how about:
m <- matrix(c(1,2,3,2,1,3,3,1,2), ncol = 3, byrow = TRUE)
perm <- c(1, 3, 2)
out <- perm[m]
dim(out) <- dim(m)
out
I hope it helps.
Best,
Dimitris
On 3/8/2011 4:05 PM, Thaler, Thorn, LAUSANNE, Applied Mathematics wrote:
Hi all,
Suppose we have the following matrix
m<- matrix(c(1,2,3,2,1,3,3,1,2), ncol = 3, byrow=T)
where in each row each number occurs only once.
I'd like to define a permutation, e.g. 1 -> 2, 2 -> 1, 3 -> 3 and apply
it to the matrix. Thus, the following matrix should result:
m.perm<- matrix(c(2,1,3,1,2,3,3,2,1), ncol = 3, byrow=T)
i.e. each 1 should map to 2 and vice verse while 3 maps to itself. What
I've done so far is:
permutateMatrix<- function(mat, perm=NULL) {
values<- 1:NCOL(mat)
if (is.null(perm)) {
perm<- sample(values)
}
newmat<- replace(mat, sapply(values, function (val) which(mat==val)),
rep(perm, each=NROW(mat)))
return(list(mat.perm=newmat, perm=perm))
}
"perm" is the permutation vector: 1 maps to the first element of perm, 2
to the second and so on. Thus, for the example we would use
perm<- c(2,1,3)
all.equal(m.perm, permutateMatrix(m, perm)$mat.perm) # TRUE
What do you think of this solution? Are there more elegant ways of doing
that? Any comments appreciated.
Thanks + BR,
Thorn
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--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
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