similar to: time demean model matrix

List: Thank you for the replies to my post yesterday. Gabor and Phil also gave useful replies on how to improve the function by relying on mapply rather than the explicit for loop. In general, I try and use the family of apply functions rather than the looping constructs such as for, while etc as a matter of practice. However, it seems the mapply function in this case is slower (in terms of CPU

--KsGdsel6WgEHnImy Content-Type: text/plain; charset=iso-8859-1; format=flowed Content-Disposition: inline Content-Transfer-Encoding: 8bit Hi, people. Wandering in R archives, and seeing the message attached below, I noticed that: mapply(rep,times=1:4, MoreArgs=42) still segfaults on R 2.2.0, and thought I should be a good citizen and report it, even if I do not have an actual problem

Dear list and Hadley, The new plyr package seems to provide a clean and consistent way to apply a function on several arguments. However, I don't understand why the following example does not work like the standard mapply, library(plyr) df <- data.frame(a=1:10 , b=1:10) foo1 <- function(a, b, cc=0, d=0){ a + b + cc + d } mdply(df, foo1, cc=1) # fine mdply(df, foo1, d=1) #

Hello everybody. I've been experimenting with mapply(). Does anyone else have problems with: R> mapply(rep,times=1:4, MoreArgs=42) (I get a seg fault). robin R> R.version _ platform powerpc-apple-darwin6.6 arch powerpc os darwin6.6 system powerpc, darwin6.6 status beta major 1 minor 8.0 year 2003 month 10 day 02 language R >

I'm trying to use Rprof() to identify bottlenecks and speed up a particullary slow section of code which reads in a portion of a tif file and compares each of the values to values of predictors used for model fitting. I've written up an example that anyone can run. Generally temp would be a section of a tif read into a data.frame and used later for other processing. The first portion

Hi, Here are the issues I found: Typos ----- (a) Found: It a parallel version of ?evalq?, "is" missing. (b) Found: 'parLapplyLB', 'parSapplyLB' are load-balancing versions, intended for use when applying ?FUN? to 'parLapplyLB' has no 'FUN' arg (more on this below). (c) Found: 'clusterApply' calls 'fun' on the first

Hello! I have noticed that mapply() drops names in R 2.3.1 as well as in r-devel. Here is a simple example: l <- list(a=1, b=2) k <- list(1) mapply(FUN="+", l, k) [1] 2 3 mapply(FUN="+", l, k, SIMPLIFY=FALSE) [[1]] [1] 2 [[2]] [1] 3 Help page does not indicate that this should happen. Argument USE.NAMES does not have any effect here as it used only in a bit special

This script below pulls yahoo data via a function in quantmod, then massages the data around to forumalate a 3D graph with RGL library, attached is a ggplot to show the data i'm trying to create a surface with in separate line geoms . the issue is that the 3D graph looks very ugly and cut up because of the limited quantities of points on the front month expirations.. can anyone tell me whats

Dear expeRts, I am trying to wrap up a package "utilities" (for my internal use). After adding a function datNAtreat that uses mapply, R CMD check gives WARNINGs for "S3 generic/method consistency", "checking replacement functions" and?"checking foreign function calls", all of which are accompanied by the following error message: Error in .try_quietly

Dear all I know that the R way of documenting things is to work on your project in package development mode, and document each object (such as data frames) in a *.Rd files. This should work for gurus. What about a simpler way to document things, geared for mere mortals? I was thinking of a label() or tag() function that could store and retrieve an alphanumeric comment for a given object (for

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)

Dear all, when I look at the internal of mapply() function, I see a line of code: answer <- .Call("do_mapply", FUN, dots, MoreArgs, environment(), PACKAGE = "base") Can somebody please tell me how to find the source code of 'do_mapply' Thanks, [[alternative HTML version deleted]]

I would use replicate() to do an operation with random numbers repeatedly: ``` mysim <- replicate(10, { two.mat <- matrix(rnorm(4), 2, 2) four.mat <- matrix(rnorm(16), 4, 4) list(two.mat = two.mat, four.mat = four.mat) }) ``` which should give you a matrix-list. You can slice this matrix-list just like normal, then cbind it in one step: ``` two.mat <-

Despite the spirited arguments of various R-core folks who feel that mle() doesn't need a "data" argument, and that users would be better off learning to deal with function closures, I am *still* trying to make such things work in a reasonably smooth fashion ... Is there a standard idiom for "merging" environments? i.e., suppose a function has an environment that I want

> > I am looking for a more elegant way to write below code. > > #Simulation results have different dimensions > mysim <- lapply(1:10, function(y) { > two.mat <- matrix(rnorm(4), nrow = 2) > four.mat <- matrix(rnorm(16), nrow = 4) > list(two.mat = two.mat, four.mat = four.mat) #results with different dimensions > }) > > #Collect different

Dear all with image I can plot only one set of values in one plot. Do somebody have any insight how to put those 2 matrices into one picture so that in one cell in image picture are both values from mat[1,1] and mat2[1,1]. mat<-matrix(1:4, 2,2) mat2<-matrix(4:1,2,2) x <-1:2 y <-1:2 image(x, y, mat) image(x, y, mat2) The only way I found is to mix x or y for both matrices let

[redirected from R-devel: this really belongs on R-help] I have two sets of time-series that I imported from Excel using RODBC and placed in "securities" and "factors". What I need to do is generate t-scores for each security-factor pair. I tried the following: t1 <- t.test(securities[,3:42], factors[,2:41], var.equal=TRUE) ANSWER: If securities and factors

Dear all, I have a problem with the "subset()" function. I spent all day yesterday with a collegue to solve it and we did not find a satisfying solution (even in the archived mails), so I ask for your help. Let's say (for a simple example) a matrix mat: R> mat cola colb colc [1,] 1 4 7 [2,] 2 5 8 [3,] 3 6 9 My goal is to select the lines of the matrix on the basis of the

I have a data frame whose first colum contains the names of the variables and whose second colum contains the values to assign to them: : kkk <- data.frame(vars=c("var1", "var2", "var3"), vals=c(10, 20, 30), stringsAsFactors=F) If I do : assign(kkk$vars[1], kkk$vals[1]) it works : var1 [1] 10 However, if I try with mapply

Hello, I'm attempting to create a data frame with correlations between every pair of variables in a data frame, so that I can then sort by the value of the correlation coefficient and see which pairs of variables are most strongly correlated. The sm2vec function in the corpcor library works very nicely as shown here: library(Hmisc) library(corpcor) # Create example data x1 = runif(50) x2 =