R help - Mar 2011 - Replacing values in a data.frame/matrix

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

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
Tel: +31/(0)10/7043478
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