How to concatenate two matrices in a loop?
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Betha Shirisha
on 2 Mar 2015
Commented: Image Analyst
on 29 Mar 2018
I' m running a for loop say 'n' times,everytime a 5x2 matrix will be generted ,i have to join all these matrices such that at last run i have to get a matrix of size 5x2n .How to solve this?
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Accepted Answer
James Tursa
on 2 Mar 2015
Edited: James Tursa
on 2 Mar 2015
Another way using 3-dimensional syntax:
n = number of individual matrices you have;
R = number of rows per matrix;
C = number of columns per matrix;
z = zeros(R,C,n); % Pre-allocate n pages of R x C matrices
for k=1:n
m = individual R x C matrix;
z(:,:,k) = m; % Assign to the k-th page
end
z = reshape(z,R,C*n); % Reshape into R x (C*n) matrix
2 Comments
Oce@n
on 28 Mar 2018
This works for double arrays. For cell arrays you should try something else (see below)
Image Analyst
on 29 Mar 2018
You just need to make 3 changes for cells:
- Use cell() instead of zeros()
- Use z{k} instead of z(:,:,k), and
- Get rid of the reshape().
Try this:
n = 3; % number of individual matrices you have;
R = 4; % number of rows per matrix;
C = 5; % number of columns per matrix;
z = cell(n, 1); % Pre-allocate n cells of R x C matrices
for k = 1 : n
m = rand(R, C); % individual R x C matrix;
z{k} = m; % Put into the k-th cell
end
celldisp(z) % Display z in the command window.
More Answers (2)
Image Analyst
on 2 Mar 2015
Maybe not the most efficient way, but for small numbers it won't make any difference. Try this - I think it's what you asked for, where the number of columns might vary.
n = 6; % 6 iterations
m = []; % Initialize to null.
for k = 1 : n
% Get the number of columns for this iteration
% and make up a sample matrix with that many columns.
numberOfColumns = randi([5,12], 1); % Random number between 2 and 12
% Get a matrix with random numbers that is
% 5 rows tall by numberOfColumns columns wide.
thisMatrix = randi(9, 5, numberOfColumns);
% Append onto m
m = [m, thisMatrix];
end
% Print final m to the command window.
m
2 Comments
Brendan Hamm
on 2 Mar 2015
n = 10; % Change n as necessary
z = zeros(5,2*n); % pre-allocate space
for ii = 1:2:2*n-1
x = rand(5,2); % Just to make sore data
z(:,[ii,ii+1]) = x; % Assign to the appropriate location
end
2 Comments
Brendan Hamm
on 3 Mar 2015
To generalize this example:
m = 12;
n = 10; % Change n as necessary
z = zeros(5,2*m); % pre-allocate space
for ii = 1:m:m*n-1
x = rand(5,m); % Just to make sore data
z(:,ii:ii+m-1) = x; % Assign to the appropriate location
end
If your matrix x is independent of a loop, then you don't need a loop to concatenate it. In your question this is what you ask for.
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