If you don't have any NAs, then one way is:
n <- 3
r <- 5
s <- 6
raw <- lapply(seq_len(n), function(i){
matrix(rnorm(r * s), ncol = r)
})
Reduce("+", raw) / length(raw)
I hope it helps.
Best,
Dimitris
On 11/5/2012 11:32 AM, ONKELINX, Thierry wrote:
> Dear all,
>
> I have a list of n matrices which all have the same dimension (r x s). What
> would be a fast/elegant way to calculate the element wise average? So
> result[1, 1] <- mean(c(raw[[1]][1, 1] , raw[[2]][1, 1], raw[[...]][1, 1],
> raw[[n]][1, 1]))
>
> Here is my attempt.
>
> #create a dummy dataset
> n <- 3
> r <- 5
> s <- 6
> raw <- lapply(seq_len(n), function(i){
> matrix(rnorm(r * s), ncol = r)
> })
>
> #do the calculation
> result <- array(dim = c(dim(raw[[1]]), length(raw)))
> for(i in seq_along(raw)){
> result[,,i] <- raw[[i]]
> }
> result <- apply(result, 1:2, mean)
>
>
> Best regards,
>
> Thierry
>
> ir. Thierry Onkelinx
> Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
> Forest
> team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
> Kliniekstraat 25
> 1070 Anderlecht
> Belgium
> + 32 2 525 02 51
> + 32 54 43 61 85
> thierry.onkel...@inbo.be
> www.inbo.be
>
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> asking him to perform a post-mortem examination: he may be able to say what
> the experiment died of.
> ~ Sir Ronald Aylmer Fisher
>
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> ~ Roger Brinner
>
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> ensure that a reasonable answer can be extracted from a given body of data.
> ~ John Tukey
>
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--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
Web: http://www.erasmusmc.nl/biostatistiek/
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