Reduce matrix by adding every n - matlab

Question

Not sure how to explain it so here it goes as an example:
A=[1 0 0 1 4 4 4 4
0 0 0 0 2 3 2 2
0 0 0 0 0 0 0 1
2 3 4 5 2 3 4 1 ]
result:
b=[ 1 1 13 12
5 9 5 6];
Each of the elements is computed by adding a N size submatrix inside the original, in this case N=2.
so b(1,1) is A(1,1)+A(1,2)+A(2,1)+A(2,2), and b(1,4) is A(1,7)+A(2,7)+A(1,8)+A(2,8).
Visually and more clearly:
A=[|1 0| 0 1| 4 4| 4 4|
|0 0| 0 0| 2 3| 2 2|
____________________
|0 0| 0 0| 0 0| 0 1|
|2 3| 4 5| 2 3| 4 1| ]
b is the sum of the elements on those squares, in this example of size 2.
I can imagine how to make it with loops, but its just feels vectorizable. Any ideas of how it could be done?
Assume that the matrix A has sizes that are multipliers of N.

Answer 1

If you have the Image processing toolbox blockproc could be an option as well:
B = blockproc(A,[2 2],#(x) sum(x.data(:)))

Answer 2

Method 1:
Using mat2cell and cellfun
n = 2;
AC = mat2cell(A,repmat(n,size(A,1)/n,1),repmat(n,size(A,2)/n,1));
out = cellfun(#(x) sum(x(:)), AC)
Method 2:
Using permute and reshape
n = 2;
[rows,cols] = size(A);
out = reshape(sum(sum(permute(reshape(A,n,rows/n,n,[]),[1 3 2 4]))),rows/n,[]);
PS: Here is a close question related to this one, which you might find useful. That question is to find mean while this one is to find sum.

Answer 3

Here are two alternative methods:
Method #1 - im2col
Another method using the image processing toolbox is to use im2col with the distinct flag and sum over all of the resulting columns. You would then need to reshape the matrix back to the right size:
n = 2;
B = im2col(A, [n n], 'distinct');
C = reshape(sum(B, 1), size(A,1)/n, size(A,2)/n);
We get for C:
>> C
C =
1 1 13 12
5 9 5 6
Method #2 - accumarray and kron
We can generate an index matrix with kron which we can use as bins into accumarray and invoke sum as the custom function. We would again have to reshape the matrix back to the right size:
n = 2;
M = reshape(1:prod([size(A,1)/n, size(A,2)/n]), size(A,1)/n, size(A,2)/n);
ind = kron(M, ones(n));
C = reshape(accumarray(ind(:), A(:), [], #sum), size(A,1)/n, size(A,2)/n);
Again we get for C:
C =
1 1 13 12
5 9 5 6