is 360-bearingdiff. Also the
     strongestantenna and secondstrongestantenna have to be next to each other
     (e.g. value0 with value60, value240 with value300, value0 with value300)
     or the results should be NA. I have been trying to use a series of if,
     else statements to produce these bearing, but all I am producing is
     errors. Any suggestion would be appreciated.
     Again THANKS for you efforts.
     Mike
     ---- Original message ----
     >Date: Fri, 23 Jul 2010 23:01:56 -0700
     >From: Joshua Wiley <[191]jwiley.psych at gmail.com>
     >Subject: Re: [R] Trouble retrieving the second largest value from each
     row of a data.frame
     >To: [192]mpward at illinois.edu
     >Cc: [193]r-help at r-project.org
     >
     >Hi,
     >
     >Here is a little function that will do what you want and return a nice
     output:
     >
     >#Function To calculate top two values and return
     >my.finder <- function(mydata) {
     > my.fun <- function(data) {
     > strongest <- which.max(data)
     > secondstrongest <- which.max(data[-strongest])
     > strongestantenna <- names(data)[strongest]
     > secondstrongantenna <- names(data[-strongest])[secondstrongest]
     > value <- matrix(c(data[strongest], data[secondstrongest],
     > strongestantenna, secondstrongantenna), ncol =4)
     > return(value)
     > }
     > dat <- apply(mydata, 1, my.fun)
     > dat <- t(dat)
     > dat <- as.data.frame(dat, stringsAsFactors = FALSE)
     > colnames(dat) <- c("strongest",
"secondstrongest",
     > "strongestantenna", "secondstrongantenna")
     > dat[ , "strongest"] <- as.numeric(dat[ ,
"strongest"])
     > dat[ , "secondstrongest"] <- as.numeric(dat[ ,
"secondstrongest"])
     > return(dat)
     >}
     >
     >
     >#Using your example data:
     >
     >yourdata <- structure(list(value0 = c(-13007L, -12838L, -12880L,
-12805L,
     >-12834L, -11068L, -12807L, -12770L, -12988L, -11779L), value60     
c(-11707L,
     >-13210L, -11778L, -11653L, -13527L, -11698L, -14068L, -11665L,
     >-11736L, -12873L), value120 = c(-11072L, -11176L, -11113L, -11071L,
     >-11067L, -12430L, -11092L, -11061L, -11137L, -12973L), value180     
c(-12471L,
     >-11799L, -12439L, -12385L, -11638L, -12430L, -11709L, -12373L,
     >-12570L, -12537L), value240 = c(-12838L, -13210L, -13089L, -11561L,
     >-13527L, -12430L, -11607L, -11426L, -13467L, -12973L), value300     
c(-13357L,
     >-13845L, -13880L, -13317L, -13873L, -12814L, -13025L, -12805L,
     >-13739L, -11146L)), .Names = c("value0", "value60",
"value120",
     >"value180", "value240", "value300"),
class = "data.frame", row.names      c("1",
     >"2", "3", "4", "5",
"6", "7", "8", "9", "10"))
     >
     >my.finder(yourdata) #and what you want is in a nicely labeled data
frame
     >
     >#A potential problem is that it is not very efficient
     >
     >#Here is a test using a matrix of 100, 000 rows
     >#sampled from the same range as your data
     >#with the same number of columns
     >
     >data.test <- matrix(
     > sample(seq(min(yourdata), max(yourdata)), size = 500000, replace     
TRUE),
     > ncol = 5)
     >
     >system.time(my.finder(data.test))
     >
     >#On my system I get
     >
     >> system.time(my.finder(data.test))
     > user system elapsed
     > 2.89 0.00 2.89
     >
     >Hope that helps,
     >
     >Josh
     >
     >
     >
     >On Fri, Jul 23, 2010 at 6:20 PM, <[194]mpward at illinois.edu>
wrote:
     >> I have a data frame with a couple million lines and want to
retrieve
     the largest and second largest values in each row, along with the label of
     the column these values are in. For example
     >>
     >> row 1
     >> strongest=-11072
     >> secondstrongest=-11707
     >> strongestantenna=value120
     >> secondstrongantenna=value60
     >>
     >> Below is the code I am using and a truncated data.frame.
?Retrieving
     the largest value was easy, but I have been getting errors every way I
     have tried to retrieve the second largest value. ?I have not even tried to
     retrieve the labels for the value yet.
     >>
     >> Any help would be appreciated
     >> Mike
     >>
     >>
     >>> data<-data.frame(value0, value60, value120, value180,
value240,
     value300)
     >>> data
     >> ? value0 value60 value120 value180 value240 value300
     >> 1 ?-13007 ?-11707 ? -11072 ? -12471 ? -12838 ? -13357
     >> 2 ?-12838 ?-13210 ? -11176 ? -11799 ? -13210 ? -13845
     >> 3 ?-12880 ?-11778 ? -11113 ? -12439 ? -13089 ? -13880
     >> 4 ?-12805 ?-11653 ? -11071 ? -12385 ? -11561 ? -13317
     >> 5 ?-12834 ?-13527 ? -11067 ? -11638 ? -13527 ? -13873
     >> 6 ?-11068 ?-11698 ? -12430 ? -12430 ? -12430 ? -12814
     >> 7 ?-12807 ?-14068 ? -11092 ? -11709 ? -11607 ? -13025
     >> 8 ?-12770 ?-11665 ? -11061 ? -12373 ? -11426 ? -12805
     >> 9 ?-12988 ?-11736 ? -11137 ? -12570 ? -13467 ? -13739
     >> 10 -11779 ?-12873 ? -12973 ? -12537 ? -12973 ? -11146
     >>> #largest value in the row
     >>> strongest<-apply(data, 1, max)
     >>>
     >>>
     >>> #second largest value in the row
     >>> n<-function(data)(1/(min(1/(data[1, ]-max(data[1, ]))))+
(max(data[1,
     ])))
     >>> secondstrongest<-apply(data, 1, n)
     >> Error in data[1, ] : incorrect number of dimensions
     >>>
     >>
     >> ______________________________________________
     >> [195]R-help at r-project.org mailing list
     >> https://stat.ethz.ch/mailman/listinfo/r-help
     >> PLEASE do read the posting guide
     [196]http://www.R-project.org/posting-guide.html
     >> and provide commented, minimal, self-contained, reproducible code.
     >>
     >
     >
     >
     >--
     >Joshua Wiley
     >Ph.D. Student, Health Psychology
     >University of California, Los Angeles
     >[197]http://www.joshuawiley.com/
     ------------------------------
     Message: 28
     Date: Sat, 24 Jul 2010 22:21:21 +0100
     From: Paul Smith <[198]phhs80 at gmail.com>
     To: [199]r-help at r-project.org
     Subject: Re: [R] (no subject)
     Message-ID:
     <AANLkTi=qKpbPZutDNr9n8af1jGFsLfTqT1C7Nm8Ko+[200]oH at
mail.gmail.com>
     Content-Type: text/plain; charset=UTF-8
     2010/7/23 w ? <[201]hw_joyce_cn at hotmail.com>:
     > I use the constrOptim to maximize a function with four constriants but
     the answer does not leave from the starting value and there is only one
     outer iteration. The function is defined as follows:
     > tm<-function(p){
     > p1<-p[1]; p2<-p[2]; p3<-p[3];
     > p4<-1-p1-p2-p3;
     > p1*p2*p3*p4}
     >
     > ##the constraints are p1>=0; p2>=0; p3>=0 and p4>=0 i.e.
p1+p2+p3<=1
     > start<-c(0.9999991, 0.0000001, 0.0000001)
     > dev<-rbind(diag(3), -diag(3), rep(-1, 3))
     > bvec<-c(rep(0, 3), rep(-1, 4))
     > constrOptim(start, tm, NULL, ui=dev, ci=bvec,
control=list(maxit=10000))
     >
     > Am i missing something obviously that cause the problem or there is
some
     bugs in constrOptim. Could you please help me out
     Wenwen,
     I believe that the reason why constrOptim behaves as described is
     related to the fact that
     (p1, p2, p3) = (1, 0, 0)
     is a stationary point and you use it as a starting point. Try a
     different starting point.
     If the objective is to maximize, then you should use the following
     command:
     constrOptim(start, tm, NULL, ui=dev, ci=bvec, control=list(maxit=10000,
     fnscale=-1))
     (Notice fnscale=-1.)
     Finally, whenever you ask something on this list, please use a
     meaningful title for your message, as it will dramatically increase
     the chances of you getting an answer.
