Sort 3d matrix according to another 3d matrix

Hi,

I have two 3d matrices (A and B) with the same size (m,n,o).

I want to sort matrix B along the third dimension based on the sort index of matrix A:

[~, idx] = sort(A,3,'descend')

If I use the following syntax it doesn't really work:

B = B(:,:,idx)

I get a (m,n,m*n*o) matrix instead of a sorted (m,n,o) matrix.

Any ideas?

Answer 1

[Massimo Zanetti]

According to Walter's comment, the previous solution was wrong. Here is a way to sort B given the sorted A along the 3-rd dimension:

m=3; n=4; o=2;
A=randi(9,m,n,o)
B(:,:,1)=[1,11,111,1111;2,22,222,2222;3,33,333,3333];
B(:,:,2)=[4,44,444,4444;5,55,555,5555;6,66,666,6666];

[~,ix] = sort(A,3,'descend');

[C,R,~]=meshgrid(1:n,1:m,1:o);
lix = sub2ind([m,n,o],R,C,ix);
B(lix)

Looking for a more general solution...

[Walter Roberson]

No, the indices are relative to the layer, not absolute.

[Christian Maschner]

That's it! Thank you

[Massimo Zanetti]

Walter, you are right. Checking for solution.

[Guillaume]

A completely generic solution (ignoring potential edge cases such as A, B being vectors, etc):

function [sA, sB] = dualsort(A, B, dim, direction)
  %dualsort, sort A according to dim and direction and sort B according to the order of A:
  [sA, idx] = sort(A, dim, direction);
  gridargs = arrayfun(@(s) 1:s, size(A), 'UniformOutput', false);
  [gridargs{:}] = ndgrid(gridargs{:});
  gridargs{dim} = idx;
  sB = B(sub2ind(size(B), gridargs{:}));
end

[Massimo Zanetti]

That is exactly the solution I have finally run into to generalize the one above. However, it seems quite surprising that apparently there is no something like "built-in" solution...

[Andrei Bobrov]

[~, ii] = sort(A,3,'descend');
[m,n,o] = size(A);
B = B((ii-1)*m*n + reshape(1:m*n,m,[]));