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

Answer 1

[Matt J]

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

Answer 2

[Catalytic]

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.

[Matt J]

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.