     Good luck,
     Paul
     ------------------------------
     Message: 29
     Date: Sat, 24 Jul 2010 19:02:57 -0400
     From: jim holtman <[202]jholtman at gmail.com>
     To: Felipe Carrillo <[203]mazatlanmexico at yahoo.com>
     Cc: [204]r-help at stat.math.ethz.ch
     Subject: Re: [R] How to generate a sequence of dates without
     hardcoding the year
     Message-ID:
     <AANLkTikBNcm2=EzKaWMGNez-b53QhYYe_9CuK4LRN=[205]3P at
mail.gmail.com>
     Content-Type: text/plain; charset=ISO-8859-1
     Is this what you want if you want to assume that the date without a
     year is this year:
     > seq(as.Date("7-1", "%m-%d"), by="week",
length=52)
     [1] "2010-07-01" "2010-07-08" "2010-07-15"
"2010-07-22" "2010-07-29"
     "2010-08-05" "2010-08-12" "2010-08-19"
     [9] "2010-08-26" "2010-09-02" "2010-09-09"
"2010-09-16" "2010-09-23"
     "2010-09-30" "2010-10-07" "2010-10-14"
     [17] "2010-10-21" "2010-10-28" "2010-11-04"
"2010-11-11" "2010-11-18"
     "2010-11-25" "2010-12-02" "2010-12-09"
     [25] "2010-12-16" "2010-12-23" "2010-12-30"
"2011-01-06" "2011-01-13"
     "2011-01-20" "2011-01-27" "2011-02-03"
     [33] "2011-02-10" "2011-02-17" "2011-02-24"
"2011-03-03" "2011-03-10"
     "2011-03-17" "2011-03-24" "2011-03-31"
     [41] "2011-04-07" "2011-04-14" "2011-04-21"
"2011-04-28" "2011-05-05"
     "2011-05-12" "2011-05-19" "2011-05-26"
     [49] "2011-06-02" "2011-06-09" "2011-06-16"
"2011-06-23"
     >
     On Sat, Jul 24, 2010 at 5:07 PM, Felipe Carrillo
     <[206]mazatlanmexico at yahoo.com> wrote:
     > Hi:
     > I have a dataframe named 'spring' and I am trying to add a new
variable
     named
     > 'IdDate'
     > This line of code works fine:
     > spring$idDate <- seq(as.Date("2008-07-01"),
as.Date("2009-06-30"),
     by="week")
     >
     > But I don't want to hardcode the year because it will be used
again the
     > following year
     > Is it possible to just generate dates with the month and day?
     >
     > I tried the code below:
     > seq(as.Date("7-1", "%B%d"),
as.Date("6-30", "%B%d"), by="week")
     >
     > and got this?error message:
     > Error in seq.int(0, to - from, by) : 'to' must be finite
     > Thanks for any pointers
     >
     >
     > Felipe D. Carrillo
     > Supervisory Fishery Biologist
     > Department of the Interior
     > US Fish & Wildlife Service
     > California, USA
     >
     >
     >
     >
     > ______________________________________________
     > [207]R-help at r-project.org mailing list
     > https://stat.ethz.ch/mailman/listinfo/r-help
     > PLEASE do read the posting guide
     [208]http://www.R-project.org/posting-guide.html
     > and provide commented, minimal, self-contained, reproducible code.
     >
     --
     Jim Holtman
     Cincinnati, OH
     +1 513 646 9390
     What is the problem that you are trying to solve?
     ------------------------------
     Message: 30
     Date: Sat, 24 Jul 2010 16:27:06 -0700 (PDT)
     From: zachmohr <[209]zachmohr at gmail.com>
     To: [210]r-help at r-project.org
     Subject: Re: [R] glm - prediction of a factor with several levels
     Message-ID:
     <[211]AANLkTinzngqO3S_p-E25N3TTWEdiQBBS4CvzErgywj1p at
mail.gmail.com>
     Content-Type: text/plain
     As far as I know, glm only works with dichotomous or count data. polr in
     the MASS package works and so does lrm {Design} for ordinal dependent
     variables. I would assume that the model produced by glm is a dichotomous
     version of your model but not sure. Only one intercept would be given
     because if you used the log link then it would have produced a dichotomous
     model instead of an ordered logistic regression. My suggestion is to use
     polr or lrm.
     On Fri, Jul 23, 2010 at 6:15 PM, blackscorpio [via R] <
     ml-node+[212]2300793-1751019155-246278 at
n4.nabble.com<ml-node%[213]2B230079
     3-1751019155-246278 at n4.nabble.com>
     > wrote:
     > Dear community,
     > I'm currently attempting to predict the occurence of an event
(factor)
     > having more than 2 levels with several continuous predictors. The
model
     > being ordinal, I was waiting the glm function to return several
     intercepts,
     > which is not the case when looking to my results (I only have one
     > intercept). I finally managed to perform an ordinal polytomous
logisitc
     > regression with the polr function, which gives several intercepts.
     > But does anyone know what was the model performed with glm and why
only
     one
     > intercept was given ?
     > Thanks a lot for your help !
     >
     >
     > ------------------------------
     > View message @
     >
     [214]http://r.789695.n4.nabble.com/glm-prediction-of-a-factor-with-several
     -levels-tp2300793p2300793.html
     > To unsubscribe from R, click here< (link removed) ==>.
     >
     >
     >
     --
     Department of Public Administration
     University of Kansas
     4060 Wesco Hall
     Office W
     Lawrence KS 66045-3177
     Phone: (785) 813-1384
     Email: [215]zachmohr at gmail.com
     --
     View this message in context:
     [216]http://r.789695.n4.nabble.com/glm-prediction-of-a-factor-with-several
     -levels-tp2300793p2301324.html
     Sent from the R help mailing list archive at Nabble.com.
     [[alternative HTML version deleted]]
     ------------------------------
     Message: 31
     Date: Sat, 24 Jul 2010 20:09:32 -0400
     From: David Winsemius <[217]dwinsemius at comcast.net>
     To: <[218]mpward at illinois.edu>
     Cc: [219]r-help at r-project.org
     Subject: Re: [R] Trouble retrieving the second largest value from each
     row of a data.frame
     Message-ID: <[220]52EA484F-C066-4ACC-B5BC-1A3A20876D9E at
comcast.net>
     Content-Type: text/plain; charset=US-ASCII; format=flowed; delsp=yes
     On Jul 24, 2010, at 4:54 PM, <[221]mpward at illinois.edu> wrote:
     > THANKS, but I have one issue and one question.
     >
     > For some reason the "secondstrongest" value for row 3 and 6
are
     > incorrect (they are the strongest) the remaining 10 are correct??
     In my run of Wiley's code I instead get identical values for rows
     2, 5, 6. Holtman's and my solutions did not suffer from that defect,
     although mine suffered from my misreading of your request, thinking
     that you wanted the top 3. The fix is trivial
     >
     > These data are being used to track radio-tagged birds, they are from
     > automated radio telemetry receivers. I will applying the following
     > formula
     >
     > diff <- ((strongest- secondstrongest)/100)
     > bearingdiff <-30-(-0.0624*(diff**2))-(2.8346*diff)
     vals <- c("value0", "value60", "value120",
"value180", "value240",
     "value300")
     value.str2 <- (match(yourdata$secondstrongestantenna, vals)-1)*60
     value.str1 <- (match(yourdata$strongestantenna, vals)-1)*60
     change.ind <- abs(match(yourdata, vals) - which(match(yourdata, vals) )
     >
     > A) Then the bearing diff is added to strongestantenna (value0     
> 0degrees) if the secondstrongestatenna is greater (eg value0 and
     > value60),
     > B) or if the secondstrongestantenna is smaller than the
     > strongestantenna,
     > then the bearingdiff is substracted from the strongestantenna.
     >
     > C) The only exception is that if value0 (0degrees) is strongest and
     > value300(360degrees) is the secondstrongestantenna then the bearing
     > is 360-bearingdiff.
     > D) Also the strongestantenna and secondstrongestantenna have to be
     > next to each other (e.g. value0 with value60, value240 with
     > value300, value0 with value300) or the results should be NA.
     After setting finalbearing with A, B, and C then:
     yourdata$finalbearing <- with(yourdata, ifelse (
     change.ind <5 & change.ind > 1 ,
     NA, finalbearing) )
     > I have been trying to use a series of if, else statements to produce
     > these bearing, but all I am producing is errors. Any suggestion
     > would be appreciated.
     >
     > Again THANKS for you efforts.
