how to index a matrix by using a index matrix that has same size?

I have a m by n data matrix and a m by n index matrix which rearranges the order of the n elements in each row of the data matrix. How can I get an indexed data matrix without using loop? Thanks.

[Ahmed A. Selman]

But matrices already are indexed arrays or vectors... right?

Answer 1

[Walter Roberson]

Use sub2ind()

[Cedric]

typo: sub2ind

To illustrate Walter's answer, if M is the matrix of data and I the matrix of column indices, you can do

 rId = (1:size(M,1)).' * ones(1,size(M,2)) ;    % Matrix of row indices.
 M_reordered = M(sub2ind(size(M), rId, I)) ;

[Walter Roberson]

Oops, yes, I was concentrating on not putting an "s" in and so left out the "2" :(

[Cedric]

I thought that you had forgotten the "s" and I ended up realizing that it was a "2" ;-)

Answer 2

[Anand]

If A is your original matrix and idx is the matrix of indices, you can use logical indexing: A(idx).

Here's an example:

>> A = rand(3)

A =

      0.8147    0.9134    0.2785
      0.9058    0.6324    0.5469
      0.1270    0.0975    0.9575

>> idx = [9 8 7;6 5 4;3 2 1]

idx =

       9     8     7
       6     5     4
       3     2     1

>> A(idx)

ans =

      0.9575    0.5469    0.2785
      0.0975    0.6324    0.9134
      0.1270    0.9058    0.8147

[Cedric]

This is linear indexing actually; the idx matrix that you defined is essentially what we get after calling sub2ind() with a third arg that is a matrix of column indices for each row (different from linear index).

Answer 3

[Piment]

to be more specific, it's something like(it's actually 300 by 5500 in my case):

A =

    0.9649    0.4854    0.9157
    0.1576    0.8003    0.7922
    0.9706    0.1419    0.9595
    0.9572    0.4218    0.6557

idx =

     1     3     2
     3     1     2
     2     1     3
     3     2     1

[Cedric]

Walter's answer and my illustration should work; just replace M with A and I with idx.

[Piment]

thank you both, it really does work.