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non zero rows per column

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babis
babis el 5 de Dic. de 2013
Editada: Alfonso Nieto-Castanon el 6 de Dic. de 2013
i have a matrix. suppose
A=[1 0 8; 0 0 2; 3 0 5; 4 8 0; 0 5 3; 6 1 3; 1 6 5; 0 7 1]
and i want to get the non zero rows per column in a new matrix. in my example that will be
B = [ 1 3 4 6 7 0 0 0; 4 5 6 7 8 0 0 0; 1 2 3 5 6 7 8 0]
( if A=(m,n) B will be B=(n,m) )
  2 comentarios
dpb
dpb el 5 de Dic. de 2013
Use the "Code" button (or insert couple spaces in front of code lines on separate line w/ blank line between it and preceding text) to format the code to be legible.
Azzi Abdelmalek
Azzi Abdelmalek el 5 de Dic. de 2013
Babis, can you explain?

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Alfonso Nieto-Castanon
Alfonso Nieto-Castanon el 5 de Dic. de 2013
Editada: Alfonso Nieto-Castanon el 6 de Dic. de 2013
If I understand your question correctly this should do:
[a,b]=sort(A>0,1,'descend');
B=a'.*b';

Más respuestas (2)

dpb
dpb el 5 de Dic. de 2013
"Deadahead" solution...
B=zeros(size(A))';
for i=1:size(A,2)
ix=A(:,i)~=0;
B(i,ix)=find(A(:,i));
end
  2 comentarios
babis
babis el 5 de Dic. de 2013
i think that this is really close to what i want, thank you
dpb
dpb el 5 de Dic. de 2013
Editada: dpb el 5 de Dic. de 2013
It reproduces you example (w/ the exception of the extra row of zeros which I presumed was an error). If that is indeed wanted, then just augment the end result. You can, of course, with care to keep parens nested properly, do away with the intermediary I used for clarity of exposition. So what is on "really close" about it instead of "dead on"?
It should be reasonably easy to accumarray or otherwise vectorize it w/ the idea given altho it's not convenient here at the moment...

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José-Luis
José-Luis el 5 de Dic. de 2013
your_mat = ndgrid(1:size(A,1),1:size(A,2));
your_mat(A==0) = 0;
your_mat(your_mat==0) = Inf;
your_mat = sort(your_mat);
your_mat(your_mat==Inf) = 0;

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