permutation within rows of a matrix

On Wed, 2011-11-16 at 14:29 -0500, R. Michael Weylandt wrote:

Suppose your matrix is called X.

? sample
X[sample(nrow(X)),]

That will shuffle the rows at random, not permute within the rows.

Here is an alternative, first using one of my packages (permute -
shameful promotion ;-) !:

mat<- matrix(sample(0:1, 100, replace = TRUE), ncol = 10)

require(permute)
perms<- shuffleSet(10, nset = 10)
## permute mat
t(sapply(seq_len(nrow(perms)),
          function(i, perms, mat) mat[i, perms[i,]],
          mat = mat, perms = perms))

If you don't want to use permute, then you can do this via standard R
functions

perms<- t(replicate(nrow(mat), sample(ncol(mat))))
## permute mat
t(sapply(seq_len(nrow(perms)),
          function(i, perms, mat) mat[i, perms[i,]],
          mat = mat, perms = perms))

HTH

G

Michael

On Wed, Nov 16, 2011 at 11:45 AM, Juan Antonio Balbuena<balbuena at uv.es>  wrote:

Hello
This is probably a basic question but I am quite new to R.

I need to permute elements within rows of a binary matrix, such as

     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
  [1,]    0    0    0    0    1    0    0    0    0     0
  [2,]    0    0    1    1    0    0    0    1    1     0
  [3,]    0    1    0    0    0    0    1    0    0     0
  [4,]    0    0    0    0    0    0    1    1    0     0
  [5,]    0    0    0    1    0    0    0    0    1     0
  [6,]    0    0    1    1    0    0    0    0    0     1
  [7,]    0    0    0    0    0    0    0    0    0     0
  [8,]    1    1    0    1    0    0    0    1    0     1
  [9,]    1    0    0    1    0    1    0    1    0     0
[10,]    0    0    0    0    0    0    0    1    0     1


That is, elements within each row are permuted freely and independently from
the other rows.

I see that is is workable by creating a array for each row, performing
sample and binding the arrays again, but I wonder whether there is a more
efficient way of doing the trick.

Any help will be much appreciated.



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