storing multiple matrices from a loop

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

[KSSV]

Can you show us the epxected output for the above?

[Luqmann Mahama]

DBIBC(1)=[1 1 1 1 1; 0 0 0 0 0; 0 0 0 0 0 ; 0 0 0 0 0 ; 0 0 0 0 0]

DBIBC(2)=[1 1 1 1 1; 0 1 1 1 1; 0 0 0 0 0 ; 0 0 0 0 0 ; 0 0 0 0 0]

DBIBC(3)=[1 1 1 1 1; 0 1 1 1 1; 0 0 1 1 0 ; 0 0 0 0 0 ; 0 0 0 0 0]

DBIBC(4)=[1 1 1 1 1; 0 1 1 1 1; 0 0 1 1 0 ; 0 0 0 1 0 ; 0 0 0 0 0]

DBIBC(5)=[1 1 1 1 1; 0 1 1 1 1; 0 0 0 0 0 ; 0 0 0 0 0 ; 0 0 0 0 1]

Answer 1

[Daniel Abajo]

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

[Luqmann Mahama]

Thank you so much. You’re a savior. IT WORKS!!!!

Answer 2

[Daniel Abajo]

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)