     >
     > Mike
     >
     > ---- Original message ----
     >> Date: Fri, 23 Jul 2010 23:01:56 -0700
     >> From: Joshua Wiley <[222]jwiley.psych at gmail.com>
     >> Subject: Re: [R] Trouble retrieving the second largest value from
     >> each row of a data.frame
     >> To: [223]mpward at illinois.edu
     >> Cc: [224]r-help at r-project.org
     >>
     >> Hi,
     >>
     >> Here is a little function that will do what you want and return a
     >> nice output:
     >>
     >> #Function To calculate top two values and return
     >> my.finder <- function(mydata) {
     >> my.fun <- function(data) {
     >> strongest <- which.max(data)
     >> secondstrongest <- which.max(data[-strongest])
     >> strongestantenna <- names(data)[strongest]
     >> secondstrongantenna <- names(data[-strongest])[secondstrongest]
     >> value <- matrix(c(data[strongest], data[secondstrongest],
     >> strongestantenna, secondstrongantenna), ncol =4)
     >> return(value)
     >> }
     >> dat <- apply(mydata, 1, my.fun)
     >> dat <- t(dat)
     >> dat <- as.data.frame(dat, stringsAsFactors = FALSE)
     >> colnames(dat) <- c("strongest",
"secondstrongest",
     >> "strongestantenna", "secondstrongantenna")
     >> dat[ , "strongest"] <- as.numeric(dat[ ,
"strongest"])
     >> dat[ , "secondstrongest"] <- as.numeric(dat[ ,
"secondstrongest"])
     >> return(dat)
     >> }
     >>
     >>
     >> #Using your example data:
     >>
     >> yourdata <- structure(list(value0 = c(-13007L, -12838L,
-12880L,
     >> -12805L,
     >> -12834L, -11068L, -12807L, -12770L, -12988L, -11779L), value60    
>> c(-11707L,
     >> -13210L, -11778L, -11653L, -13527L, -11698L, -14068L, -11665L,
     >> -11736L, -12873L), value120 = c(-11072L, -11176L, -11113L,
-11071L,
     >> -11067L, -12430L, -11092L, -11061L, -11137L, -12973L), value180   
>> c(-12471L,
     >> -11799L, -12439L, -12385L, -11638L, -12430L, -11709L, -12373L,
     >> -12570L, -12537L), value240 = c(-12838L, -13210L, -13089L,
-11561L,
     >> -13527L, -12430L, -11607L, -11426L, -13467L, -12973L), value300   
>> c(-13357L,
     >> -13845L, -13880L, -13317L, -13873L, -12814L, -13025L, -12805L,
     >> -13739L, -11146L)), .Names = c("value0",
"value60", "value120",
     >> "value180", "value240", "value300"),
class = "data.frame",
     >> row.names = c("1",
     >> "2", "3", "4", "5",
"6", "7", "8", "9", "10"))
     >>
     >> my.finder(yourdata) #and what you want is in a nicely labeled data
     >> frame
     >>
     >> #A potential problem is that it is not very efficient
     >>
     >> #Here is a test using a matrix of 100, 000 rows
     >> #sampled from the same range as your data
     >> #with the same number of columns
     >>
     >> data.test <- matrix(
     >> sample(seq(min(yourdata), max(yourdata)), size = 500000, replace  
>> TRUE),
     >> ncol = 5)
     >>
     >> system.time(my.finder(data.test))
     >>
     >> #On my system I get
     >>
     >>> system.time(my.finder(data.test))
     >> user system elapsed
     >> 2.89 0.00 2.89
     >>
     >> Hope that helps,
     >>
     >> Josh
     >>
     >>
     >>
     >> On Fri, Jul 23, 2010 at 6:20 PM, <[225]mpward at
illinois.edu> wrote:
     >>> I have a data frame with a couple million lines and want to
     >>> retrieve the largest and second largest values in each row,
along
     >>> with the label of the column these values are in. For example
     >>>
     >>> row 1
     >>> strongest=-11072
     >>> secondstrongest=-11707
     >>> strongestantenna=value120
     >>> secondstrongantenna=value60
     >>>
     >>> Below is the code I am using and a truncated data.frame.
     >>> Retrieving the largest value was easy, but I have been getting
     >>> errors every way I have tried to retrieve the second largest
     >>> value. I have not even tried to retrieve the labels for the
value
     >>> yet.
     >>>
     >>> Any help would be appreciated
     >>> Mike
     >>>
     >>>
     >>>> data<-
     >>>> data.frame(value0, value60, value120, value180, value240,
value300)
     >>>> data
     >>> value0 value60 value120 value180 value240 value300
     >>> 1 -13007 -11707 -11072 -12471 -12838 -13357
     >>> 2 -12838 -13210 -11176 -11799 -13210 -13845
     >>> 3 -12880 -11778 -11113 -12439 -13089 -13880
     >>> 4 -12805 -11653 -11071 -12385 -11561 -13317
     >>> 5 -12834 -13527 -11067 -11638 -13527 -13873
     >>> 6 -11068 -11698 -12430 -12430 -12430 -12814
     >>> 7 -12807 -14068 -11092 -11709 -11607 -13025
     >>> 8 -12770 -11665 -11061 -12373 -11426 -12805
     >>> 9 -12988 -11736 -11137 -12570 -13467 -13739
     >>> 10 -11779 -12873 -12973 -12537 -12973 -11146
     >>>> #largest value in the row
     >>>> strongest<-apply(data, 1, max)
     >>>>
     >>>>
     >>>> #second largest value in the row
     >>>> n<-function(data)(1/(min(1/(data[1, ]-max(data[1,
]))))+
     >>>> (max(data[1, ])))
     >>>> secondstrongest<-apply(data, 1, n)
     >>> Error in data[1, ] : incorrect number of dimensions
     >>>>
     >>>
     >>> ______________________________________________
     >>> [226]R-help at r-project.org mailing list
     >>> https://stat.ethz.ch/mailman/listinfo/r-help
     >>> PLEASE do read the posting guide
     [227]http://www.R-project.org/posting-guide.html
     >>> and provide commented, minimal, self-contained, reproducible
code.
     >>>
     >>
     >>
     >>
     >> --
     >> Joshua Wiley
     >> Ph.D. Student, Health Psychology
     >> University of California, Los Angeles
     >> [228]http://www.joshuawiley.com/
     >
     > ______________________________________________
     > [229]R-help at r-project.org mailing list
     > https://stat.ethz.ch/mailman/listinfo/r-help
     > PLEASE do read the posting guide
     [230]http://www.R-project.org/posting-guide.html
     > and provide commented, minimal, self-contained, reproducible code.
     ------------------------------
     Message: 32
     Date: Sat, 24 Jul 2010 17:57:10 -0700
     From: Joshua Wiley <[231]jwiley.psych at gmail.com>
     To: [232]mpward at illinois.edu
     Cc: [233]r-help at r-project.org
     Subject: Re: [R] Trouble retrieving the second largest value from each
     row of a data.frame
     Message-ID:
     <AANLkTikG3SgD+af50n2dsHMA3+[234]fa9gCQLuYBAaMS_P8w at
mail.gmail.com>
     Content-Type: text/plain; charset=UTF-8
     On Sat, Jul 24, 2010 at 5:09 PM, David Winsemius
     <[235]dwinsemius at comcast.net> wrote:
     >
     > On Jul 24, 2010, at 4:54 PM, <[236]mpward at illinois.edu>
wrote:
     >
     >> THANKS, but I have one issue and one question.
     >>
     >> For some reason the "secondstrongest" value for row 3
and 6 are
     incorrect
     >> (they are the strongest) the remaining 10 are correct??
     >
     > In my run of Wiley's code I instead get identical values for rows
2, 5,
     6.
     Yes, my apologies; I neglected a [-strongest] when extracting the
     second highest value. I included a corrected form below; however,
     Winsemius' code is cleaner, not to mention easier to generalize, so I
     see no reason not to use that option. You might consider using a
     different object name than 'diff' since it is also the name of a
     function.
     Josh
     #######
     my.finder <- function(mydata) {
     my.fun <- function(data) {
     strongest <- which.max(data)
     secondstrongest <- which.max(data[-strongest])
     strongestantenna <- names(data)[strongest]
     secondstrongantenna <- names(data[-strongest])[secondstrongest]
     value <- matrix(c(data[strongest], data[-strongest][secondstrongest],
     strongestantenna, secondstrongantenna), ncol =4)
     return(value)
     }
     dat <- apply(mydata, 1, my.fun)
     dat <- t(dat)
     dat <- as.data.frame(dat, stringsAsFactors = FALSE)
     colnames(dat) <- c("strongest", "secondstrongest",
     "strongestantenna", "secondstrongantenna")
     dat[ , "strongest"] <- as.numeric(dat[ ,
"strongest"])
     dat[ , "secondstrongest"] <- as.numeric(dat[ ,
"secondstrongest"])
     return(dat)
     }
     > Holtman's and my solutions did not suffer from that defect,
although
     mine
     > suffered from my misreading of your request, thinking that you wanted
     the
     > top 3. The fix is trivial
     >>
     >> These data are being used to track radio-tagged birds, they are
from
     >> automated radio telemetry receivers. ?I will applying the
following
     formula
     >>
     >> ?diff <- ((strongest- secondstrongest)/100)
     >> ?bearingdiff <-30-(-0.0624*(diff**2))-(2.8346*diff)
     >
     > vals <- c("value0", "value60",
"value120", "value180", "value240",
     > "value300")
     > value.str2 <- (match(yourdata$secondstrongestantenna, vals)-1)*60
     > value.str1 <- (match(yourdata$strongestantenna, vals)-1)*60
     > change.ind <- abs(match(yourdata, vals) - which(match(yourdata,
vals) )
     >
     >>
     >> A) Then the bearing diff is added to strongestantenna (value0     
0degrees)
     >> if the secondstrongestatenna is greater (eg value0 and value60),
     >
     >> B) or if the secondstrongestantenna is smaller than the
     strongestantenna,
     >> then the bearingdiff is substracted from the strongestantenna.
     >
     >>
     >> C) The only exception is that if value0 (0degrees) is strongest
and
     >> value300(360degrees) is the secondstrongestantenna then the
bearing is
     >> 360-bearingdiff.
