how to velocize it (avoid loop is possible)

1 visualizzazione (ultimi 30 giorni)
E=[0 5 6 9 2;0 0 1 3 1;0 5 4 2 4]'
E = 5×3
0 0 0 5 0 5 6 1 4 9 3 2 2 1 4
filtro=ones(size(E));
for i=1:width(E)
bu=find(E(:,i),1,'first');
filtro(1:bu,i)=0;
end
filtro
filtro = 5×3
0 0 0 0 0 0 1 0 1 1 1 1 1 1 1

Risposta accettata

Matt J
Matt J il 3 Feb 2024
Modificato: Matt J il 4 Feb 2024
E=[0 5 6 9 2;0 0 1 3 1;0 5 4 2 4]';
[~,I]=max(logical(E),[],1);
filtro=(1:height(E))'>I
filtro = 5×3 logical array
0 0 0 0 0 0 1 0 1 1 1 1 1 1 1

Più risposte (1)

Catalytic
Catalytic il 4 Feb 2024
I don't know if "velocize" (not a word) is supposed to mean "accelerate" or "vectorize". The two are not the same.
If you're looking for the fastest possible code, there's no way to know in advance because it depends on the sparsity of E. For very dense E, your loop will probably be faster than @Matt J's answer.
E=rand(5000,3000)>0.1;
tic
filtro=true(size(E));
for i=1:width(E)
bu=find(E(:,i),1,'first');
filtro(1:bu,i)=0;
end
toc
Elapsed time is 0.016652 seconds.
tic
[~,I]=max(logical(E),[],1);
filtro=(1:height(E))'>I;
toc
Elapsed time is 0.031441 seconds.
  1 Commento
Matt J
Matt J il 4 Feb 2024
True, but be mindful of the flipside:
E=rand(5000,3000)>0.7;
tic
filtro=true(size(E));
for i=1:width(E)
bu=find(E(:,i),1,'first');
filtro(1:bu,i)=0;
end
toc
Elapsed time is 0.051811 seconds.
tic
[~,I]=max(logical(E),[],1);
filtro=(1:height(E))'>I;
toc
Elapsed time is 0.020633 seconds.

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