similar to: rowVars

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

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

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.)

Is it possible to compute the variance of every column in a matrix without a loop? -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- 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

I've uploaded a package colSums_1.0.tar.gz to CRAN /src/contrib/Devel. It contains functions colSums, colMeans, colVars, colStdevs, rowSums, rowMeans, rowVars, and rowStdevs. These do simple, fast arithmetic on columns/rows of a matrix, or more generally across dimensions of an array, e.g. colSums(m) = apply(m, 2, sum) but faster. They should be compatible with the corresponding S-Plus

Hi together, I have some data in a matrix structure - say 1000 rows with 10 columns. And I like to do some calculations (like max, avg or min) on row level. The only solution I found so fare was using a loop like for (i in 1:1000) { max[i] <- max(matrix[I,]) } It looks like that this is not very fast. Does an other way exists? Kind regards Ulrich

Hi, I have what I think is an easy question. I have a data frame, called stockdata, of stock prices that looks like this: date ticker close 1 01/02/1998 GE 24.667 2 01/05/1998 GE 25.104 3 01/06/1998 GE 24.771 4 01/07/1998 GE 24.979 5 01/08/1998 GE 24.750 6 01/02/1998 HIT 71.125 7 01/05/1998 HIT 72.313

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),

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

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

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

Hello all, I am trying to run a two sample t-test on a matrix which is a 196002*22 matrix. I want to run the t-test, row-wise, with the first 11 columns being a part of the first group and columns 12-22 being a part of the second group. I tried running something like (temp.matrix being my 196002*22 matrix) t.test(temp.matrix[,1:11],temp.matrix[,12:22],paired=TRUE) or somthing like

After upgrading to R-2.3.1 on Linux Redhat, I was suprised by this: R> x <- c(721.077, 592.291, 372.208, 381.182) R> sum(x) - 2066.758 [1] 4.547474e-13 Now I understand that floating point arithmetic is not precise, but 1) the result is exactly 0 in R-2.2.1 (patched) on the same machine, 2) .Machine$double.eps = 2.2e-16, so the error seems quite large. Also note I get the same

Hi, all, I've been converting code from S-Plus ("S" for short) to R for a few weeks. Here are some differences I've found, aside from the big well-known ones (scoping, models, data storage) and the contents of Kurt Hornik's FAQ section 3.3.3. Let me start with the ones that seem like serious bugs or deficiencies: 1) LETTERS[c(NA,2)] in S is

Hi, all, I've been converting code from S-Plus ("S" for short) to R for a few weeks. Here are some differences I've found, aside from the big well-known ones (scoping, models, data storage) and the contents of Kurt Hornik's FAQ section 3.3.3. Let me start with the ones that seem like serious bugs or deficiencies: 1) LETTERS[c(NA,2)] in S is

In general this is what "apply" does. In your example you could use rowMeans(ar) as well, I think. Reid Huntsinger -----Original Message----- From: Olivier Martin [mailto:olivier.martin at inrialpes.fr] Sent: Tuesday, July 02, 2002 10:55 AM To: r-help Subject: [R] mean and array Hi all, I have an array, one say ar, with 3 dimensions. dim(ar) >200 3 4 I would like to

Hi, Can anyone help me with the following. I have been using R for Monte Carlo simulations and got some results I couldn't explain. Therefor I performed following short test: -------------- mean.sds <- NULL sample.sizes <- 3:30 for(N in sample.sizes){ dum <- NULL for(I in 1:5000){ x <- rnorm(N,0,1) dum <- c(dum,sd(x)) } mean.sds<- c(mean.sds,mean(dum)) }

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