storing multiple matrices from a loop
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Luqmann Mahama
le 2 Juin 2020
Commenté : Luqmann Mahama
le 2 Juin 2020
I have a 5x5 matrix and what the program is supposed to do is to run through the columns(i) and whenever it finds a row in the ith column that is zero. It makes the whole row zero [0 0 0 0 0]. And store the different matrices for every(i) column in a matrix DBIBC(i). So far the loop only runs for the DBIBC(1) and stores only that in the array.
rw finds the rows in the column that are zero.
Please help me.
BIBC = [1 1 1 1 1;0 1 1 1 1;0 0 1 1 0;0 0 0 1 0;0 0 0 0 1] ;
DBIBC = BIBC;
N=length(BIBC);
for i=1:N
rw = find(DBIBC(:,i)==0)
DBIBC(rw,:)= 0
DBIBC(i)
B{i} = DBIBC;
end
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Daniel Abajo
le 2 Juin 2020
Hi,
You need to re-inizialize the temporary matrix DBIBC for each iteration, if not the first iteration makes 2-5 rows 0 and thats all....
BIBC = [1 1 1 1 1;0 1 1 1 1;0 0 1 1 0;0 0 0 1 0;0 0 0 0 1] ;
N=length(BIBC);
for i=1:N
DBIBC = BIBC;
rw = find(DBIBC(:,i)==0)
DBIBC(rw,:)= 0
DBIBC(i)
B{i} = DBIBC;
end
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Daniel Abajo
le 2 Juin 2020
Actually is shorter, faster and smart the following code, see that the isequal returns true...
BIBC = [1 1 1 1 1;0 1 1 1 1;0 0 1 1 0;0 0 0 1 0;0 0 0 0 1] ;
DBIBC=repmat(BIBC,1,1,size(BIBC,2));
DBIBC2=permute(repmat(BIBC,1,1,size(BIBC,2)),[1,3,2]);
DBIBC(find(DBIBC2==0))=0;
N=length(BIBC);
for i=1:N
DBIBC = BIBC;
rw = find(DBIBC(:,i)==0)
DBIBC(rw,:)= 0
DBIBC(i)
B{i} = DBIBC;
end
isequal(reshape(cell2mat(B),size(BIBC,1),size(BIBC,2),[]),DBIBC)
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