[R] get means of elements of 5 matrices in a list

Marc Schwartz marc_schwartz at me.com
Tue Apr 27 17:19:56 CEST 2010


On Apr 27, 2010, at 10:05 AM, David Freedman wrote:

> 
> I've got a list of 5 matrices that are each 5 x 6.  I'd like to end up with a
> 5 x 6 matrix that contains the mean value of the 5 original matrices.  I can
> do this by brute force, but there must be a better way than making each
> matrix into a vector and then remaking a matrix
> 
> thanks very much for any help
> david freedman
> 
> ll=list(structure(c(9.7, 17.6, 20.8, 24.1, 33.8, 14.5, 25.7, 29.8, 
> 33.6, 44.8, 21.8, 32.6, 37.5, 40.9, 53.3, 16.7, 26.1, 29.5, 32.7, 
> 42.6, 26.2, 34.3, 37, 39.8, 47.1, 31.9, 40.3, 43.3, 46.2, 54.1
> ), .Dim = 5:6), structure(c(9.4, 17.7, 20.7, 24.1, 33.7, 14.5, 
> 25.7, 29.8, 33.6, 44.8, 21.8, 32.7, 37.5, 40.9, 53.1, 16, 26, 
> 29.5, 32.7, 42.7, 25.6, 34.1, 37, 39.8, 47.1, 31.9, 40.3, 43.3, 
> 46.1, 54.1), .Dim = 5:6), structure(c(9.6, 17.7, 20.8, 24.4, 
> 34.3, 14.5, 25.7, 29.8, 33.6, 44.8, 21.8, 32.6, 37.5, 40.9, 53.2, 
> 16.7, 26.1, 29.5, 32.8, 42.8, 26.2, 34.2, 36.8, 39.9, 47.1, 31.9, 
> 40.3, 43.3, 46.1, 54.8), .Dim = 5:6), structure(c(9.7, 17.6, 
> 20.7, 24.1, 33.8, 14.5, 25.7, 29.8, 33.6, 44.8, 21.8, 32.6, 37.5, 
> 41.1, 52.6, 16.7, 26.1, 29.5, 32.8, 42.8, 26.1, 34.3, 37, 40, 
> 47.1, 31.9, 40.3, 43.3, 46.2, 54.9), .Dim = 5:6), structure(c(8.6, 
> 17.6, 20.7, 24.1, 33.8, 14.5, 25.6, 29.8, 33.6, 44.8, 21.8, 32.6, 
> 37.5, 40.9, 52.5, 16, 26, 29.4, 32.8, 42.8, 25.6, 34.2, 37, 40.1, 
> 47.1, 31.9, 40.3, 43.1, 46.1, 54.1), .Dim = 5:6))
> 
> ll
> x=rbind(as.vector(ll[[1]]),as.vector(ll[[2]]),as.vector(ll[[3]]),as.vector(ll[[4]]),as.vector(ll[[5]]));
> x
> x2=apply(x,2,mean); matrix(x2,byrow=F,nrow=5); 


How about:

> matrix(rowMeans(sapply(ll, unlist)), nrow = length(ll))
      [,1]  [,2]  [,3]  [,4]  [,5]  [,6]
[1,]  9.40 14.50 21.80 16.42 25.94 31.90
[2,] 17.64 25.68 32.62 26.06 34.22 40.30
[3,] 20.74 29.80 37.50 29.48 36.96 43.26
[4,] 24.16 33.60 40.94 32.76 39.92 46.14
[5,] 33.88 44.80 52.94 42.74 47.10 54.40


HTH,

Marc Schwartz



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