search for: pyne

Dear R helpers I have following input f = c(257, 520, 110). I need to generate a decreasing sequence (decreasing by 100) which will give me an input (in a tabular form) like 257, 157, 57 520, 420, 320, 220, 120, 20 110, 10 I tried the following R code f = c(257, 520, 110) yy = matrix(data = NA, nrow = 3, ncol = 6) for (i in 1:3)      {      value = NULL      for (j in 1 : 6)           {

...see it in the Network Neighborhood, but whenever I try to browse any shares it asks for a password. No matter what I try, I can't get it to authenticate (even thought the shares are marked as public). Any suggestions on troubleshooting will be greatly appreciated. --Richard -- Richard B. Pyne Software Engineer ShopSite, Inc.

Dear R helpers, xx = data.frame(country = c("USA", "UK", "Canada"), x = c(10, 50, 20), y = c(40, 80, 35), z = c(70, 62, 10)) > xx        country      x     y    z 1      USA        10    40  70 2      UK          50   80   62 3     Canada    20   35   10 I need to arrange this as a new data.frame as follows - country       type     values USA           

Dear R helpers Suppose I have a vector as vect1 = as.character(c("ABC", "XYZ", "LMN", "DEF")) > vect1 [1] "ABC" "XYZ" "LMN" "DEF" I want to reverse the order of this vector as vect2 = c("DEF", "LMN", "XYZ", "ABC") Kindly guide Regards Vincy [[alternative HTML

...I am simply not able to figure out how acf is used? Thanking you in advance. Regards Vincy --- On Wed, 8/24/11, Prof Brian Ripley <ripley@stats.ox.ac.uk> wrote: From: Prof Brian Ripley <ripley@stats.ox.ac.uk> Subject: Re: [R] Autocorrelation using library(tseries) To: "Vincy Pyne" <vincy_pyne@yahoo.ca> Cc: r-help@r-project.org Received: Wednesday, August 24, 2011, 9:08 AM Your understanding is wrong.  For a start, there is no function acf() in package tseries: it is in stats. And the autocorrelation at lag one is not the correlation omitting the first and last...

Dear R list, I have one very elementary question regrading correlation between two variables. x = c(44,46,46,47,45,43,45,44) y = c(44,43,41,41,46,48,44,43) > cov(x, y) [1] -2.428571 However, if I try to calculate the covariance using the formula as covariance = sum((x-mean(x))*(y-mean(y)))/8       # no of of paired obs. = 8 or     covariance = sum(x*y)/8-(mean(x)*mean(y)) gives

.................         1/31/2010     429      103         When I tried                                     mydat.sort <- mydat[order(mydat$date)] > mydat.sort <- mydat[order(mydat$date)] Error in `[.data.frame`(mydat, order(mydat$date)) :  undefined columns selected Kindly guide Vincy Pyne [[alternative HTML version deleted]]

Dear R helpers I have a dataframe as df = data.frame(x = c(1, 14, 3, 21, 11), y = c(102, 500, 40, 101, 189)) > df    x   y 1  1 102 2 14 500 3  3  40 4 21 101 5 11 189 # Actually I am having dataframe having multiple columns. I am just giving an example. I need to subtract all the rows of df by the first row of df i.e. I need to subtract each element of 'x' column by 1. Likewise I

Dear R helpers I have a data which gives Month-wise and Rating-wise Rates. So the input file is something like month           rating           rate January        AAA             9.04 February      AAA             9.07 .......................................... .......................................... Decemeber     AAA            8.97  January           BBB           11.15 February        

Hi, I have one basic doubt. Suppose X ~ N(50,10). I need to calculate Probability X = 50. dnorm(50, 50, 10) gives me [1] 0.03989423 My understanding is (which is bit statistical or may be mathematical) on a continuous scale, Probability of the type P(X = .....) are nothing but 1/Infinity i.e. = 0. So as per my understanding P(X = 50) should be 0, but even excel also gives 0.03989422. Obviously

Dear R helpers Wish you all a very Happy and Prosperous New Year 2011. I have following query. country = c("US", "France", "UK", "NewZealand", "Germany", "Austria", "Italy", "Canada") Through some other R process, the result.csv file is generated as result.csv      var1   var2  var3  var4    var5    var6   var7  

Dear R helpers, I have recently installed R version 2.15.0 I just wanted to calculate mean(16, 18) Surprisingly I got answer as > mean(16, 18) [1] 16 > mean(18, 16) [1] 18 > mean(14, 11, 17, 9, 5, 18) [1] 14 So instead of calculating simple Arithmetic average, mean command is generating first element as average. I restarted the machine, changed the machine, but still the

Dear R helpers Suppose x  <- c(1:3) y  <- matrix(1:12, ncol = 3, nrow = 4) > y      [,1] [,2] [,3] [1,]    1    5    9 [2,]    2    6   10 [3,]    3    7   11 [4,]    4    8   12 I wish to multiply 1st column of y by first element of x i.e. 1, 2nd column of y by 2nd element of x i.e. 2 an so on. Thus the resultant matrix should be like > z      [,1]   [,2]    [,3] [1,]    1   

Dear R helpers, (At the outset I sincerely apologize if I have not put forward my following query properly, though I have tried to do so.) Following is a curtailed part of my R - code where I am trying to generate say 100 random no.s for each of the products under consideration. library(plyr) n = 100 my_code = function(product, output_avg, output_stdev)     { BUR_mc = rnorm(n, output_avg,

After upgrading my dovecot installation about a month ago, I have started seeing "Requested CRAM-MD5 scheme, but we have only MD5-CRYPT" message from dovecot in my logs. Any help in finding and correcting the cause will be greatly appreciated. --Richard

Dear R helpers Suppose I have a vector as vect_1 = c("AAA", "AA", "A", "BBB", "BB", "B", "CCC") vect_1_id = c(1:length(vect_1)) Through some process I obtain vect_2_id = c(2, 3, 7), then I need a new vector say vect_2 which will give me vect2 = ("AA", "A", "CCC")  i.e. I need the subset of

Dear R helpers Suppose, x = c(0,  1,  2,  3) y = c("A", "B", "C", "D") z = c(1, 3) For given values of z, I need to the values of y. So I should get "B" and "D". I tried doing y[x][z] but it gives > y[x][z] [1] "A" "C" Kindly guide. Regards Vincy [[alternative HTML version deleted]]

Dear R helpers, I have one basic doubt about the value of pi. In school, we have learned that pi = 22/7 (which is = 3.142857). However, if I type pi in R, I get pi = 3.141593. So which value of pi should be considered? Regards Vincy [[alternative HTML version deleted]]

...you in advance and sincerely apologize for any mis-communication if it had occurred in my earlier mail. Regards Vincy --- On Fri, 8/27/10, David Winsemius <dwinsemius@comcast.net> wrote: From: David Winsemius <dwinsemius@comcast.net> Subject: Re: [R] Band-wise Sum To: "Vincy Pyne" <vincy_pyne@yahoo.ca> Cc: r-help@r-project.org Received: Friday, August 27, 2010, 2:36 PM On Aug 27, 2010, at 9:49 AM, Vincy Pyne wrote: > Hi > > I have a large credit portfolio (exceeding 50000 borrowers). For particular process I need to add up the exposures based on the...

Dear R helpers, I have one trivial problem while writing an output file in csv format. I have two dataframes say df1 and df2 which I am reading from two different csv files. df1 has column names as date, r1, r2, r3 while the dataframe df2 has column names as date, 1w, 2w. (the dates in both the date frames are identical also no of elements in each column are equal say = 10). I merge these