Extract first and last nonzero elemenents inside subarrays avoiding mat2cell

2 vues (au cours des 30 derniers jours)
Can somone please give me a hand?
I have a matrix A such as:
A = [0 0 1 2 0 5 0 0 6 0 3 0 ; 0 3 0 2 6 0 0 0 0 0 0 0];
I'd like to check where are the first and last nonzero values along my subarrays of 6 elements on each row. So the expected Output shoul be:
Out = [0 0 1 0 0 1 0 0 1 0 1 0 ; 0 1 0 0 1 0 0 0 0 0 0 0];
I know I can easilly use mat2cell to perform this but I'd like to avoid using this funciton.
Thank's for the help,
Santos

Réponse acceptée

Bruno Luong
Bruno Luong le 22 Mar 2021
Modifié(e) : Bruno Luong le 22 Mar 2021
A = [0 0 1 2 0 5 0 0 6 0 3 0 ; 0 3 0 2 6 0 0 0 0 0 0 0];
b = reshape(A.',6,[]) ~= 0;
[m,n] = size(b);
[~,istart] = max(b,[],1);
[~,istop] = max(flip(b,1),[],1);
i = [istart; m+1-istop] + (0:n-1)*m;
B = zeros(size(b));
B(i) = b(i);
Out = reshape(B,flip(size(A))).'

Plus de réponses (1)

DGM
DGM le 22 Mar 2021
I arbitrarily chose to avoid loops and find() and ended up with this ugly thing. It works, but I wouldn't call it elegant. I added some dummy values to the test array to demonstrate that it handles cases where there are no matches
A = [0 0 1 2 0 5 0 0 6 0 3 0 ; 0 0 0 0 0 0 0 0 0 0 0 0; 0 3 0 2 6 0 0 0 0 0 0 0];
A=reshape(A',6,[])'
matches=[sum(cumprod(A == 0,2),2)+1; 6-sum(cumprod(A == 0,2,'reverse'),2)];
matches=[[1:6 1:6]; matches']';
matches=matches(matches(:,2)>0 & matches(:,2)<7,:);
B=zeros(size(A));
B(sub2ind(size(A),matches(:,1),matches(:,2)))=1;
B=reshape(B',12,[])'
this yields:
B =
0 0 1 0 0 1 0 0 1 0 1 0
0 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 1 0 0 0 0 0 0 0
Again, I added the extra row
  1 commentaire
Santos García Rosado
Santos García Rosado le 22 Mar 2021
Thank you DGM. Check Bruno's answer. You might like the code he came up with.

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