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Return subscripts of common rows for multi-dimensional matrix?

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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.
  4 Kommentare
Guillaume
Guillaume am 13 Jul. 2018
And you don't care about which is the row in B that match a A row in that layer?
Andrew Poissant
Andrew Poissant am 13 Jul. 2018
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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Guillaume
Guillaume am 13 Jul. 2018
Bearbeitet: Guillaume am 13 Jul. 2018
[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)))

Weitere Antworten (1)

dpb
dpb am 13 Jul. 2018
ix=mod(find(all(ismember(A,B),2)),size(A,3));
  4 Kommentare
Andrew Poissant
Andrew Poissant am 13 Jul. 2018
Thank you for the answer but I went with dpb's answer because yours was returning 0s.
dpb
dpb am 13 Jul. 2018
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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