[R] Permuting rows of a matrix

[Petr Savicky]

On Thu, Feb 10, 2011 at 01:45:46AM -0500, Diogo Almeida wrote:
> Hi,
> 
> I need to permute the rows of a matrix, where each row is independently rearranged. A simple solution is this:
> 
> shuffled <- datamatrix <- matrix(1:24, ncol = 4)
> for (i in 1:nrow(datamatrix)) { shuffled[i, ] <- sample(datamatrix[i, ]) }
> 
> > datamatrix
>      [,1] [,2] [,3] [,4]
> [1,]    1    7   13   19
> [2,]    2    8   14   20
> [3,]    3    9   15   21
> [4,]    4   10   16   22
> [5,]    5   11   17   23
> [6,]    6   12   18   24
> > shuffled
>      [,1] [,2] [,3] [,4]
> [1,]    7   19   13    1
> [2,]    2    8   14   20
> [3,]   15    3    9   21
> [4,]   22   10   16    4
> [5,]    5   11   17   23
> [6,]   24    6   12   18
> 
> However, I need to perform quite a lot of these permutations, and I was wondering whether anyone knows of a faster way of doing this. Something that might not involve a "for" loop, for instance?

Hi.

There is also a solution using apply(), which is t(apply(datamatrix, 1, sample)),
however, this is not optimized for speed.

Let me provide some test results for the suggestion from the previous email.

  shuffle1 <- function(A) {
    for (i in 1:nrow(A)) { A[i, ] <- sample(A[i, ]) }
    A
  }

  shuffle2 <- function(A) {
    B <- t(A)
    ind <- order(c(col(B)), runif(length(B)))
    matrix(B[ind], nrow=nrow(A), ncol=ncol(A), byrow=TRUE)
  }

  m <- 6
  n <- 4
  datamatrix <- matrix(1:(m*n), ncol = n)
  system.time(replicate(10000, shuffle1(datamatrix)))

     user  system elapsed 
    0.828   0.005   0.834 

  system.time(replicate(10000, shuffle2(datamatrix)))

     user  system elapsed 
    0.528   0.001   0.530 

  m <- 50
  n <- 20
  datamatrix <- matrix(1:(m*n), ncol = n)
  system.time(replicate(10000, shuffle1(datamatrix)))

     user  system elapsed 
    7.787   0.084   7.872 

  system.time(replicate(10000, shuffle2(datamatrix)))

     user  system elapsed 
    3.554   0.054   3.608 

Petr Savicky.