How do I find rows that match a list of vectors without using a loop?
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
David Haydock
am 13 Apr. 2023
Kommentiert: David Haydock
am 13 Apr. 2023
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?
0 Kommentare
Akzeptierte Antwort
Matt J
am 13 Apr. 2023
Bearbeitet: Matt J
am 13 Apr. 2023
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
3 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Creating and Concatenating Matrices finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!