similar to: Why does 'apply(.., 1, .., ..)' transpose result

I am sure this si a simple problem but the solution is evading me. I have a list of matrices all with the same number of columns but different number of rows. The first two columns label the row. The labels are allways the same for the same row numbers, just some matricies have more rows. For example using 3 column matrices... > q.1 <- function(r){return(cbind(seq(0, 10, by=1)[1:r],

I have an R script with the following applicable lines: xshort <- window(s, start=st, end=ed) . . . xshort <- ts(xshort, frequency=1, start=1) . . . m1 <- m2 <- m3 <- m4 <- m5 <- m6 <- NULL m1 <- tslm(xshort ~ trend) I get an error: Error in get(dataname) : object 'xshort' not found When I do traceback() I get: 3: get(dataname) 2: tslm(xshort ~

thanks to all the help so far I have code now that works on windows (albeit very clunky) that can read the pertinent info from the shape header and return a list with all the necessary parameters to read info out of the shape file. Now come the big design questions 1) I have considered building a map object, basically a list that would hold the info from the shape file and a data frame as part of

It seems obvious to me that the empty string "" is length 0. cheers Worik [[alternative HTML version deleted]]

I have trouble with this: as.Date("Sep-1981", format="%b-%Y") Returns "NA" >From documentation for strftime '%b' Abbreviated month name in the current locale. (Also matches full name on input.) '%Y' Year with century. What am I doing wrong? cheers Worik [[alternative HTML version deleted]]

I am interested in the package 'rgp' But.... > install.packages("rgp") Installing package(s) into ‘/home/worik/R/x86_64-pc-linux-gnu-library/2.12’ (as ‘lib’ is unspecified) Warning message: In getDependencies(pkgs, dependencies, available, lib) : package ‘rgp’ is not available It is in cran: http://cran.r-project.org/web/packages/rgp/index.html How do I install it?

When I say: > Sys.time() [1] "2012-11-12 21:30:14 NZDT" > But that is not what my clock on the wall and my system say. Cannot show you my clock but... worik@lemy:/tmp$ date Tue Nov 13 10:32:20 NZDT 2012 Sys.time() is returning GMT $version.string [1] "R version 2.14.1 (2011-12-22)" > Sys.timezone() [1] "NZDT" > I'm a little lost! Worik

I have two pairs of related vectors x1,y1 and x2,y2 I wish to do a test for differences in means of x1 and y1, ditto x2 and y2. I am getting odd results. I am not sure I am using 'pt' properly... I have not included the raw vectors as they are long. I am interested if I am using R properly... > c(length(x1), length(y1), length(x2), length(y2)) [1] 3436 1619 2677 2378 First

Is it possible to get a line number with an error report? I have a long script and an error: Error in `[.xts`(x, xsubset) : subscript out of bounds It would be very helpful, and save a lot of time, if there was some indication in the error message which line the error was. I can find it using binary search but that is a painful process. cheers Worik [[alternative HTML version deleted]]

I think I am whistling in the wind, but is there a modern symbolic debugger for R programmes? I am working through some one else's code, thousands of lines, that has the occasional bug in it, and a lot in my understanding of it. I cannot make setBreakpoint or findLineNum work. I get "No source refs found.". I am starting to loose my mind! A debugger where I could set

Friends I am trying to format a number to a string so 2189.745 goes to "2,189.35" and 309283.929 goes to "309,283.93" I have tried to use formatC(X, big.mark=",",drop0trailing=FALSE, format="f") but it does not get the number of decimals correct. Specifying digits does not work as that is significant digits. I could use a switch statement switching on

I hope you can help with this one. I have two matricies: 1. A species abundance matrix: <http://r.789695.n4.nabble.com/file/n4643979/2species_matrix.jpg> 2. A species trait score matrix: <http://r.789695.n4.nabble.com/file/n4643979/2trait_matrix.jpg> The trait matrix lists trait scores for each species as listed in the species abundance matrix. I would like to create a script

Dear all, Happy New Year for all of you! I hope we have an year of essential freedom for everyone! I am trying to manipulate a matrix in order to know in which column and in which row a number is allocated. But, when we use the function "which" it returns the position of the number in the "vector representation of the matrix". For example: >

Friends I have an xts that I wish to access. Browse[2]> DATA.ba[[p]]["2012-03-20 00:59:57","bid"] bid 2012-03-20 00:59:57 1.4993 So far so good. Now putting the index into a variable: Browse[2]> Time [1] "2012-03-20 00:59:57 NZDT" Browse[2]> DATA.ba[[p]][Time, "bid"] bid Where has it gone? Looking closer....

If I create a vector thusly > v1 <- runif(20, min=0, max=1) > v1 [1] 0.9754443 0.6306228 0.3238158 0.3175769 0.6791534 0.6956507 0.3840803 [8] 0.1421328 0.8592398 0.4388306 0.9472040 0.4727435 0.5645302 0.7391616 [15] 0.6116199 0.2727754 0.2657867 0.5261744 0.8764804 0.2032126 And I want to create a logical vector the same length that is true if v1<.3 or v1 > .7 how do I do it

Friends I am looking at Rcpp and I am a bit stuck on a simple matter. (I am calling R from c++, if there is a better way...) Given this simple example using the TTR package and the SMA function which returns a simple moving average.... Rcpp::NumericVector rv; for(int i = 0; i < 100; i++){ rv.push_back(rand()); } Rcpp::Environment TTR("package:TTR");

I have a large programme that after running half an hour or so fails with an error Error in x[value] <- NA : only 0's may be mixed with negative subscripts How can I find out where that error occurs? If I have to do a binary search using error messages it will take a long time! Is there some way I can generate a stack trace? cheers Worik [[alternative HTML version deleted]]

How can I increment the value of a Date class? I want to add a day, month or year to a date. cheers Worik [[alternative HTML version deleted]]

This programme for(T in 1:3){ for(j in 1:(5-1)){ for(k in (j+1):5){ for(l in (j+2):5){ print(paste("1 JKL:", j,k,l,sep=" ")) } } } } Prints out (among other things) [1] "1 JKL: 4 5 6" That is for(l in (j+2):5) sets l to 6 one more than the upper limit. cheers Worik [[alternative HTML version deleted]]