# [R] Componentwise means of a list of matrices?

Stephan Kolassa Stephan.Kolassa at gmx.de
Tue Dec 30 16:51:34 CET 2008

```Hi Marc,

thanks, that works great! With yet another apply(), I can also make it
accept psych::winsor() and trimming:

some.sort.of.apply <- function ( foo, FUN, ... ) {
matrix(apply(sapply(foo, c),1,FUN,...), dim(foo[[1]]))
}
some.sort.of.apply ( foo, FUN=winsor,trim=0.3 )

Thank you all for your help!
Stephan

Marc Schwartz schrieb:
> on 12/30/2008 08:33 AM Stephan Kolassa wrote:
>> Dear useRs,
>>
>> I have a list, each entry of which is a matrix of constant dimensions.
>> Is there a good way (i.e., not using a for loop) to apply a mean to each
>> matrix entry *across list entries*?
>>
>> Example:
>>
>> foo <- list(rbind(c(1,2,3),c(4,5,6)),rbind(c(7,8,9),c(10,11,12)))
>> some.sort.of.apply(foo,FUN=mean)
>>
>> I'm looking for a componentwise mean across the two entries of foo,
>> i.e., the following output:
>>
>>      [,1] [,2] [,3]
>> [1,]    4    5    6
>> [2,]    7    8    9
>>
>> [NB. My "real" application involves trimming and psych::winsor(), so
>> anything that generalizes to this would be extra good.]
>>
>> I've been looking at apply and {s,l,m,t}apply, by, with and aggregate
>> and searched the list archives... any ideas?
>>
>> Thanks a lot,
>> Stephan
>
>
> How about something like this:
>
>> matrix(rowMeans(sapply(foo, c)), dim(foo[[1]]))
>      [,1] [,2] [,3]
> [1,]    4    5    6
> [2,]    7    8    9
>
>
> Essentially, first create matching elementwise rows:
>
>> sapply(foo, c)
>      [,1] [,2]
> [1,]    1    7
> [2,]    4   10
> [3,]    2    8
> [4,]    5   11
> [5,]    3    9
> [6,]    6   12
>
>
> Then, get the row means:
>
>> rowMeans(sapply(foo, c))
> [1] 4 7 5 8 6 9
>
>
> Then finally restructure the result to a matrix, using the dimensions of
> foo[[1]].
>
> This of course makes the assumption that each matrix is of the same size
> and does not otherwise check for that.
>
> HTH,
>
> Marc Schwartz
>

```