# How do I find rows that match a list of vectors without using a loop?

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David Haydock 2023 年 4 月 13 日
コメント済み: David Haydock 2023 年 4 月 13 日
Let's say I have a list of length 2 vectors that can occur, which I put in a matrix as rows:
possible = [1 2; 1 3; 1 4; 1 5;
2 1; 2 3; 2 4; 2 5;
3 1; 3 2; 3 4; 3 5;
4 1; 4 2; 4 3; 4 5;
5 1; 5 2; 5 3; 5 4];
and I have a set of observed vectors of length 2 that did occur:
observed = [1 3;
1 5;
4 2;
4 3;
3 2;
4 2]; % ... and so on
I need to go through the rows in the list of possible length 2 vectors, and get the index of where each row occurs in observed, like this:
for c = 1:size(possible, 1)
[~, index{c}] = ismember(observed, possible(c,:),'rows');
end
Whilst this approach does work, it proves to be very slow for my approach, as I have many observed matrices to run through, and many possible matrices to run through as well.
Is there a way of making this more efficient? Perhaps by using something other than a for loop?

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### 採用された回答

Matt J 2023 年 4 月 13 日

It would be advisable to obtain the indices as a logical matrix rather than as subscripts. This can be done looplessly with pdist2 as below.
observed = [1 3;
1 5;
4 2;
4 3;
3 2;
4 2]; % ... and so on
possible = [1 2; 1 3; 1 4; 1 5;
2 1; 2 3; 2 4; 2 5;
3 1; 3 2; 3 4; 3 5;
4 1; 4 2; 4 3; 4 5;
5 1; 5 2; 5 3; 5 4];
index=pdist2(observed,possible)==0
index = 6×20 logical array
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
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Matt J 2023 年 4 月 13 日
Yes.
David Haydock 2023 年 4 月 13 日
Thank you so much. I just implemented it into my code and I am not joking when I say it will have saved me a day of processing time. Thank you so much.

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