using a matrix as an index to another matrix

Simple case:

>> x = [ 10 8 ; 4 3 ]

x =

      10     8
       4     3

>> [y,i] = sort(x,2 )

y =

       8    10
       3     4

i =

       2     1
       2     1

>> x(i)

ans =

       4    10
       4    10

We see x(i) is not equiv. to y. Can I use x & i to derive y ... ?

Answer 1

[Jan]

3 versions with a speed comparison:

function speedtest
x = rand(2000, 1000);

tic;
for k = 1:5
  % Method 1: SUB2IND:
  [y, idx2] = sort(x, 2);
  sx        = size(x);
  index     = sub2ind(sx, repmat((1:sx(1)).', 1, sx(2)), idx2);
  y2        = x(index);
end
toc

tic;
for k = 1:5
  % Method 2: Simplified loop, row-wise
  [y, idx2] = sort(x, 2);
  y3 = zeros(size(x));
  for r = 1:size(x,1) 
    y3(r, :) = x(r, idx2(r, :));
  end
end
toc

tic;
for k = 1:5
  % Method 3: Simplified loop, column-wise
  xt         = x.';
  [yt, idx1] = sort(xt, 1);
  y4         = zeros(size(xt));
  for r = 1:size(x,1)
     y4(:, r) = xt(idx1(:, r), r);
  end
  y4 = y4.';
end
toc

isequal(y, y2, y3, y4)

Matlab 2009a/64, Win7, 2 cores of an i5 in a VM:

 Elapsed time is 2.174286 seconds.  % SUB2IND
 Elapsed time is 2.512037 seconds.  % Loop, rowwise
 Elapsed time is 0.706579 seconds.  % Loop, columnwise

The cloumn-wise loop method is faster for [200, 10000] and [10000, 200] inputs also.

[Andy]

thanks, and if I want to return a matrix similar to y, but instead of having all (ordered) values present, just to keep highest N on each row, and fill with zero / nan all the other values. The highest N values on each row, should also keep their original column position (they had in x). Is this something easy to be built on the above code?

[Jan]

@Andrei: This is a different question. Prefer to open a new thread for a new problem in the future.

N = 4;   % Keep the 4 largest values at their positions:
xt         = x.';
[ys, idx1] = sort(xt, 1, 'descend');
y          = NaN(size(xt));
for r = 1:size(xt, 2)
   y(idx1(1:4, r), r) = ys(1:4, r);
end
y = y.';

Answer 2

[Andy]

I could do this:

    clc; 
    a = [1 4 3 2; 6 8 7 9]

      [b,i] = sort(a,2,'descend')

      b = zeros(size(a,1),size(a,2));

      for r = 1:size(a,1) 
          for c = 1:size(a,2)
              b(r,c) = a(r, i(r,c));
          end
      end
      b

a =

       1     4     3     2
       6     8     7     9

b =

       4     3     2     1
       9     8     7     6

i =

       2     3     4     1
       4     2     3     1

b =

       4     3     2     1
       9     8     7     6

nonetheless, it's so not elegant ... and not too high performance for large sizes.