non zero rows per column

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babis
babis am 5 Dez. 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 Kommentare
dpb
dpb am 5 Dez. 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 am 5 Dez. 2013
Babis, can you explain?

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Akzeptierte Antwort

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

Weitere Antworten (2)

dpb
dpb am 5 Dez. 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 Kommentare
babis
babis am 5 Dez. 2013
i think that this is really close to what i want, thank you
dpb
dpb am 5 Dez. 2013
Bearbeitet: dpb am 5 Dez. 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 am 5 Dez. 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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