The result is linear in A so its a matter of finding the matrix to multiply it
by:
matrix(c(rep(1,3), rep(0,7)), 3, 9, byrow = TRUE) %*% A
On 1/30/06, Camarda, Carlo Giovanni <[EMAIL PROTECTED]> wrote:
> Dear R-users,
> I'm struggling in R in order to "squeeze" a matrix without using a
> for-loop.
> Although my case is a bit more complex, the following example should
> help you to understand what I would like to do, but without the slow
> for-loop.
> Thanks in advance,
> Carlo Giovanni Camarda
>
>
> A <- matrix(1:54, ncol=6) # my original matrix
> A.new <- matrix(nrow=3, ncol=6) # a new matrix which I'll fill
> # for-loop
> for(i in 1:nrow(A.new)){
> B <- A[i:(i+2), ] # selecting the rows
> C <- apply(B,2,sum) # summing by columns
> A.new[i,] <- C # inserting in the new matrix
> }
>
>
> +++++
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