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Return subscripts of common rows for multi-dimensional matrix?

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Andrew Poissant
Andrew Poissant 2018-7-13
评论： dpb 2018-7-13
I have a 8x2 matrix, A, and a 133x2x5 matrix, B. I want to return the the layer in B in which a row in A matches a row in B. How can I do that? I tried using intersect and ismember but have not had any luck thus far. Having a hard time with the matrix being multi-dimensional.
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Guillaume
Guillaume 2018-7-13
And you don't care about which is the row in B that match a A row in that layer?
Andrew Poissant
Andrew Poissant 2018-7-13
I do not care about that. It can be any row in B in any layer, as long as it matches to a row in A.

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采纳的回答

Guillaume
Guillaume 2018-7-13
编辑：Guillaume 2018-7-13
[row, layer] = ind2sub([size(B, 1), size(B, 3)], find(ismember(reshape(permute(B, [1 3 2]), [], size(B, 2)), A, 'rows')))
If you want just the layers in which any row matches any row of A:
layer = unique(layer)
edit: By the way the logic of this is to reshape B into a two column matrix by vertically concatenating the layers. Then use the traditional ismember(..., 'rows') and finally convert the matched rows back into (row, layer) coordinate.
Another way, avoiding the sub2ind would be:
layer = unique(ceil(find(ismember(reshape(permute(B, [1 3 2]), [], size(B, 2)), A, 'rows')) / size(B, 1)))

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dpb
dpb 2018-7-13
ix=mod(find(all(ismember(A,B),2)),size(A,3));
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Andrew Poissant
Andrew Poissant 2018-7-13
Thank you for the answer but I went with dpb's answer because yours was returning 0s.
dpb
dpb 2018-7-13
It is dpb's and it's supposed to be zero...the "fixup" is
ix(ix==0)=size(A,3);
I posted it as much as a lark as anything... :) G's is a much more legible and therefore maintainable approach.

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