similar to: leap year and order function

Hi, I'm new to R. I'm trying to plot my data into a circle. my data sort of looks like 12,12,4,5,6,5,11,10,3,9,9,9,12,12,2 total of 15 numbers. I'm trying to add all the same numbers, such that, there are 4 of 12s,1 of 11, 1 of 10, 3 of 9s, and such.... so the circle plot would have 4 parts of 12, 1 part of 11, 1 part of 10, 3 part of 9, and such... I tried

Hi, I am R beginner and am trying to figure out how to generate a complete list of species for every point, visit, and year. The code below is close but does not give me a list of species for every point, visit, and year in my data set. spplist<-unique(sumPtCt$Species) spplength<-length(spplist) Pointlist<-unique(sumPtCt$Point) Pointlength<-length(Pointlist)

Hi all, I was wondering if you can help me with the following situation: I have a data frame that includes weather station data for 30 years in the form: YEAR, MONTH, DAY, TEMP 1970, 01, 01, -15 ... 1999, 12, 31, -21 I would like to add another variable "JULIAN" that assigns the integers 1 to 365 (and 1 to 366 for leap years) for each day of a year over multiple years. Here is what

Dear all, I am working with vfs recycle and it is a real great thing. But something I missed all the years is a param like : recycle:purge=MAXDAYS. (where MAXDAYS is, Age of which files and directories are stored In the recycle before they are purged) I do this with my own scripts but it would be a great feature for me as admin to set this param without the need of scripting myself. EDV

When reading a data set into R using readLines, na.strings="-99.99" is ignored. Is there an equivalent command for readLines? Alternatively, either immediately after reading into R or once the data set has been converted to a data frame, what is the appropriate command (or appropriate use of is.na) to convert my -99.99s to NAs? Thanks, Ben Osborne -- Botany Department University of

I am running the following command. for( i in 1:378){ for (j in 1:5) { *********Year [((i-1)*5)+j,1]<- yearly[i,2];********* Year [((i-1)*5)+j,2] <- j; Year [((i-1)*5)+j,3] <- yearly[i,(j+2)]; j <- (j+1); } i <- (i+1) } The array referenced yearly[i,2] contains conference names i.e. BIG12, SEC, BIG10. The loop executes fine except for the fact that

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 I have a stock market trading values time series. What's the best way to extract the "last day of month" values. I looked at function window() but doesn't appear suitable for this since it expects regular dates. Thank you. lukas - -- Lukas Kubin lukas.kubin at permonik.com phone: 00420603836180 -----BEGIN PGP SIGNATURE-----

v="IBM" library(quantmod) v v1=getSymbols(v) to.yearly(v1) =============================== when i pass the value through a variable in to.yearly() function it shows the error msg like "Error in try.xts(x) : Error in UseMethod("as.xts") : no applicable method for 'as.xts' applied to an object of class "character"" i need the result of OHLC

I'm wondering if the page application is broken in 1.2.5 The following: exten => 2001,1,Page(SIP/3254105) does strange stuff. The caller's phone immediately drops into the call, while the callee's phone is still ringing. I'd think it was a SIP messaging issue, except that the Dial() command is working fine, which makes me wonder if it's a bug in the Page appplication.

Hi everyone,I have a data daily data (x) for 10 years starting from 04-01-1995 to 03-31-2005. I was able to get the yearly sum for the ten years using aggregate(x, years, sum). But this gave me the yearly sum for 1995 (Apr- Dec); 1996 (Jan-Dec) ---------2005 (Jan-Mar). But I want to get the aggregates for Apr-1995 to Mar 1996, Apr 1996- mar 1997 and so on. your help will be higly appreciated.

I have some data set which is available from 2005-2010 . I would like plot monthly maps out of it. So how I should write loop that can plot this yearly data to for every month and write title for individual months too. -- View this message in context: http://r.789695.n4.nabble.com/Plotting-monthly-maps-from-yearly-data-tp4391704p4391704.html Sent from the R help mailing list archive at

Dear List, ? I want to fit a CPH survival model with a time-dependent cumulative exposure variable. More specifically, I have exposure measurements for all cohort members at discrete times (yearly measurements) during follow-up. ? I suppose that when estimating the model parameters the most correct would be to have the cumulative exposure for everybody at risk at each death time, as I cannot

Hi, I want to switch from our Windows NT server ( which works as our PDC ) to Linux - Samba. Could you advice a step-by-step guide about this process, or is that possible, at all? Can I use Samba as a BDC, convert the user DB and than promote it to PDC? Thanks in advance! Regards; Istvan

Hi R community I copied a bit of my R code that gets some data from a database. You won't be able to run the code, but I am a beginner so you will probably understand what going on. I would like to make a variable I can refer to inside the sqlQuery. Instead of writing the start date and time (ex SP.lokaldatotid >= '2005-01-01 00:00:00') inside the query I would like to define it

Dear R helpers Suppose val1 = c(10, 20, 35, 80, 12) val2 = c(3, 8, 11, 7) I want to select either val1 or val2 depending on value of third quantity val3. val3 assumes either of the values "Monthly" or "Yearly". If val3 = "Monthly", then val = val1 and if val3 = "Yearly", then val = val2. I tried the ifelse statement as ifelse(val3 =

Dear useRs,while trying to plot the yearly curves of 1000 stations and overlapping each set of curves with mean curve and then saving it automatically in a pdf file, i tried the following commands >Path = "C:\\R\\003.pdf">pdf(file=Path) for (i in seq(1:1000) >a<-lapply(seq_along(tcp), function(x) tcp[[x]][,-1]) >b<-lapply(seq_along(a), function(a)

Sorry, this is my first time to post. I have a big data set: first colume is date (ex: 2008-2-150, the second is time (10:30:00), and the following columes are variaty measurement data. Every 30 min, I have one data. I want to find an effecient way to calculate the hourly, daily, monthly and yearly average, and plot them, and eventually use these average data to do further analysis. Thanks!

Someone told me that if you have a CentOS or Fedora server, you can pay a Red Hat yearly fee and get them to support it (because the environments are so similar). Can anyone here substantiate this claim? -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://lists.centos.org/pipermail/centos/attachments/20091211/03c9856b/attachment.html>

This is a follow-up to the thread ending with: http://rubyforge.org/pipermail/vpim-talk/2008/000120.html I too am in search of some ruby parser for icalendar which properly handles timezones on the datetimes in the icalendar RFC. As I understand it there are actually three types of times. 1) UTC times with a string form of yyyymmddThhmmssZ note the trailing Z indicates zulu time aka utc. 2)

can you recommend a good manual for R that starts with a data set and gives demonstrations on what can be done using R? I downloadedR Langauage definition and An introduction to R but haven't found them overly useful. I'd really like to be able to follow some tutorials using a dataset or many datasets. The datasets I have available on R are Data sets in package 'datasets':