similar to: compute variance of every column in a matrix without a loop

Hi people, If there is a fn "colMeans" why isn't there a "colSDs" etc Thanks, Phil. -- Philip Rhoades Pricom Pty Limited (ACN 003 252 275) GPO Box 3411 Sydney NSW 2001 Australia Mobile: +61:0411-185-652 Fax: +61:2:8923-5363 E-mail: pri at chu.com.au -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

I need to calculate a large number of t statistics, and would like to do so via matrix operations. So far I have figured out a way to calculate the mean of each row of the matrix: d <- matrix(runif(100000,1,10), 1000, 10) # some test data s <- rep(1,ncol(d)) # a sum vector to use for matrix multiplication means <- (d%*%s)/ncol(d) This is at least 1 order of magnitude faster than

I am using the R 2.2.1 in a Windows XP environment. I have a dataframe with 12 columns and 1,000 rows. (Some of the rows have 1 or fewer values.) I am trying to use rowVars to calculate the variance of each row. I am getting the following message: ?Error in na.remove.default(x) : length of 'dimnames' [1] not equal to array extent? Is there a good work-around?

Mark Leeds <mleeds at mlp.com> wrote (to S-News): > does anyone know the R equivalent of the SPlus rowVars function ? Andy Liaw <andy_liaw at merck.com> replied: > More seriously, I seem to recall David Brahms at one time had created an R > package with these dimensional summary statistics, using C code. (And I > pointed him to the `two-pass' algorithm for variance.)

Hi, Currently, the "base" has colSums, colMeans. It seems that it would be useful to extend this to also include colMin, colMax (of course, rowMin and rowMax, as well) in order to facilitate faster computations for large vectors (compared to using apply). Has this been considered before? Please forgive me if this has already been discussed before. Thanks, Ravi Ravi Varadhan, Ph.D.

Hi. I am trying to use the linear optimizer from package lpSolve in R 2.4.1 on Windows XP (Version 5.1). The problem I am trying to solve has 2843 variables (2841 integer, 2 continuous) and 8524 constraints, and I have 2 Gb of memory. After I load the input data into R, I have at most 1.5 Gb of memory available. If I start the lp with significantly less memory available (say 1 Gb), I get

Hi. I have defined a function, f1, that calls another function, f2. Inside f1 an intermediate variable called nm1 is created; it is a matrix. f2 takes a matrix argument, and I defined f2 (schematically) as follows: f2<-function(nmArg1=nm1,...){nC<-ncol(nmArg1); ... } so that it expects nm1 as the default value of its argument. f1 is defined (schematically) as:

Dear group, I have a matrix like the following: Name Sample1 sample2 sample3 sample4 ..... sample(n) nm1 10.5 13.5 30 31 nm2 8 11 34 29 nm3 9 10.3 27.8 35 nm(j) I want to be able to calculate correlation between all pairs of names. For example (nm1,nm2),

My computer is a pentium 4 running at 2.4 GHz. My R is 1.7.1 I have written a program in R that calls C++. The program spends most of the time in C++ ( > 90% ). R basically deals with output and input. How slower can this be compared with the program I would get from rewriting everything in C++? Thank you.

To whom it may concern: Given that my sample size is n, my mean is 100, and my sd is 10, I need to use a random number generator (which I believe is the function rnorm(5,100,10)), but I need to repeat it a large number of times, and then plot the sampling distributions of the sample means, sd's, and variances of those generated sets. I'm having a real hard time trying to figure out how

Hi all I have a 3d array containing missing values. > Xa , , 1 [,1] [,2] [1,] 1 3 [2,] NA 4 , , 2 [,1] [,2] [1,] 5 7 [2,] NA NA , , 3 [,1] [,2] [1,] 9 11 [2,] 10 12 I want to replace the missing values with the mean, but the mean of each 'page' in the array (wrong terminology I'm sure). So - for the array above - [2,1,2] and

Hello, everybody Say I have nm1 <- c(rep(1,10), rep(0,10)) then I can do: diff(nm1) to see where I have shift in value but what if I have nm2 <- c(rep("SPZ8", 10), rep("SPX9", 10)) how can I produce the same ouput as diff(nm1) does, that is zeros everywhere except for one place where SPZ8 changes to SPX9 (there should be 1 there)? What if I have a matrix of characters

Subject: Eliminate repeated components from a vector X-Mailer: VM 7.00 under 21.4 (patch 6) "Common Lisp" XEmacs Lucid Reply-To: fjmolina at lbl.gov FCC: /home/f/.xemacs/mail/sent Does anyone know how I can eliminate repeated elements from a vector?

Hello, colnames seems to be not optimized well for data.frame. It escapes processing for data.frame in if (is.data.frame(x) && do.NULL) return(names(x)) but only when do.NULL true. This makes huge difference when do.NULL false. Minimal edit to `colnames`: if (is.data.frame(x)) { nm <- names(x) if (do.NULL || !is.null(nm)) return(nm) else

Subject: activity of CPU used by R X-Mailer: VM 7.00 under 21.4 (patch 6) "Common Lisp" XEmacs Lucid Reply-To: fjmolina at lbl.gov FCC: ~/mail/sent My R processes never use more than 50% of my CPU activity. Is there any way I can make them use more of it? -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

Hey Everyone, I'm pretty new at R and wanted to try and make some graphs with dummy data before using it to analyze my own. I successfully made a bar graph and error bars, but I can't figure out how to align them properly (currently they are not centered on the bars and some of them aren't even close). Here's the code that I'm using: > marks <- sample(4:10, size=50,

Dear Listers: I used cronbach{psy} to evaluate the internal consistency and some set of variables gave me alpha=-1.1003, while other, alpha=-0.2; alpha=0.89; and so on. I am interested in knowing how to interpret 1. negative value 2. negative value less than -1. I also want to re-mention my previous question about how to evaluate the consistency of a set of variables and about the total

Is there any fast way to check that an object is a member of a list? -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch

Is there a function like "is.there" such that is.there ( a ) returns TRUE is object 'a' is in the current environment and FALSE otherwise? Thank you. Francisco J. Molina

Full_Name: Doug Grove Version: 1.7.0 OS: Linux Submission from: (NULL) (209.31.211.56) Hi, Minor mistake in the documentation on the colSums page. In the ARGUMENTS section it states for 'dims' that: For `col*', the sum or mean is over dimensions `dims+1, ...'; for `row*' it is over dimensions `1:dims'. These two are reversed. Thanks, Doug Grove