Subsetting an n-d array with no loop
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I have an n-d array that I want to subset along specific dimensions accordingly.
For instance, consider the following array,
Consider also a permutation or selection of specific indices from the last dimension:
I want to extract an n-d array B that contains all the diagonal elements of the 1st and last dimensions (applying for that last dimension the σ transformation) :
One option, rather brute-force, is to do the following:
A = rand(10,25,7,10);
sigma = [5,5,4,1,1,9,8,7,2,4];
for i=1:10
B(i,:,:)= A (i,:,:,sigma(i))
end
However, if I deal with very large arrays, this requires a loop which really slows down my code.
I have tried subsetting like below, but Matlab doesn't understand that the index for the 1st dimension runs with the index for last dimension.
C = A(1:10,:,:,sigma(1:10))
If I do this, I end up with a 
matrix, which would be the following
matrix, which would be the following
Is there a way of subsetting an n-d array along specific dimensions simultaneously without a loop ?
Accepted Answer
Vilém Frynta
on 12 Apr 2023
Moved: Walter Roberson
on 12 Apr 2023
Could this work for you? It may be too "sample specific" instead working for every example.
This script generates a 4D array A of random numbers, and an array sigma of indices (which you provided). It then uses these indices to extract elements from A along the fourth dimension, and saves the results in a new array B, creating a new 3D array that only contains the elements of A selected by the sigma array.
sub2ind function needs same length vectors, so it's been workarounded a bit.
A = rand(10,25,7,10);
sigma = [5,5,4,1,1,9,8,7,2,4];
idx = sub2ind(size(A), repmat((1:10)',25*7,1), ...
repmat(reshape(repmat((1:25)',1,7),[],1),10,1), ...
repmat(reshape(repmat((1:7)',25,1),[],1),10,1), ...
repmat(sigma',25*7,1));
B = reshape(A(idx), [10, 25, 7]);
disp(B)
6 Comments
Francois Miguet
on 12 Apr 2023
Moved: Walter Roberson
on 12 Apr 2023
Adam Danz
on 12 Apr 2023
Moved: Walter Roberson
on 12 Apr 2023
Even more generic:
A = rand(10,25,7,10);
sigma = [5,5,4,1,1,9,8,7,2,4];
[idI,idJ,idK] = ndgrid(1:size(A,1),1:size(A,2),1:size(A,3));
idA = {idI(:),idJ(:),idK(:),repmat(sigma',[prod(size(A,[2,3])),1])};
B = reshape(A(sub2ind(size(A),idA{:})),size(A,[1,2,3]));
disp(B)
Vilém Frynta
on 12 Apr 2023
Moved: Walter Roberson
on 12 Apr 2023
@Francois Miguet glad I could help!
If my answer was helpful to you, I'd be happy if you could formally accept it.
Vilém Frynta
on 12 Apr 2023
Moved: Walter Roberson
on 12 Apr 2023
@Francois Miguet you have accepted your own answer. what i meant was my comment :p
Francois Miguet
on 12 Apr 2023
Edited: Adam Danz
on 13 Apr 2023
@Vilém Frynta weirdly it is the only option I see...
Walter Roberson
on 12 Apr 2023
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