Find Indexes above zero in a matrix and replace thoses indexes with the value of a vector with the same length.
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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)
2 件のコメント
José-Luis
2016 年 9 月 7 日
You want to store the indexes or just get the matrix with the replaced values?
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Andrei Bobrov
2016 年 9 月 7 日
編集済み: Andrei Bobrov
2016 年 9 月 7 日
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);
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