how to velocize it (avoid loop is possible)

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

 采纳的回答

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

更多回答(1 个)

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 个评论

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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