     >
     >
     >> D) Also the strongestantenna and secondstrongestantenna have to be
next
     to
     >> each other (e.g. value0 with value60, value240 with value300,
value0
     with
     >> value300) or the results should be NA.
     >
     > After setting finalbearing with A, B, and C then:
     > yourdata$finalbearing <- with(yourdata, ifelse (
     > ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?change.ind <5 & change.ind >
1 ,
     > ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? NA, finalbearing) )
     >
     >> I have been trying to use a series of if, else statements to
produce
     these
     >> bearing, but all I am producing is errors. Any suggestion would be
     >> appreciated.
     >
     >
     >>
     >> Again THANKS for you efforts.
     >>
     >> Mike
     >>
     >> ---- Original message ----
     >>>
     >>> Date: Fri, 23 Jul 2010 23:01:56 -0700
     >>> From: Joshua Wiley <[237]jwiley.psych at gmail.com>
     >>> Subject: Re: [R] Trouble retrieving the second largest value
from each
     >>> row of ?a data.frame
     >>> To: [238]mpward at illinois.edu
     >>> Cc: [239]r-help at r-project.org
     >>>
     >>> Hi,
     >>>
     >>> Here is a little function that will do what you want and
return a nice
     >>> output:
     >>>
     >>> #Function To calculate top two values and return
     >>> my.finder <- function(mydata) {
     >>> my.fun <- function(data) {
     >>> ?strongest <- which.max(data)
     >>> ?secondstrongest <- which.max(data[-strongest])
     >>> ?strongestantenna <- names(data)[strongest]
     >>> ?secondstrongantenna <-
names(data[-strongest])[secondstrongest]
     >>> ?value <- matrix(c(data[strongest], data[secondstrongest],
     >>> ? ? ? ? ? ? ? ? ? ?strongestantenna, secondstrongantenna),
ncol =4)
     >>> ?return(value)
     >>> }
     >>> dat <- apply(mydata, 1, my.fun)
     >>> dat <- t(dat)
     >>> dat <- as.data.frame(dat, stringsAsFactors = FALSE)
     >>> colnames(dat) <- c("strongest",
"secondstrongest",
     >>> ? ? ? ? ? ? ? ? ? "strongestantenna",
"secondstrongantenna")
     >>> dat[ , "strongest"] <- as.numeric(dat[ ,
"strongest"])
     >>> dat[ , "secondstrongest"] <- as.numeric(dat[ ,
"secondstrongest"])
     >>> return(dat)
     >>> }
     >>>
     >>>
     >>> #Using your example data:
     >>>
     >>> yourdata <- structure(list(value0 = c(-13007L, -12838L,
-12880L,
     -12805L,
     >>> -12834L, -11068L, -12807L, -12770L, -12988L, -11779L), value60
>>> c(-11707L,
     >>> -13210L, -11778L, -11653L, -13527L, -11698L, -14068L, -11665L,
     >>> -11736L, -12873L), value120 = c(-11072L, -11176L, -11113L,
-11071L,
     >>> -11067L, -12430L, -11092L, -11061L, -11137L, -12973L),
value180      >>> c(-12471L,
     >>> -11799L, -12439L, -12385L, -11638L, -12430L, -11709L, -12373L,
     >>> -12570L, -12537L), value240 = c(-12838L, -13210L, -13089L,
-11561L,
     >>> -13527L, -12430L, -11607L, -11426L, -13467L, -12973L),
value300      >>> c(-13357L,
     >>> -13845L, -13880L, -13317L, -13873L, -12814L, -13025L, -12805L,
     >>> -13739L, -11146L)), .Names = c("value0",
"value60", "value120",
     >>> "value180", "value240",
"value300"), class = "data.frame", row.names     
>>> c("1",
     >>> "2", "3", "4", "5",
"6", "7", "8", "9", "10"))
     >>>
     >>> my.finder(yourdata) #and what you want is in a nicely labeled
data
     frame
     >>>
     >>> #A potential problem is that it is not very efficient
     >>>
     >>> #Here is a test using a matrix of 100, 000 rows
     >>> #sampled from the same range as your data
     >>> #with the same number of columns
     >>>
     >>> data.test <- matrix(
     >>> sample(seq(min(yourdata), max(yourdata)), size = 500000,
replace      TRUE),
     >>> ncol = 5)
     >>>
     >>> system.time(my.finder(data.test))
     >>>
     >>> #On my system I get
     >>>
     >>>> system.time(my.finder(data.test))
     >>>
     >>> ?user ?system elapsed
     >>> ?2.89 ? ?0.00 ? ?2.89
     >>>
     >>> Hope that helps,
     >>>
     >>> Josh
     >>>
     >>>
     >>>
     >>> On Fri, Jul 23, 2010 at 6:20 PM, ?<[240]mpward at
illinois.edu> wrote:
     >>>>
     >>>> I have a data frame with a couple million lines and want
to retrieve
     the
     >>>> largest and second largest values in each row, along with
the label
     of the
     >>>> column these values are in. For example
     >>>>
     >>>> row 1
     >>>> strongest=-11072
     >>>> secondstrongest=-11707
     >>>> strongestantenna=value120
     >>>> secondstrongantenna=value60
     >>>>
     >>>> Below is the code I am using and a truncated data.frame.
?Retrieving
     the
     >>>> largest value was easy, but I have been getting errors
every way I
     have
     >>>> tried to retrieve the second largest value. ?I have not
even tried to
     >>>> retrieve the labels for the value yet.
     >>>>
     >>>> Any help would be appreciated
     >>>> Mike
     >>>>
     >>>>
     >>>>> data<-data.frame(value0, value60, value120,
value180, value240,
     value300)
     >>>>> data
     >>>>
     >>>> ?value0 value60 value120 value180 value240 value300
     >>>> 1 ?-13007 ?-11707 ? -11072 ? -12471 ? -12838 ? -13357
     >>>> 2 ?-12838 ?-13210 ? -11176 ? -11799 ? -13210 ? -13845
     >>>> 3 ?-12880 ?-11778 ? -11113 ? -12439 ? -13089 ? -13880
     >>>> 4 ?-12805 ?-11653 ? -11071 ? -12385 ? -11561 ? -13317
     >>>> 5 ?-12834 ?-13527 ? -11067 ? -11638 ? -13527 ? -13873
     >>>> 6 ?-11068 ?-11698 ? -12430 ? -12430 ? -12430 ? -12814
     >>>> 7 ?-12807 ?-14068 ? -11092 ? -11709 ? -11607 ? -13025
     >>>> 8 ?-12770 ?-11665 ? -11061 ? -12373 ? -11426 ? -12805
     >>>> 9 ?-12988 ?-11736 ? -11137 ? -12570 ? -13467 ? -13739
     >>>> 10 -11779 ?-12873 ? -12973 ? -12537 ? -12973 ? -11146
     >>>>>
     >>>>> #largest value in the row
     >>>>> strongest<-apply(data, 1, max)
     >>>>>
     >>>>>
     >>>>> #second largest value in the row
     >>>>> n<-function(data)(1/(min(1/(data[1, ]-max(data[1,
]))))+
     (max(data[1, ])))
     >>>>> secondstrongest<-apply(data, 1, n)
     >>>>
     >>>> Error in data[1, ] : incorrect number of dimensions
     >>>>>
     >>>>
     >>>> ______________________________________________
     >>>> [241]R-help at r-project.org mailing list
     >>>> https://stat.ethz.ch/mailman/listinfo/r-help
     >>>> PLEASE do read the posting guide
     >>>> [242]http://www.R-project.org/posting-guide.html
     >>>> and provide commented, minimal, self-contained,
reproducible code.
     >>>>
     >>>
     >>>
     >>>
     >>> --
     >>> Joshua Wiley
     >>> Ph.D. Student, Health Psychology
     >>> University of California, Los Angeles
     >>> [243]http://www.joshuawiley.com/
     >>
     >> ______________________________________________
     >> [244]R-help at r-project.org mailing list
     >> https://stat.ethz.ch/mailman/listinfo/r-help
     >> PLEASE do read the posting guide
     >> [245]http://www.R-project.org/posting-guide.html
     >> and provide commented, minimal, self-contained, reproducible code.
     >
     >
     --
     Joshua Wiley
     Ph.D. Student, Health Psychology
     University of California, Los Angeles
     [246]http://www.joshuawiley.com/
     ------------------------------
     Message: 33
     Date: Sat, 24 Jul 2010 21:27:17 -0400
     From: David Winsemius <[247]dwinsemius at comcast.net>
     To: "[248]r-help at r-project.org list" <[249]r-help at
r-project.org>
     Cc: [250]mpward at illinois.edu
     Subject: Re: [R] Trouble retrieving the second largest value from each
     row of a data.frame
     Message-ID: <[251]2B19FDC3-4358-4731-87C7-89399E5DD75E at
comcast.net>
     Content-Type: text/plain; charset=US-ASCII; format=flowed; delsp=yes
     On Jul 24, 2010, at 8:09 PM, David Winsemius wrote:
     >
     > On Jul 24, 2010, at 4:54 PM, <[252]mpward at illinois.edu>
wrote:
     >
     >> THANKS, but I have one issue and one question.
