similar to: extracting rows of a dataframe

Hi I want to write a little function that takes a matrix X of size m-by-n, and a list L of length "m", whose elements are matrices all of which have the same number of columns but possibly a different number of rows. I then want to get a sort of dumbed-down kronecker product in which X[i,j] is replaced by X[i,j]*L[[j]] where L[[j]] is the j-th of the "m" matrices. For

Hello there, I would like to simulate X --Normal (20, 5) Y-- Normal (40, 10) and the correlation between X and Y is 0.6. How do I do it in R? Thank you very much Lisa Wang Msc. Princess Margaret Hospital Toronto, Ca

I have found a strange scoping problem in Sweave. The following Rnw file doesn't produce the same output in Sweave as it does if I produce an R file using Stangle and execute that: \documentclass[12pt]{article} \begin{document} <<R>>= election <- data.frame(A=1:3, B=9:7, C=rep(0,3)) partytotal <- rep(0, ncol(election)) for (i in 1:ncol(election)) { partytotal[i] <-

I need to make a bunch of PDF files of histograms. I tried gatelist = unique(mdf$ArrivalGate) for( gate in gatelist) { outfile = paste("../", airport, "/", airport, "taxiHistogram", gate, ".pdf", sep="") pdf(file = outfile, width = 10, height=8, par(lwd=1)) title=paste("Taxi time for Arrival Gate", gate, "by

Hi, I've been trying to get this working for ages, but it causes R to hang. Here is my C code saved as test1.c # include <R.h> # include <Rmath.h> void test1 (double *x, double *result) { result[0] = rgamma(*x, 2.0); } This was compiled using R CMD SHLIB test1.c & loaded in R using: dyn.load("test1.dll") out <-

Hello! I am writing some functions and I repeatedly acces internal factor codes. I figured out that internal codes are 1:n where 1 represents 1st level, 2 2nd level etc. This is not documented and I wonder if this is on purpose and subject to change or would be a good idea to add this to documentation for factor? I can prepair a pacth if this is accepted. -- Lep pozdrav / With regards,

When a graph is saved as PostScript, the bounding box is often too big. A consequence is that when the graph is included in a LaTeX document, the spacing does not look good. Is this a recognized problem? Is someone working on it? Could I help? David

Full_Name: Myung Geun Kim Version: 2.4.0 OS: Window XP Submission from: (NULL) (210.110.8.105) For the data c1 of size 14, stem provides the following result. ************************************************************** >c1 [1] 14 39 70 11 38 20 37 15 41 74 74 34 48 51 ZZangi>stem(c1) The decimal point is 1 digit(s) to the right of the | 0 | 145 2 | 04789 4 | 181 6 | 044

I want to do the following: DEVw=500 DEVh=350 fname="my_plot" dev.print(file=fname, device=FOO, width=DEVw, height=DEVh, bg="transparent") How do I do this such that I can specify FOO to be one of several choices? (GDD, PNG, postscript, etc.) If I make FOO a character variable, then "dev.print" complains. I tried a simpled "substitute" but

What can I do? Thanks a lot! [[alternative HTML version deleted]]

Dear R community, I'm fitting a complex mixed-effects model that requires numerous iterations and function evaluations. I note that nlminb accepts a list of control parameters, including eval.max. Is there a way to change the default eval.max value for nlminb when it is being called from lme? Thanks for any thoughts, Andrew -- Andrew Robinson Department of Mathematics and Statistics

## While in R v. 2.3.1 (the mid-July patch for Windows) ## on a Windows XP machine, this call to order() works fine... order(1:10,decreasing = TRUE) ## [1] 10 9 8 7 6 5 4 3 2 1 ## ...however, the argument name 'decreasing' ## must be typed in toto (note the missing 'g' ## in the following): > order(1:10,decreasin = TRUE) ## Error in order(na.last, decreasing, ...) :

I want to know how to get trace of product of matrices **faster** when the matrices are really big. Unfortunately the matrices are not symmetric. If anybody know how to get the trace of it, please help me. An example is as below. n <- 2500 a <- matrix(rnorm(n*n),n,n) b <- matrix(rnorm(n*n),n,n) tr1 <- sum(diag(a %*% b)) tr2 <- sum(diag(a %*% b %*% a %*% b)) Thanks, Yongwan Chun

Dear all, seems that easy question but cannot find the function for that. How to get the elements of the minor diagonal of the matrix? Thanks a lot. [[alternative HTML version deleted]]

OK, my last question didn't get any replies so I am going to try and ask a different way. When I generate contrasts with contr.sum() for a 3 level categorical variable I get the 2 orthogonal contrasts: > contr.sum( c(1,2,3) ) [,1] [,2] 1 1 0 2 0 1 3 -1 -1 This provides the contrasts <1-3> and <2-3> as expected. But I also want it to create <1-2> (i.e.

Question: I installed R verison 2-12.0 on my Windows 7 (64 bit) PC. When I was installing it, it did not ask me anything about 32 vs. 64 bit. So, if I run R now - is it running as a 32-bit or a 64-bit? thank you! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

Hi, R users, Here is an example. k <- c(1,2,3,4,5) i <- c(0,1,3,2,1) if k=1, then j=0 from i if k=2, then j=0, 1 from i if k=3, then j=0, 1, 2, 3 from i if k=4, then j=0, 1, 2 from i if k=5, then j=0, 1 from i so i'd like to create a list like below. > list k j 1 1 0 2 2 0 3 2 1 4 3 0 5 3 1 6 3 2 7 3 3 8 4 0 9 4 1 10 4 2 11 5 0 12 5 1 I tried expand.grid, but I

Hi, Given some example data: dat <- seq(4, 7, by = 0.05) x <- sample(dat, 30) y <- sample(dat, 30) error <- x - y I have broken the rage of x into 10 groups and I can calculate the bias (mean(error)) for each of these 10 groups: groups <- cut(x, breaks = 10) max.bias <- aggregate(error, list(group = groups), mean) max.bias group x 1 (4,4.3] -0.7750000 2

This is a very simple question but I couldn't form a site search quesry that would return a reasonable result set. Say I have a vector: x <- c(0,2,3,4,5,-1,-2) I want to replace all of the values in 'x' with the log of x. Naturally this runs into problems since some of the values are negative or zero. So how can I replace all of the positive elements of x with the log(x) and the

Dear All, Is this right? > floor((5.05-floor(5))*100) [1] 4 I would expect 5, or am I wrong? Thanks and regards, W --------------------------------- [[alternative HTML version deleted]]