Reshape a 'column' matrix into a 'row' matrix

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Hi, I hope you are well.
I have a matrix 'X' that is generated through a series of parameters, that in the end of the day is a column-wise concatenated matrix such as:
X = [A; B; C; ...; P], where A, B, C, ..., P are 'p' equally-sized [m - 1, q] matrices.
My goal is to reshape this matrix in a way that the new matrix 'X' is:
X = [A, B, C, ..., P]
Ideally, this would be done without loops or manually, so that the operation could be vectorised. A representative example to solve 'similar' to my case matrix would be:
m = 8;
p = 2;
q = 10;
X = rand(p * (m - 1), q);
X(:, [1, q]) = 0;
Which would be a random matrix made of 2 (p) submatrices of [7, 10] ([m - 1, q]) elements. The final solution that I am seeking would be, for this particular case:
Y = [X(1 : m - 1, :), X(m : 2 * (m - 1), :)];
Some help on how to do this transformation for a generic number of matrices 'p' of dimension [m - 1, q] via the reshape command, circshift, rot90, transpose or similar would be greatly appreciated, so that the transformation was compact.
Thanks in advance and regards,
Moreno, M.

采纳的回答

DGM
DGM 2022-4-10
There are other ways this could be done, but just using reshape() and permute() is often the fastest:
m = 5;
p = 2;
q = 5;
X = rand(p * (m - 1), q);
X(:, [1, q]) = 0
X = 8×5
0 0.4353 0.6008 0.1466 0 0 0.9317 0.9804 0.1811 0 0 0.9719 0.7923 0.6327 0 0 0.3685 0.4549 0.4899 0 0 0.9231 0.8038 0.1596 0 0 0.3666 0.9581 0.8240 0 0 0.8689 0.9033 0.3245 0 0 0.3822 0.8846 0.7878 0
% given example
Y = [X(1 : m - 1, :), X(m : 2 * (m - 1), :)]
Y = 4×10
0 0.4353 0.6008 0.1466 0 0 0.9231 0.8038 0.1596 0 0 0.9317 0.9804 0.1811 0 0 0.3666 0.9581 0.8240 0 0 0.9719 0.7923 0.6327 0 0 0.8689 0.9033 0.3245 0 0 0.3685 0.4549 0.4899 0 0 0.3822 0.8846 0.7878 0
% using reshape()
Y = reshape(permute(reshape(X.',q,m-1,p),[1 3 2]),[],m-1).'
Y = 4×10
0 0.4353 0.6008 0.1466 0 0 0.9231 0.8038 0.1596 0 0 0.9317 0.9804 0.1811 0 0 0.3666 0.9581 0.8240 0 0 0.9719 0.7923 0.6327 0 0 0.8689 0.9033 0.3245 0 0 0.3685 0.4549 0.4899 0 0 0.3822 0.8846 0.7878 0
  1 个评论
Moreno, M.
Moreno, M. 2022-4-11
Hi, thanks for the reply and for the answer. I have just tested it and it works perfectly for arbitrary parameters.
Regards,
Moreno, M.

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