     >>
     >> For some reason the "secondstrongest" value for row 3
and 6 are
     >> incorrect (they are the strongest) the remaining 10 are correct??
     >
     > In my run of Wiley's code I instead get identical values for rows
     > 2, 5, 6. Holtman's and my solutions did not suffer from that
defect,
     > although mine suffered from my misreading of your request, thinking
     > that you wanted the top 3. The fix is trivial
     >>
     >> These data are being used to track radio-tagged birds, they are
     >> from automated radio telemetry receivers. I will applying the
     >> following formula
     >>
     >> diff <- ((strongest- secondstrongest)/100)
     >> bearingdiff <-30-(-0.0624*(diff**2))-(2.8346*diff)
     >
     > vals <- c("value0", "value60",
"value120", "value180", "value240",
     > "value300")
     > value.str2 <- (match(yourdata$secondstrongestantenna, vals)-1)*60
     > value.str1 <- (match(yourdata$strongestantenna, vals)-1)*60
     > change.ind <- abs(match(yourdata, vals) - which(match(yourdata,
     > vals) )
     OOOPs should have been
     change.ind <- abs(match(yourdata, vals) - match(yourdata, vals) )
     >
     >>
     >> A) Then the bearing diff is added to strongestantenna (value0     
>> 0degrees) if the secondstrongestatenna is greater (eg value0 and
     >> value60),
     >
     >> B) or if the secondstrongestantenna is smaller than the
     >> strongestantenna,
     >> then the bearingdiff is substracted from the strongestantenna.
     yourdata$finalbearing <- with(yourdata, ifelse
(value.str2>value.str1,
     bearingdiff+value.str1, value.str1-bearingdiff) )
     >
     >>
     >> C) The only exception is that if value0 (0degrees) is strongest
and
     >> value300(360degrees) is the secondstrongestantenna then the
bearing
     >> is 360-bearingdiff.
     >
     yourdata$finalbearing <- with(yourdata, ifelse (strongestantenna =    
"value0" & secondstrongantenna == "value300", 360-
bearingdiff,
     finalbearing) );
     >> D) Also the strongestantenna and secondstrongestantenna have to be
     >> next to each other (e.g. value0 with value60, value240 with
     >> value300, value0 with value300) or the results should be NA.
     >
     > After setting finalbearing with A, B, and C then:
     > yourdata$finalbearing <- with(yourdata, ifelse (
     > change.ind <5 & change.ind > 1 ,
     > NA, finalbearing) )
     > yourdata
     strongest secondstrongest strongestantenna secondstrongantenna
     finalbearing
     1 -11072 -11707 value120 value60
     -11086.52
     2 -11176 -11799 value120 value180
     -11190.76
     3 -11113 -11778 value120 value60
     -11126.91
     4 -11071 -11561 value120
     value240 NA
     5 -11067 -11638 value120 value180
     -11082.85
     6 -11068 -11698 value0 value60
     -11082.62
     7 -11092 -11607 value120
     value240 NA
     8 -11061 -11426 value120
     value240 NA
     9 -11137 -11736 value120 value60
     -11152.26
     10 -11146 -11779 value300 value0
     -11160.56
     >
     >> I have been trying to use a series of if, else statements to
produce
     >> these bearing,
     ifelse is the correct construct for processing vectors
     --
     David.
     >> but all I am producing is errors. Any suggestion would be
     >> appreciated.
     >
     >
     >>
     >> Again THANKS for you efforts.
     >>
     >> Mike
     >>
     >> ---- Original message ----
     >>> Date: Fri, 23 Jul 2010 23:01:56 -0700
     >>> From: Joshua Wiley <[253]jwiley.psych at gmail.com>
     >>> Subject: Re: [R] Trouble retrieving the second largest value
from
     >>> each row of a data.frame
     >>> To: [254]mpward at illinois.edu
     >>> Cc: [255]r-help at r-project.org
     >>>
     >>> Hi,
     >>>
     >>> Here is a little function that will do what you want and
return a
     >>> nice output:
     >>>
     >>> #Function To calculate top two values and return
     >>> my.finder <- function(mydata) {
     >>> my.fun <- function(data) {
     >>> strongest <- which.max(data)
     >>> secondstrongest <- which.max(data[-strongest])
     >>> strongestantenna <- names(data)[strongest]
     >>> secondstrongantenna <-
names(data[-strongest])[secondstrongest]
     >>> value <- matrix(c(data[strongest], data[secondstrongest],
     >>> strongestantenna, secondstrongantenna), ncol =4)
     >>> return(value)
     >>> }
     >>> dat <- apply(mydata, 1, my.fun)
     >>> dat <- t(dat)
     >>> dat <- as.data.frame(dat, stringsAsFactors = FALSE)
     >>> colnames(dat) <- c("strongest",
"secondstrongest",
     >>> "strongestantenna", "secondstrongantenna")
     >>> dat[ , "strongest"] <- as.numeric(dat[ ,
"strongest"])
     >>> dat[ , "secondstrongest"] <- as.numeric(dat[ ,
"secondstrongest"])
     >>> return(dat)
     >>> }
     >>>
     >>>
     >>> #Using your example data:
     >>>
     >>> yourdata <- structure(list(value0 = c(-13007L, -12838L,
-12880L,
     >>> -12805L,
     >>> -12834L, -11068L, -12807L, -12770L, -12988L, -11779L), value60
>>> c(-11707L,
     >>> -13210L, -11778L, -11653L, -13527L, -11698L, -14068L, -11665L,
     >>> -11736L, -12873L), value120 = c(-11072L, -11176L, -11113L,
-11071L,
     >>> -11067L, -12430L, -11092L, -11061L, -11137L, -12973L),
value180      >>> c(-12471L,
     >>> -11799L, -12439L, -12385L, -11638L, -12430L, -11709L, -12373L,
     >>> -12570L, -12537L), value240 = c(-12838L, -13210L, -13089L,
-11561L,
     >>> -13527L, -12430L, -11607L, -11426L, -13467L, -12973L),
value300      >>> c(-13357L,
     >>> -13845L, -13880L, -13317L, -13873L, -12814L, -13025L, -12805L,
     >>> -13739L, -11146L)), .Names = c("value0",
"value60", "value120",
     >>> "value180", "value240",
"value300"), class = "data.frame",
     >>> row.names = c("1",
     >>> "2", "3", "4", "5",
"6", "7", "8", "9", "10"))
     >>>
     >>> my.finder(yourdata) #and what you want is in a nicely labeled
data
     >>> frame
     >>>
     >>> #A potential problem is that it is not very efficient
     >>>
     >>> #Here is a test using a matrix of 100, 000 rows
     >>> #sampled from the same range as your data
     >>> #with the same number of columns
     >>>
     >>> data.test <- matrix(
     >>> sample(seq(min(yourdata), max(yourdata)), size = 500000,
replace      >>> TRUE),
     >>> ncol = 5)
     >>>
     >>> system.time(my.finder(data.test))
     >>>
     >>> #On my system I get
     >>>
     >>>> system.time(my.finder(data.test))
     >>> user system elapsed
     >>> 2.89 0.00 2.89
     >>>
     >>> Hope that helps,
     >>>
     >>> Josh
     >>>
     >>>
     >>>
     >>> On Fri, Jul 23, 2010 at 6:20 PM, <[256]mpward at
illinois.edu> wrote:
     >>>> I have a data frame with a couple million lines and want
to
     >>>> retrieve the largest and second largest values in each
row, along
     >>>> with the label of the column these values are in. For
example
     >>>>
     >>>> row 1
     >>>> strongest=-11072
     >>>> secondstrongest=-11707
     >>>> strongestantenna=value120
     >>>> secondstrongantenna=value60
     >>>>
     >>>> Below is the code I am using and a truncated data.frame.
     >>>> Retrieving the largest value was easy, but I have been
getting
     >>>> errors every way I have tried to retrieve the second
largest
     >>>> value. I have not even tried to retrieve the labels for
the
     >>>> value yet.
     >>>>
     >>>> Any help would be appreciated
     >>>> Mike
     >>>>
     >>>>
     >>>>> data<-
     >>>>> data.frame(value0, value60, value120, value180,
value240, value300)
     >>>>> data
     >>>> value0 value60 value120 value180 value240 value300
     >>>> 1 -13007 -11707 -11072 -12471 -12838 -13357
     >>>> 2 -12838 -13210 -11176 -11799 -13210 -13845
     >>>> 3 -12880 -11778 -11113 -12439 -13089 -13880
     >>>> 4 -12805 -11653 -11071 -12385 -11561 -13317
     >>>> 5 -12834 -13527 -11067 -11638 -13527 -13873
     >>>> 6 -11068 -11698 -12430 -12430 -12430 -12814
     >>>> 7 -12807 -14068 -11092 -11709 -11607 -13025
     >>>> 8 -12770 -11665 -11061 -12373 -11426 -12805
     >>>> 9 -12988 -11736 -11137 -12570 -13467 -13739
     >>>> 10 -11779 -12873 -12973 -12537 -12973 -11146
     >>>>> #largest value in the row
     >>>>> strongest<-apply(data, 1, max)
     >>>>>
     >>>>>
     >>>>> #second largest value in the row
     >>>>> n<-function(data)(1/(min(1/(data[1, ]-max(data[1,
]))))+
     >>>>> (max(data[1, ])))
     >>>>> secondstrongest<-apply(data, 1, n)
     >>>> Error in data[1, ] : incorrect number of dimensions
     >>>>>
     >>>>
     >>>> ______________________________________________
     >>>> [257]R-help at r-project.org mailing list
     >>>> https://stat.ethz.ch/mailman/listinfo/r-help
     >>>> PLEASE do read the posting guide
     [258]http://www.R-project.org/posting-guide.html
     >>>> and provide commented, minimal, self-contained,
reproducible code.
