Or even: abind(foo, along = 3)
On Tue, Dec 30, 2008 at 1:24 PM, Gabor Grothendieck <ggrothendi...@gmail.com> wrote: > Try: > > do.call(abind, c(foo, along = 3)) > > > On Tue, Dec 30, 2008 at 1:15 PM, baptiste auguie <ba...@exeter.ac.uk> wrote: >> In fact, when writing my post I tried to do exactly what you did in creating >> a 3d array from the list, and I failed miserably! This is (imho) partly >> because the syntax is not very clean or straightforward as compared to the >> apply and *ply family. A list of matrices with equal dimensions is easily >> produced by mapply(... , simplify=F), or lapply, while an array needs to be >> created in a more verbose manner (as far as i know, some version of a loop). >> >> I just remembered the abind package which makes this a bit easier, although >> the default is not quite as convenient for this purpose as I'd initially >> hoped: >> >> foo <- list(rbind(c(1,2,3),c(4,5,6)),rbind(c(7,8,9),c(10,11,12))) >> foo2 <- unlist(foo) >> dim(foo2) <- c(dim(foo[[1]]), length(foo)) >> >> library(abind) >> foo3 <- do.call(function(...) abind(..., along=3), foo) >> foo2==foo3 >> >> Best wishes, >> >> baptiste >> >> >> On 30 Dec 2008, at 18:53, hadley wickham wrote: >> >>> On Tue, Dec 30, 2008 at 10:21 AM, baptiste auguie <ba...@exeter.ac.uk> >>> wrote: >>>> >>>> I thought this was a good candidate for the plyr package, but it seems >>>> that >>>> l*ply functions are meant to operate only on separate list elements: >>>> >>>> Lists are the simplest type of input to deal with because they are >>>> already >>>> naturally >>>> divided into pieces: the elements of the list. For this reason, the l*ply >>>> functions don't >>>> need an argument that describes how to break up the data structure. >>>> >>>> (from: plyr: divide and conquer, Hadley Wickham 2008) >>>> >>>> Perhaps a new case to consider? >>> >>> Possibly, but here I would argue that the choice of data structure >>> isn't quite right - if the matrices all have the same dimension, then >>> they should be stored in an array, not a list: >>> >>> foo <- list(rbind(c(1,2,3),c(4,5,6)),rbind(c(7,8,9),c(10,11,12))) >>> foo2 <- unlist(foo) >>> dim(foo2) <- c(dim(foo[[1]]), length(foo)) >>> >>> Then you can use apply (or aaply) directly on that matrix: >>> >>> apply(foo2, c(1,2), mean) >>> apply(foo2, c(1,2), mean, trim = 0.1) >>> >>> etc. >>> >>> Hadley >>> >>> -- >>> http://had.co.nz/ >> >> _____________________________ >> >> Baptiste Auguié >> >> School of Physics >> University of Exeter >> Stocker Road, >> Exeter, Devon, >> EX4 4QL, UK >> >> Phone: +44 1392 264187 >> >> http://newton.ex.ac.uk/research/emag >> >> ______________________________________________ >> R-help@r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
