Find Indexes above zero in a matrix and replace thoses indexes with the value of a vector with the same length.

Hello, I have matrix A=(112,20) size and a vector with V=(112,1) size. I want to find the indexes in every column in A that are above zero and replace them with the value of the vector with the same index.[x z]=find(A(:,1)>0) returns the indexes and also V(x) returns the vectors with the values i want, but i want to do this for A(:,1),A(:,2)...A(:,20) and store those values in a new Matrix B=(:,20)

[José-Luis]

You want to store the indexes or just get the matrix with the replaced values?

[Alexandros Samp]

I want to get the matrix with the replaced values.

Answer 1

[Andrei Bobrov]

B = bsxfun(@times,A > 0,V) + A.*(A <= 0);

or

[ii,jj] = find(A > 0);
B = A;
B(sub2ind(size(A),ii,jj)) = V(ii);

[Alexandros Samp]

This just gives me the same Matrix as before.

[Andrei Bobrov]

corrected code