     >>>>
     >>>
     >>>
     >>>
     >>> --
     >>> Joshua Wiley
     >>> Ph.D. Student, Health Psychology
     >>> University of California, Los Angeles
     >>> [259]http://www.joshuawiley.com/
     >>
     >> ______________________________________________
     >> [260]R-help at r-project.org mailing list
     >> https://stat.ethz.ch/mailman/listinfo/r-help
     >> PLEASE do read the posting guide
     [261]http://www.R-project.org/posting-guide.html
     >> and provide commented, minimal, self-contained, reproducible code.
     >
     > ______________________________________________
     > [262]R-help at r-project.org mailing list
     > https://stat.ethz.ch/mailman/listinfo/r-help
     > PLEASE do read the posting guide
     [263]http://www.R-project.org/posting-guide.html
     > and provide commented, minimal, self-contained, reproducible code.
     ------------------------------
     Message: 34
     Date: Sat, 24 Jul 2010 21:48:12 -0400
     From: David Winsemius <[264]dwinsemius at comcast.net>
     To: David Winsemius <[265]dwinsemius at comcast.net>
     Cc: "[266]r-help at r-project.org list" <[267]r-help at
r-project.org>,
     [268]mpward at illinois.edu
     Subject: Re: [R] Trouble retrieving the second largest value from each
     row of a data.frame
     Message-ID: <[269]9B25E777-650F-4419-92F0-9319A2B753B4 at
comcast.net>
     Content-Type: text/plain; charset=US-ASCII; format=flowed; delsp=yes
     On Jul 24, 2010, at 9:27 PM, David Winsemius wrote:
     >
     > On Jul 24, 2010, at 8:09 PM, David Winsemius wrote:
     >
     >>
     >> On Jul 24, 2010, at 4:54 PM, <[270]mpward at illinois.edu>
wrote:
     >>
     >>> THANKS, but I have one issue and one question.
     >>>
     >>> For some reason the "secondstrongest" value for row
3 and 6 are
     >>> incorrect (they are the strongest) the remaining 10 are
correct??
     >>
     >> In my run of Wiley's code I instead get identical values for
rows
     >> 2, 5, 6. Holtman's and my solutions did not suffer from that
defect,
     >> although mine suffered from my misreading of your request,
thinking
     >> that you wanted the top 3. The fix is trivial
     >>>
     >>> These data are being used to track radio-tagged birds, they
are
     >>> from automated radio telemetry receivers. I will applying the
     >>> following formula
     >>>
     >>> diff <- ((strongest- secondstrongest)/100)
     >>> bearingdiff <-30-(-0.0624*(diff**2))-(2.8346*diff)
     >>
     >> vals <- c("value0", "value60",
"value120", "value180", "value240",
     >> "value300")
     >> value.str2 <- (match(yourdata$secondstrongestantenna,
vals)-1)*60
     Had a misspelling ... rather:
     match(yourdata$secondstrongantenna, vals)
     >> value.str1 <- (match(yourdata$strongestantenna, vals)-1)*60
     >> change.ind <- abs(match(yourdata, vals) - which(match(yourdata,
     >> vals) )
     >
     > OOOPs should have been
     >
     > change.ind <- abs(match(yourdata, vals) - match(yourdata, vals) )
     >
     >
     >>
     >>>
     >>> A) Then the bearing diff is added to strongestantenna (value0 
>>> 0degrees) if the secondstrongestatenna is greater (eg value0 and
     >>> value60),
     >>
     >>> B) or if the secondstrongestantenna is smaller than the
     >>> strongestantenna,
     >>> then the bearingdiff is substracted from the strongestantenna.
     >
     > yourdata$finalbearing <- with(yourdata, ifelse
     > (value.str2>value.str1, bearingdiff+value.str1, value.str1-
     > bearingdiff) )
     >
     >
     >>
     >>>
     >>> C) The only exception is that if value0 (0degrees) is
strongest
     >>> and value300(360degrees) is the secondstrongestantenna then
the
     >>> bearing is 360-bearingdiff.
     >>
     >
     > yourdata$finalbearing <- with(yourdata, ifelse (strongestantenna = 
> "value0" & secondstrongantenna == "value300", 360-
bearingdiff,
     > finalbearing) );
     >
     >
     >>> D) Also the strongestantenna and secondstrongestantenna have
to be
     >>> next to each other (e.g. value0 with value60, value240 with
     >>> value300, value0 with value300) or the results should be NA.
     >>
     >> After setting finalbearing with A, B, and C then:
     >> yourdata$finalbearing <- with(yourdata, ifelse (
     >> change.ind <5 & change.ind > 1 ,
     >> NA, finalbearing) )
     >
     Better result with proper creation of value.str2:
     yourdata
     strongest secondstrongest strongestantenna secondstrongantenna
     finalbearing
     1 -11072 -11707 value120 value60
     105.48359
     2 -11176 -11799 value120 value180
     134.76237
     3 -11113 -11778 value120 value60
     106.09061
     4 -11071 -11561 value120
     value240 NA
     5 -11067 -11638 value120 value180
     135.84893
     6 -11068 -11698 value0 value60
     14.61868
     7 -11092 -11607 value120
     value240 NA
     8 -11061 -11426 value120
     value240 NA
     9 -11137 -11736 value120 value60
     104.74034
     10 -11146 -11779 value300 value0
     285.44272
     >>
     >>> I have been trying to use a series of if, else statements to
     >>> produce these bearing,
     >
     > ifelse is the correct construct for processing vectors
     >
     > --
     > David.
     >>> but all I am producing is errors. Any suggestion would be
     >>> appreciated.
     >>
     >>
     >>>
     >>> Again THANKS for you efforts.
     >>>
     >>> Mike
     >>>
     >>> ---- Original message ----
     >>>> Date: Fri, 23 Jul 2010 23:01:56 -0700
     >>>> From: Joshua Wiley <[271]jwiley.psych at gmail.com>
     >>>> Subject: Re: [R] Trouble retrieving the second largest
value from
     >>>> each row of a data.frame
     >>>> To: [272]mpward at illinois.edu
     >>>> Cc: [273]r-help at r-project.org
     >>>>
     >>>> Hi,
     >>>>
     >>>> Here is a little function that will do what you want and
return a
     >>>> nice output:
     >>>>
     >>>> #Function To calculate top two values and return
     >>>> my.finder <- function(mydata) {
     >>>> my.fun <- function(data) {
     >>>> strongest <- which.max(data)
     >>>> secondstrongest <- which.max(data[-strongest])
     >>>> strongestantenna <- names(data)[strongest]
     >>>> secondstrongantenna <-
names(data[-strongest])[secondstrongest]
     >>>> value <- matrix(c(data[strongest],
data[secondstrongest],
     >>>> strongestantenna, secondstrongantenna), ncol =4)
     >>>> return(value)
     >>>> }
     >>>> dat <- apply(mydata, 1, my.fun)
     >>>> dat <- t(dat)
     >>>> dat <- as.data.frame(dat, stringsAsFactors = FALSE)
     >>>> colnames(dat) <- c("strongest",
"secondstrongest",
     >>>> "strongestantenna",
"secondstrongantenna")
     >>>> dat[ , "strongest"] <- as.numeric(dat[ ,
"strongest"])
     >>>> dat[ , "secondstrongest"] <- as.numeric(dat[
, "secondstrongest"])
     >>>> return(dat)
     >>>> }
     >>>>
     >>>>
     >>>> #Using your example data:
     >>>>
     >>>> yourdata <- structure(list(value0 = c(-13007L, -12838L,
-12880L,
     >>>> -12805L,
     >>>> -12834L, -11068L, -12807L, -12770L, -12988L, -11779L),
value60      >>>> c(-11707L,
     >>>> -13210L, -11778L, -11653L, -13527L, -11698L, -14068L,
-11665L,
     >>>> -11736L, -12873L), value120 = c(-11072L, -11176L, -11113L,
-11071L,
     >>>> -11067L, -12430L, -11092L, -11061L, -11137L, -12973L),
value180      >>>> c(-12471L,
     >>>> -11799L, -12439L, -12385L, -11638L, -12430L, -11709L,
-12373L,
     >>>> -12570L, -12537L), value240 = c(-12838L, -13210L, -13089L,
-11561L,
     >>>> -13527L, -12430L, -11607L, -11426L, -13467L, -12973L),
value300      >>>> c(-13357L,
     >>>> -13845L, -13880L, -13317L, -13873L, -12814L, -13025L,
-12805L,
     >>>> -13739L, -11146L)), .Names = c("value0",
"value60", "value120",
     >>>> "value180", "value240",
"value300"), class = "data.frame",
     >>>> row.names = c("1",
     >>>> "2", "3", "4",
"5", "6", "7", "8", "9",
"10"))
     >>>>
     >>>> my.finder(yourdata) #and what you want is in a nicely
labeled
     >>>> data frame
     >>>>
     >>>> #A potential problem is that it is not very efficient
     >>>>
     >>>> #Here is a test using a matrix of 100, 000 rows
     >>>> #sampled from the same range as your data
     >>>> #with the same number of columns
     >>>>
     >>>> data.test <- matrix(
     >>>> sample(seq(min(yourdata), max(yourdata)), size = 500000,
replace      >>>> TRUE),
     >>>> ncol = 5)
     >>>>
     >>>> system.time(my.finder(data.test))
     >>>>
     >>>> #On my system I get
     >>>>
     >>>>> system.time(my.finder(data.test))
     >>>> user system elapsed
     >>>> 2.89 0.00 2.89
     >>>>
     >>>> Hope that helps,
     >>>>
     >>>> Josh
     >>>>
     >>>>
     >>>>
     >>>> On Fri, Jul 23, 2010 at 6:20 PM, <[274]mpward at
illinois.edu> wrote:
     >>>>> I have a data frame with a couple million lines and
want to
     >>>>> retrieve the largest and second largest values in each
row,
     >>>>> along with the label of the column these values are
in. For
     >>>>> example
     >>>>>
     >>>>> row 1
     >>>>> strongest=-11072
     >>>>> secondstrongest=-11707
     >>>>> strongestantenna=value120
     >>>>> secondstrongantenna=value60
     >>>>>
     >>>>> Below is the code I am using and a truncated
data.frame.
