similar to: how to coerce the result of sweep to be an array if result of FUN is a string?

Dear R-help, I've encounter what seems to me a strange problem with "names<-". Suppose I define the function: fun <- function(x, f) { m <- tapply(x, f, mean) ans <- x - m[match(f, unique(f))] names(ans) <- names(x) ans } which subtract out the means of `x' grouped by `f' (which is the same as, e.g., resid(lm(x~f)) if `f' is a factor).

Hi, I need to extract the second largest element from each row of a matrix. Below is my solution, but I think there should be a more efficient way to accomplish the same, or not? set.seed(1) a <- matrix(rnorm(9), 3 ,3) sec.large <- as.vector(apply(a, 1, order, decreasing=T)[2,]) ans <- sapply(1:length(sec.large), function(i) a[i, sec.large[i]]) ans Thanks in advance for your

Hi Jean-Marc and others, I put Speex support in Sweep (a sound editor) yesterday, and had some fun with it :) I'd like to get some feedback before releasing it publically. I'm not really used to speech codecs (I work more with music) but I've been quite impressed at the file size and quality (especially when you hear what I did with some of the sample files ;) I didn't have any

A friend of mine just got bitten by the fact that sweep() will happily sweep out a STATS vector of an arbitrary length -- even one whose length is not a divisor of any of the margins -- without complaining. I know the answer to this could be "well just don't do that", but it's easy to make a mistake in which margin you're sweeping ... What would R-core think of the

I am having a hard time understanding just what 'sweep' does. The documentation states: Return an array obtained from an input array by sweeping out a summary statistic. So what does it mean "weeping out a summary statistic"? Thank you. Kevin

The call to sweep in this function which was working in 0.62.2 is giving me trouble in 62.3: prcomponents <- function(x, center=TRUE, scale=TRUE, N=nrow(x)-1) {if (center) center <- apply(x,2,mean) else center <- rep(0, ncol(x)) if (scale) scale <- sqrt(apply(x,2,var)) else scale <- rep(1, ncol(x)) s <- svd(sweep(sweep(x,2, center),2, scale,

I'm looking for an R equivalent of Beaton's (1964) Sweep algorithim for partial inversion of a matrix by pivoting. It implemented in SAS/IML as sweep(matrix, indices), described here http://support.sas.com/documentation/cdl/en/imlug/59656/HTML/default/langref_sect266.htm and here for python http://adorio-research.org/wordpress/?p=262 -- Michael Friendly Email: friendly AT yorku

Hi I had a hard-to-find bug in some of my code the other day, which I eventually traced to my misusing of sweep(). I would expect sweep() to give me a warning if the elements don't recycle nicely, but X <- matrix(1:36,6,6) sweep(X,1,1:5,"+") [,1] [,2] [,3] [,4] [,5] [,6] [1,] 2 9 16 23 30 32 [2,] 4 11 18 25 27 34 [3,] 6 13 20 22

For example, subtracting 1:2 from the rows of a two-column matrix: > t(apply(matrix(1:6,ncol=2),MARGIN=1,function(y) y - 1:2)) [,1] [,2] [1,] 0 2 [2,] 1 3 [3,] 2 4 > sweep(matrix(1:6,ncol=2),MARGIN=2,1:2,FUN="-") [,1] [,2] [1,] 0 2 [2,] 1 3 [3,] 2 4 Is there a logic to this difference, or is it just a quirk of the history of these

Hi, I'd appreciate help with this. I have a data matrix with one column, called f in the example below, a factor. I'd like to subtract the means from each of other columns for each level of the factor. That is, in the example, to go from the first matrix below to the second. I know SWEEP will take out means, but I want to do this for each level of the factor. f x 1 2 1

Dear R users, I?m student of Master in Statistic and Data analysis, in New University of Lisbon. And now i?m writting my dissertation in variance estimation.So i?m using Survey Package to compute the principal estimators and theirs variances. My data is from Incoming and Expendire Survey. This is stratified Multi-stage Survey care out by National Statistic Institute of Mozambique. My domain of

Hey fellas: I would like to integrate the following function: integrand <- function (x,t) { exp(-2*t)*(2*t)^x/(10*factorial(x)) } with respect to the t variable, from 0 to 10. The variable x here works as a parameter: I would like to integrate the said function for each value of x in 0,1,..,44. I have tried Vectorize to no avail. Thanks in advance, jose romero

I'm trying to convert an S-Plus program to R.  Since I'm a SAS programmer I'm not facile is either S-Plus or R, so I need some help.  All I did was convert the underscores in S-Plus to the assignment operator <-.  Here are the first few lines of the S-Plus file:   sshc _ function(rc, nc, d, method, alpha=0.05, power=0.8,              tol=0.01, tol1=.0001, tol2=.005, cc=c(.1,2),

Given a long zoo matrix, the goal is to "sweep" out a statistic from the entire length of the sequences. longzoomatrix<-zoo(matrix(rnorm(720),ncol=6),as.yearmon(outer(1900,seq(0,length=120)/12,"+"))) cnames<-c(12345,23456,34567,45678,56789,67890) colnames(longzoomatrix)<-cnames longzoomatrix[1:24,] 12345 23456 34567 45678

Hi All, I am currently doing the following to compute summary statistics of aggregated data: a = aggregate(warpbreaks$breaks, warpbreaks[,-1], mean) b = aggregate(warpbreaks$breaks, warpbreaks[,-1], sum) c = aggregate(warpbreaks$breaks, warpbreaks[,-1], length) ans = cbind(a, b[,3], c[,3]) This seems unnecessarily complex to me so I tried > aggregate(warpbreaks$breaks, warpbreaks[,-1],

rc<-list(c( 123,321,234,543,654,768,986,987,246,284),c("Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec")) # the matrix has rownames that are used as identifiers and columns # of time. 1 years worth of data. Thats the native format

Hi, I'm updating speex support in sweep to RC2 and I'd like some advice on encoder options, and would appreciate some testing of the user interface. a screenshot of the updated speex encoding dialog: http://www.vergenet.net/~conrad/tmp/sw_speexRC2_1.png you can grab a tarball of sweep for testing here: http://www.vergenet.net/~conrad/tmp/sweep-0.8.0-speexRC2.tar.gz general

I''ve got a Rails project that has been on the shelf for a few months, and I''m picking it back up. I''ve frozen Rails at 1.1.4 in vendor/, and did the "rails ." and "rake rails:update" to get it up to a current Rails level. The code initially seems to be working fine, but when I go to my login method, I get a 500 error thrown. Looking in the

Dear list, I have a matrix, spc, that should row-wise be interpolated: E.g. spc <- matrix (1:1e6, ncol = 250, nrow = 4000, byrow = TRUE) spc [1:10, 1:10] shifts <- seq_len (nrow (spc)) wl <- seq_len (ncol (spc)) interpolate <- function (spc.row, shift, wl) spline (wl + shift, spc.row, xout = wl, method = "natural")$y interpolate (spc [1,], shift = shifts [1], wl = wl)

Hello, This is my first post to the help request list, so I'm going to err on the side of giving too much information. I'm working on writing a simulation in which agents will make repeated production and exchange decisions with randomly chosen partners. The idea is, all agents can produce two goods which they want to consume, they choose a value t in [0,10] which sets their production