similar to: preliminary Speex support in Sweep

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

Hi, I would like to use sweep to "sweep out" proportions and confidence intervals for an array, however when I supply a function which returns a string (containing something like "9% (3-18%)") I get back a list instead of an array, here is a simplified example: # example showing that sweep does not return an array with same dimensions as STATS as advertised

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

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

Hi all, I have recently added Speex support to my app (http://sbooth.org/Max/) and am in the process of creating a GUI for the user to control the codec parameters. I am new to Speex and as such I am not really sure which parameters are more important than others, and deserve prominent placement, etc. Is there a recommended GUI for people to follow or use as an example? Stephen

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

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

On Tue, Jan 17, 2006 at 02:19:17PM -0500, me@sbooth.org wrote: > Hi all, > > I have recently added Speex support to my app (http://sbooth.org/Max/) and > am in the process of creating a GUI for the user to control the codec > parametere. I am new to Speex and as such I am not really sure which > parameters are more important than others, and deserve prominent > placement,

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)

Interrogating some (of my own) code in another package. >norm.meth <- getMethod("normalize", "MatrixLike") >message("str(norm.meth)") >str(norm.meth) >message("show(norm.meth at .Data)") >show(norm.meth at .Data) Last show() displays this: function (object, ...) { .local <- function (object, method = c("median",

I'm attempting to reflect the information for use with corresponding fields in GUI (in a different package), to provide default values, argname as key for UI label lookups, etc. So I want something much more like the formals of the implementation: { "object", "method": c("median", "vs", "tukey"),

Dear List: I am wondering if there is a more efficient way to compute the following. For the example I am using the star data frame in the mlmRev package. This has 80 schools and includes grades K, 1, 2, and 3. First I compute the grade level mean in each school using tapply as: tapply(star$math, list(star$sch,star$gr), mean, na.rm=T) This results in a table of means by school for each grade.

Submitted for perusal, comment, improvements, and/or critique. The presentation is in 3 sections: motivation, code, and comment. Motivation: As a new-comer to R from matrix oriented Gauss and Mata, I miss the tools for using a vector (and operator) to ‘sweep’ across a matrix. Here is how these work. If M is I rows by J columns, then one entry corresponding to

Hi, Anyone got some slick ideas on how to sweep an AR session store? I have a few ideas how I can do it, just want to see if there is an elegant solution that someone is already using. Bob Silva http://www.railtie.net/ -------------- next part -------------- An HTML attachment was scrubbed... URL:

Hi! Excuse the delay, but I had to deal with a corrupted NTFS file system that ate many important files on an USB stick... The FLAC version of the original is almost 6MB and it can be downloaded slowly from this time-limited link: https://sbr5vjid0jgmce4q.myfritz.net:40262/nas/filelink.lua?id=0ba5a10529a6fe7b On the meaning of a logarithmic sweep: If you use foobar2000 and the