     >>>>> Retrieving the largest value was easy, but I have been
getting
     >>>>> errors every way I have tried to retrieve the second
largest
     >>>>> value. I have not even tried to retrieve the labels
for the
     >>>>> value yet.
     >>>>>
     >>>>> Any help would be appreciated
     >>>>> Mike
     >>>>>
     >>>>>
     >>>>>> data<-
     >>>>>> data.frame(value0, value60, value120, value180,
value240, value300)
     >>>>>> data
     >>>>> value0 value60 value120 value180 value240 value300
     >>>>> 1 -13007 -11707 -11072 -12471 -12838 -13357
     >>>>> 2 -12838 -13210 -11176 -11799 -13210 -13845
     >>>>> 3 -12880 -11778 -11113 -12439 -13089 -13880
     >>>>> 4 -12805 -11653 -11071 -12385 -11561 -13317
     >>>>> 5 -12834 -13527 -11067 -11638 -13527 -13873
     >>>>> 6 -11068 -11698 -12430 -12430 -12430 -12814
     >>>>> 7 -12807 -14068 -11092 -11709 -11607 -13025
     >>>>> 8 -12770 -11665 -11061 -12373 -11426 -12805
     >>>>> 9 -12988 -11736 -11137 -12570 -13467 -13739
     >>>>> 10 -11779 -12873 -12973 -12537 -12973 -11146
     >>>>>> #largest value in the row
     >>>>>> strongest<-apply(data, 1, max)
     >>>>>>
     >>>>>>
     >>>>>> #second largest value in the row
     >>>>>> n<-function(data)(1/(min(1/(data[1,
]-max(data[1, ]))))+
     >>>>>> (max(data[1, ])))
     >>>>>> secondstrongest<-apply(data, 1, n)
     >>>>> Error in data[1, ] : incorrect number of dimensions
     >>>>>>
     >>>>>
     >>>>> ______________________________________________
     >>>>> [275]R-help at r-project.org mailing list
     >>>>> https://stat.ethz.ch/mailman/listinfo/r-help
     >>>>> PLEASE do read the posting guide
     [276]http://www.R-project.org/posting-guide.html
     >>>>> and provide commented, minimal, self-contained,
reproducible code.
     >>>>>
     >>>>
     >>>>
     >>>>
     >>>> --
     >>>> Joshua Wiley
     >>>> Ph.D. Student, Health Psychology
     >>>> University of California, Los Angeles
     >>>> [277]http://www.joshuawiley.com/
     >>>
     >>> ______________________________________________
     >>> [278]R-help at r-project.org mailing list
     >>> https://stat.ethz.ch/mailman/listinfo/r-help
     >>> PLEASE do read the posting guide
     [279]http://www.R-project.org/posting-guide.html
     >>> and provide commented, minimal, self-contained, reproducible
code.
     >>
     >> ______________________________________________
     >> [280]R-help at r-project.org mailing list
     >> https://stat.ethz.ch/mailman/listinfo/r-help
     >> PLEASE do read the posting guide
     [281]http://www.R-project.org/posting-guide.html
     >> and provide commented, minimal, self-contained, reproducible code.
     >
     > ______________________________________________
     > [282]R-help at r-project.org mailing list
     > https://stat.ethz.ch/mailman/listinfo/r-help
     > PLEASE do read the posting guide
     [283]http://www.R-project.org/posting-guide.html
     > and provide commented, minimal, self-contained, reproducible code.
     ------------------------------
     Message: 35
     Date: Sat, 24 Jul 2010 22:51:50 -0400
     From: paaventhan jeyaganth <[284]paaveenthan at hotmail.com>
     To: r <[285]r-help at r-project.org>
     Subject: [R] c-statiscs 95% CI for cox regression model
     Message-ID: <[286]BLU140-W10DC8DB498004130C2840DB4A50 at phx.gbl>
     Content-Type: text/plain
     Dear all,
     how can i do the calculate the C-statistics
     95% confidences interval for the cox regression model.
     Thanks very much for your any help.
     Paaveenthan
     _________________________________________________________________
     [[elided Hotmail spam]]
     [[alternative HTML version deleted]]
     ------------------------------
     Message: 36
     Date: Sat, 24 Jul 2010 22:48:59 -0500
     From: Dirk Eddelbuettel <[287]edd at debian.org>
     To: Frank E Harrell Jr <[288]f.harrell at vanderbilt.edu>
     Cc: "[289]r-help at r-project.org" <[290]r-help at
r-project.org>
     Subject: Re: [R] UseR! 2010 - my impressions
     Message-ID: <[291]20100725034859.GA11668 at eddelbuettel.com>
     Content-Type: text/plain; charset=us-ascii
     On Sat, Jul 24, 2010 at 08:55:01AM -0500, Frank E Harrell Jr wrote:
     > On 07/23/2010 06:50 PM, Ravi Varadhan wrote:
     > I want to echo what Ravi said. The talks were terrific (thanks to
     > the program committee and the speakers) and Kate Mullen and her team
     > did an extraordinary job in putting the conference together and
     > running it. I am proud to have been a part of it. Thank you all!
     Not much to add to this, so I just leave it at "Yup!".
     Thanks for useR! 2010. A job well done, and then some.
     --
     Dirk Eddelbuettel | [292]edd at debian.org |
     [293]http://dirk.eddelbuettel.com
     ------------------------------
     Message: 37
     Date: Sat, 24 Jul 2010 23:19:54 -0500
     From: Frank E Harrell Jr <[294]f.harrell at Vanderbilt.Edu>
     To: paaventhan jeyaganth <[295]paaveenthan at hotmail.com>
     Cc: r <[296]r-help at r-project.org>
     Subject: Re: [R] c-statiscs 95% CI for cox regression model
     Message-ID: <[297]4C4BBB6A.9090400 at vanderbilt.edu>
     Content-Type: text/plain; charset="ISO-8859-1"; format=flowed
     On 07/24/2010 09:51 PM, paaventhan jeyaganth wrote:
     >
     > Dear all,
     > how can i do the calculate the C-statistics
     > 95% confidences interval for the cox regression model.
     > Thanks very much for your any help.
     > Paaveenthan
     install.packages('Hmisc')
     require(Hmisc
     ?rcorr.cens (there is an example at the bottom)
     Frank
     --
     Frank E Harrell Jr Professor and Chairman School of Medicine
     Department of Biostatistics Vanderbilt University
     ------------------------------
     Message: 38
     Date: Sun, 25 Jul 2010 07:57:35 +0200
     From: Michael Haenlein <[298]haenlein at escpeurope.eu>
     To: [299]r-help at r-project.org
     Subject: [R] Equivalent to go-to statement
     Message-ID:
     <[300]AANLkTimX1jOLHX6AkfzDqQEJR4LK5G_-yFfDhZK_U5_i at
mail.gmail.com>
     Content-Type: text/plain
     Dear all,
     I'm working with a code that consists of two parts: In Part 1 I'm
     generating
     a random graph using the igraph library (which represents the
     relationships
     between different nodes) and a vector (which represents a certain
     characteristic for each node):
     library(igraph)
     g <- watts.strogatz.game(1, 100, 5, 0.05)
     z <- rlnorm(100, 0, 1)
     In Part 2 I'm iteratively changing the elements of z in order to reach
a
     certain value of a certain target variable. I'm doing this using a
while
     statement:
     while (target_variable < threshold) {## adapt z}
     The problem is that in some rare cases this iterative procedure can take
     very long (a couple of million of iterations), depending on the specific
     structure of the graph generated in Part 1. I therefore would like to
     change
     Part 2 of my code in the sense that once a certain threshold number of
     iterations has been achieved, the iterative process in Part 2 stops and
     goes
     back to Part 1 to generate a new graph structure. So my idea is as
     follows:
     - Run Part 1 and generate g and z
     - Run Part 2 and iteratively modify z to maximize the target variable
     - If Part 2 can be obtained in less than X steps, then go to Part 3
     - If Part 2 takes more than X steps then go back to Part 1 and start again
     I think that R does not have a function like "go-to" or
"go-back".
