How to sort rows of a 2D array based on another array

Hi,

I would like to ask how can I sort rows of one 2D array based on values in another 2D array with same dimensions without using loop. For example: Sort rows of B based on rows of A

A = [1 2 3;2 1 3;3 2 1]

B = [0 1 2;0 1 2;0 1 2]

the result is newB = [0 1 2;1 0 2;2 1 0]

My gimmicky solution using loop is:

[~,idx] = sort(A,2)

for i=1:height(B)

B(i,:) = B(i,idx(i,:))

end

Is there cleaner slution?

Thanks.

Vojta

Answer 1

[DGM]

Here's one example. I'm sure there are others.

A = [1 2 3;2 1 3;3 2 1];
B = [0 1 2;0 1 2;0 1 2];

% get sorting and size info
sz = size(B);
[~,cidx] = sort(A,2);

% sort in a loop
C = zeros(sz);
for r = 1:sz(1)
    C(r,:) = B(r,cidx(r,:));
end
C

C = 3×3

0 1 2 1 0 2 2 1 0

% use sub2ind()
ridx = repmat((1:sz(1)).',[1 sz(2)]);
idx = sub2ind(sz(1:2),ridx,cidx);
D = B(idx)

D = 3×3

0 1 2 1 0 2 2 1 0

[Vojtech Vanecek]

Thanks a lot. I did not know about the sub2ind function.

Answer 2

[Vilém Frynta]

Zdravím kolego, // Hello,

I have tried something. I can't say it's "cleaner solution", but it's without for loop. You basically unfold matrix into one vector, and then you do not need a loop.

% original variables
A = [1 2 3;2 1 3;3 2 1];
B = [0 1 2;0 1 2;0 1 2];

% modified variables -- make them into a vector
newB = B';
newB = newB(1:numel(B));

newIdx = A';
newIdx = newIdx(1:numel(B));

final_B = newB(newIdx);
final_B = reshape(final_B,[3 3])'

final_B = 3×3

0 1 2 1 0 2 2 1 0

[Vojtech Vanecek]

Dobrý den/Hi,

thanks for the idea. Unfolding the matrix into a vector makes it easier.