Replace specific values in a matrix with zeros

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Luan Nguyen Thanh
Luan Nguyen Thanh 2017-11-18
评论： Luan Nguyen Thanh 2017-11-18
Assume, I have two matrices like this:
A =
35 1 6 26 19 24
3 32 7 21 23 25
31 9 2 22 27 20
8 28 33 17 10 15
30 5 34 12 14 16
B = [3 2 4 5 2 1];
I want to produce a matrix C below from matrix A & B like this:
C =
35 1 6 26 19 24
3 32 7 21 23 0
31 0 2 22 0 0
0 0 33 17 0 0
0 0 0 12 0 0
where the ith element in matrix B determines "the number of elements in matrix A's ith column" that will bring to the ith column of the matrix C. For example, B(1) = 3 means that the 1st column of matrix C is equal to
[A(1:B(1),1) ; zeros(5-B(1),1)];
So is there any way for me to produce matrix C from A & B without using any "loop" command?

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采纳的回答

Stephen23
Stephen23 2017-11-18
编辑：Stephen23 2017-11-18
In one line:
>> A = [...
35 1 6 26 19 24
3 32 7 21 23 25
31 9 2 22 27 20
8 28 33 17 10 15
30 5 34 12 14 16];
>> B = [3,2,4,5,2,1];
>> C = A.*bsxfun(@le,(1:size(A,1)).',B)
C =
35 1 6 26 19 24
3 32 7 21 23 0
31 0 2 22 0 0
0 0 33 17 0 0
0 0 0 12 0 0

更多回答（2 个）

Bshara Murr
Bshara Murr 2017-11-18
编辑：Bshara Murr 2017-11-18
Hello, yes try this:
A =[35 1 6 26 19 24; 3 32 7 21 23 25; 31 9 2 22 27 20; 8 28 33 17 10 15; 30 5 34 12 14 16;];
B = [3 2 4 5 2 1];
[x y]= size(A);
C = zeros(x,y);
for i = 1:y
for j = 1:B(i)
C(j,i) = A(j,i);
end
end
I just used the matrices you entered.
KL
KL 2017-11-18
编辑：KL 2017-11-18
Try this,
[r c] = size(A);
C = zeros(r,c);
indx = bsxfun(@le,repmat((1:r).',1,c),repmat(B,r,1)) %find indicies
C(indx) = A(indx)
C =
35 1 6 26 19 24
3 32 7 21 23 0
31 0 2 22 0 0
0 0 33 17 0 0
0 0 0 12 0 0

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