     Does anybody know of a convenient way of doing this?
     Thanks very much for your help,
     Michael
     [[alternative HTML version deleted]]
     ------------------------------
     Message: 39
     Date: Sat, 24 Jul 2010 23:24:40 -0700 (PDT)
     From: Vipul Agarwal <[301]iitkvipul at gmail.com>
     To: [302]r-help at r-project.org
     Subject: [R] Outlier Problem in Survreg Function
     Message-ID: <[303]1280039080326-2301422.post at n4.nabble.com>
     Content-Type: text/plain; charset=us-ascii
     Hi Everyone,
     I have recently started using r and working on survival analysis using the
     function survreg.
     I am facing a trange problem. One of the covariates in my analysis has
     outliers because of which survreg is giving incorrect results. Howevere
     when
     I am removing the outliers or scaling down the values of the covariate by
     a
     factor of 2 it is giving correct results. Below is a ditribution of the
     ariable and the results
     Min. 1st Qu. Median Mean 3rd Qu. Max.
     0 30000 54500 95450 123000 1650000
     Survreg Resuts
     survreg(formula = Surv(TIME_TO_FAILURE, CENSOR_DEFAULT) ~ ADVANCE,
     data = data)
     Coefficients:
     (Intercept) ADVANCE
     0.000000 -6.385336
     Scale= 0.9785933
     Loglik(model)= -40227366 Loglik(intercept only)= -914141
     Chisq= -78626451 on 1 degrees of freedom, p= 1
     n=198099 (885 observations deleted due to missingness)
     Survreg Results after scaling down the variable by 10
     survreg(formula = Surv(TIME_TO_FAILURE, CENSOR_DEFAULT) ~ ADVANCE_SCALED,
     data = data)
     Coefficients:
     (Intercept) ADVANCE_SCALED
     4.132962e+00 -2.181577e-05
     Scale= 0.9428758
     Loglik(model)= -909139.4 Loglik(intercept only)= -914141
     Chisq= 10003.19 on 1 degrees of freedom, p= 0
     n=198099 (885 observations deleted due to missingness)
     Survreg Results Afte removing the outliers(5% of the obs)
     data <- subset(data, data$ADVANCE <= 200000)
     > survreg(Surv(TIME_TO_FAILURE, CENSOR_DEFAULT) ~ ADVANCE , data = data
)
     Call:
     survreg(formula = Surv(TIME_TO_FAILURE, CENSOR_DEFAULT) ~ ADVANCE,
     data = data)
     Coefficients:
     (Intercept) ADVANCE
     4.224298e+00 -3.727171e-06
     Scale= 0.9601186
     Loglik(model)= -822521.9 Loglik(intercept only)= -825137.1
     Chisq= 5230.49 on 1 degrees of freedom, p= 0
     n=177332 (444 observations deleted due to missingness)
     Please let me know if someone else has faced the same problem and what is
     the way around to deal with it ? Should I scale down the variable or
     remove
     the outliers?
     --
     View this message in context:
     [304]http://r.789695.n4.nabble.com/Outlier-Problem-in-Survreg-Function-tp2
     301422p2301422.html
     Sent from the R help mailing list archive at Nabble.com.
     ------------------------------
     Message: 40
     Date: Sun, 25 Jul 2010 02:43:24 -0400
     From: Gabor Grothendieck <[305]ggrothendieck at gmail.com>
     To: Michael Haenlein <[306]haenlein at escpeurope.eu>
     Cc: [307]r-help at r-project.org
     Subject: Re: [R] Equivalent to go-to statement
     Message-ID:
     <[308]AANLkTimG-oCXkzLqxpca6HnTbnErb4vbBEQKhYyOkaWs at
mail.gmail.com>
     Content-Type: text/plain; charset=ISO-8859-1
     On Sun, Jul 25, 2010 at 1:57 AM, Michael Haenlein
     <[309]haenlein at escpeurope.eu> wrote:
     > Dear all,
     >
     > I'm working with a code that consists of two parts: In Part 1
I'm
     generating
     > a random graph using the igraph library (which represents the
     relationships
     > between different nodes) and a vector (which represents a certain
     > characteristic for each node):
     >
     > library(igraph)
     > g <- watts.strogatz.game(1, 100, 5, 0.05)
     > z <- rlnorm(100, 0, 1)
     >
     > In Part 2 I'm iteratively changing the elements of z in order to
reach a
     > certain value of a certain target variable. I'm doing this using a
while
     > statement:
     >
     > while (target_variable < threshold) {## adapt z}
     >
     > The problem is that in some rare cases this iterative procedure can
take
     > very long (a couple of million of iterations), depending on the
specific
     > structure of the graph generated in Part 1. I therefore would like to
     change
     > Part 2 of my code in the sense that once a certain threshold number of
     > iterations has been achieved, the iterative process in Part 2 stops
and
     goes
     > back to Part 1 to generate a new graph structure. So my idea is as
     follows:
     >
     > - Run Part 1 and generate g and z
     > - Run Part 2 and iteratively modify z to maximize the target variable
     > - If Part 2 can be obtained in less than X steps, then go to Part 3
     > - If Part 2 takes more than X steps then go back to Part 1 and start
     again
     >
     > I think that R does not have a function like "go-to" or
"go-back".
     >
     > Does anybody know of a convenient way of doing this?
     >
     > Thanks very much for your help,
     >
     goto's can be replaced with loops. In this case create a double loop
     such that the outer loop does not repeat if the inner loop finished
     due to reaching the target:
     target_variable <- -Inf
     while(target_variable < threshold) {
     ...
     iter <- 0
     while(target_variable < threshold && iter < max_iter) {
     ... update iter and target_variable ...
     }
     }
     ------------------------------
     _______________________________________________
     [310]R-help at r-project.org mailing list
     https://stat.ethz.ch/mailman/listinfo/r-help
     PLEASE do read the posting guide
     [311]http://www.R-project.org/posting-guide.html
     and provide commented, minimal, self-contained, reproducible code.
     End of R-help Digest, Vol 89, Issue 25
     **************************************

   [312][rKWLzcpt.zNp8gmPEwGJCA00] 

  
[@from=dllmain&rcpt=r%2Dhelp%40r%2Dproject%2Eorg&msgid=%3C20100726113540%2EH
   M%2E0000000000000bg%40dllmain%2Ewwl737%2Ehanmail%2Enet%3E]

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imail.org
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metro.net
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r-project.org
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r-project.org
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at sina.com
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at SI-MSEV02.US.SINET.SI.EDU
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si.edu
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mail.gmail.com
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r-project.org
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illinois.edu
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illinois.edu
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r-project.org
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illinois.edu
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r-project.org
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r-project.org
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comcast.net
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comcast.net
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r-project.org
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r-project.org
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illinois.edu
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at comcast.net
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illinois.edu
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illinois.edu
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r-project.org
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debian.org
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vanderbilt.edu
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r-project.org
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at eddelbuettel.com
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debian.org
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Vanderbilt.Edu
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hotmail.com
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vanderbilt.edu
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escpeurope.eu
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r-project.org
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at mail.gmail.com
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at n4.nabble.com
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at gmail.com
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escpeurope.eu
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r-project.org
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http://mail2.daum.net/hanmail/mail/MailComposeFrame.daum?TO=AANLkTimG-oCXkzLqxpca6HnTbnErb4vbBEQKhYyOkaWs
at mail.gmail.com
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escpeurope.eu
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r-project.org
 311. http://www.r-project.org/posting-guide.html
 312. mailto:dllmain at hanmail.net

Hi all,


   After writing data in MMF(Memory Map File) using SEXP structure, can i
   reference in R?
   If input data is larger than 2GB, Can i reference MMF Data in R?


   my work environment :
   R version : 2.11.1
   OS : WinXP Pro sp3

   Thanks and best regards.

   Park, Young-Ju

   from Korea.

   [1][rKWLzcpt.zNp8gmPEwGJCA00] 

  
[@from=dllmain&rcpt=r%2Dhelp%40r%2Dproject%2Eorg&msgid=%3C20100726173212%2EH
   M%2E0000000000000bk%40dllmain%2Ewwl737%2Ehanmail%2Enet%3E]

References

   1. mailto:dllmain at hanmail.net