R help - Jul 2008 - Computing row means for sets of 2 columns

Is there a better or more efficent approach than this without the use of t() ? 
 
> (m <- matrix(1:40, ncol=4))      [,1] [,2] [,3] [,4] [1,]    1   11   21
31 [2,]    2   12   22   32 [3,]    3   13   23   33 [4,]    4   14   24   34
[5,]    5   15   25   35 [6,]    6   16   26   36 [7,]    7   17   27   37 [8,] 
8   18   28   38 [9,]    9   19   29   39[10,]   10   20   30   40
> (groups <- rep(1:2, each=2))[1] 1 1 2 2
> (m.mean <- t(aggregate(t(m), by=list(groups), mean)))        [,1]
[,2]Group.1    1    2V1         6   26V2         7   27V3         8   28V4      
9   29V5        10   30V6        11   31V7        12   32V8        13   33V9    
14   34V10       15   35
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m <- matrix(1:40, ncol=4);
groups <- rep(1:2, each=2);
uGroups <- unique(groups);
mMeans <- matrix(NA, nrow=nrow(m), ncol=length(uGroups));
for (gg in seq(along=uGroups)) {
  mMeans[,gg] <- rowMeans(m[,groups == uGroups[gg], drop=FALSE]);
}

(Preallocation of result matrix is more memory efficient than using
cbind() or similar!)

/Henrik

On Sun, Jul 13, 2008 at 6:03 PM, Daren Tan <daren76 at hotmail.com>
wrote:
>
> Is there a better or more efficent approach than this without the use of
t() ?
>
>> (m <- matrix(1:40, ncol=4))      [,1] [,2] [,3] [,4] [1,]    1   11 
21   31 [2,]    2   12   22   32 [3,]    3   13   23   33 [4,]    4   14   24  
34 [5,]    5   15   25   35 [6,]    6   16   26   36 [7,]    7   17   27   37
[8,]    8   18   28   38 [9,]    9   19   29   39[10,]   10   20   30   40
>> (groups <- rep(1:2, each=2))[1] 1 1 2 2
>> (m.mean <- t(aggregate(t(m), by=list(groups), mean)))        [,1]
[,2]Group.1    1    2V1         6   26V2         7   27V3         8   28V4      
9   29V5        10   30V6        11   31V7        12   32V8        13   33V9    
14   34V10       15   35
> _________________________________________________________________
> Easily edit your photos like a pro with Photo Gallery.
>
>        [[alternative HTML version deleted]]
>
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