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

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Moreno, M. on 10 Apr 2022
Commented: Moreno, M. on 11 Apr 2022
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.
Moreno, M.

DGM on 10 Apr 2022
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
Moreno, M. on 11 Apr 2022
Hi, thanks for the reply and for the answer. I have just tested it and it works perfectly for arbitrary parameters.
Regards,
Moreno, M.