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

Question

David Haydock 2023-4-13

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https://ww2.mathworks.cn/matlabcentral/answers/1946578-how-do-i-find-rows-that-match-a-list-of-vectors-without-using-a-loop

评论： David Haydock 2023-4-13

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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Answer 1

Matt J 2023-4-13

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https://ww2.mathworks.cn/matlabcentral/answers/1946578-how-do-i-find-rows-that-match-a-list-of-vectors-without-using-a-loop#answer_1215173

编辑：Matt J 2023-4-13

在 MATLAB Online 中打开

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 2023-4-13

Yes.

David Haydock 2023-4-13

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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