问题
I have a 40x16 matrix or 8 5x16 one below the other i.e. aligned vertically. I want to get a 5x128 matrix from that such that I align the 8 5x16 matrices horizontally. is there an efficient/quicker (rather than the hardcoded for loops) way to do this?
I want the individual 5x16 matrices intact.
回答1:
This should work. Suppose your matrix is A
(40x16).
Here's a way using reshape
:
m = 5; n = 8; p = 16;
B = reshape(permute(reshape(A', p, m, n), [2 1 3]), m, n*p);
B will have your eight 5x16 matrices next to each other, intact.
And here's a way without reshape
:
m = 5; n = 8;
B = cell2mat(arrayfun(@(i) A(m*(i-1)+1:m*i, :), 1:n, 'UniformOutput', false));
回答2:
Consider using the reshape
function: doc@mathworks.
回答3:
You can use MAT2CELL to divide the big matrix into smaller ones, then combine along the dimension you want:
A = rand(8*5,16);
blkSz = 5;
C = mat2cell(A, blkSz*ones(1,size(A,1)/blkSz), size(A,2));
C = cat(2,C{:})
回答4:
Reshape a 3-by-4 matrix into a 2-by-6 matrix. A = 1 4 7 10 2 5 8 11 3 6 9 12
B = reshape(A,2,6)
B = 1 3 5 7 9 11 2 4 6 8 10 12 B = reshape(A,2,[])
B = 1 3 5 7 9 11 2 4 6 8 10 12
来源:https://stackoverflow.com/questions/11046415/how-to-reshape-matlab-matrices-for-this-example