How to repeat a rectangular matrix in matlab?

How to repeat a rectangular matrix in matlab?

Not using loops, just matlab's build-in commands.

Thanks a lot!

Answer 1

[Star Strider]

This works:

a = [1 1 1 1; 2 2 2 2];

A = zeros(6);

for k1 = 1:2:size(A,1)-1
    A(k1:k1+1, k1:k1+3) = a;
end
A       % Show Result

[rui]

Hi there, is there any way that I can avoid using loop? Thanks.

[Star Strider]

Not that I can think of. Every approach requires a loop because of the way you want your matrix.

Here’s a more efficient way to do it, but it still requires a loop:

a = [1 1 1 1; 2 2 2 2];
A = [a zeros(2,4)];
A1 = A;
for k1 = 1:2
    A1 = [A1; circshift(A, [2 2]*k1)];
end
A1      % Show Result

[rui]

The matrix that I am using actually has millions of entries, I'm worried that using loop will significantly prolong the process, but thanks again.

[Star Strider]

Loops really aren’t evil! The second version is more efficient than the first, but there is no built-in function that will do what you want. I didn’t preallocate the result matrices in my second example because they’re small. Preallocating your millions-dimension result matrices as in my first example as:

A = zeros(6,8);

will significantly speed up the process. I would favour my first example because of that.

[rui]

I used circshift, sparse and loops in my code but it still takes a very long time to obtain the results, and I've been searching all day but with no luck, so I guess that loop is inevitable in this situation. Thank you so much for your help, I really appreciate it!

[Star Strider]

My pleasure!

The circshift approach (that I used) expands the matrix with each step. That takes more time, because MATLAB has to allocate new memory each time.

I suggest using my first approach and preallocating the matrix. That eliminates the problem of expanding the matrix at each step, eliminates the call to circshift, and produces the same result.

Answer 2

[Azzi Abdelmalek]

A=[1 2 ; 3 4]
B=repmat(A,3,2)

[rui]

Hi there, A is a square matrix, but I want to know how to repeat a rectangular matrix, and automatically fill other positions with zeros, thanks.

[Azzi Abdelmalek]

You didn't say anything about how do you want to shift your matrix? it's not just repeating a matrix.

[rui]

My mistake. Shifting the matrix by 2 positions in 2-dimensions. At the beginning there is a=[1 1 1 1;2 2 2 2], after doing it 3 times without using any loops (just using build-in commands), it becomes a 6-by-8 matrix as shown in the picture. Those extra positions are filled by 0s automatically. Thanks again.

Answer 3

[Andrei Bobrov]

for your case:

t = zeros(6,2);
out = [kron(eye(3),a(:,1:2)),t]+[t,kron(eye(3),a(:,3:4))];

variant

m = 3;
k=2;

s = size(a);
n = (m-1)*k+s(2);
m1 = m*s(1);
out = zeros(m1,n);
t = sub2ind([m1,n],1:s(1):m1,1:k:k*m);
t2 = bsxfun(@plus,(0:s(2)-1)*m1,(0:s(1)-1)');
out(bsxfun(@plus,t,t2(:))) = a(:,:,ones(m,1));

[rui]

This looks quite awesome, thanks!