substituting a column into a 3D matrix

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ily
ily 2011-8-20
I have a matrix R = zeros(4,5,3) and I want to make the last column of each matrix a different number. such that given C is a vector
for n = 1: end R(:,end, n) = C(n)
I managed a solution where I have a column vector A which if I could substitute into R(:, end, :), but it says mismatch dimensions.
essentially, I want to substitute a column vector into the columns of a 3D matrix R without using a for loop.

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采纳的回答

Fangjun Jiang
Fangjun Jiang 2011-8-20
It's not clear when you mention "last column" for a 3-D matrix. Hope you mean one of the following two cases.
R=zeros(4,5,3);
A=1:20;
R(:,:,3)=reshape(A,size(R,1),size(R,2))
R=zeros(4,5,3);
A=1:12;
R(:,5,:)=reshape(A,size(R,1),size(R,3))
  1 个评论
ily
ily 2011-8-22
thanks. If we assumed the third dimension was a page. Then, I was trying to add a different vector into the last column of each page. I was trying to this without the for loop.
The second solution was very helpful. I understand the key is to make the vectors into a 2-D matrix.

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更多回答（1 个）

Jan
Jan 2011-8-21
R = zeros(4,5,3);
A = [1, 2, 3];
R(:, end, :) = reshape(repmat(A, 4, 1), [4, 1, 3]);

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