[R] Componentwise means of a list of matrices?

[Gabor Grothendieck]

Try:

do.call(abind, c(foo, along = 3))


On Tue, Dec 30, 2008 at 1:15 PM, baptiste auguie <ba208 at 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 <ba208 at 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
>
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