adding matrices inside a big matrix

Hello everybody,

I am wandering if there is a way for not using loops to add many small matrices inside a big matrix. For example I have a big 40000x240000 matrix and I want to add all the numbers of 4x4 matrices inside that matrix, resulting in a 10000x60000 matrix.

thank you all for your answers

Answer 1

[Sean de Wolski]

And this fun answer:

   sum4x4 = reshape(sum(reshape(reshape(sum(reshape(B,4,[])),size(B,1)/4,[])',4,[])),size(B)./4);
sum4x4 = reshape(sum4x4,size(sum4x4,2),size(sum4x4,1))'

where B is your matrix.

[Fangjun Jiang]

+1, that is fun! Hope your brain is not re-shaped!

[j_solar]

yeah, upps, hadn't changed magic(16). Did it now and it works, however it is slower than the mat2cell option. Not much but a bit slower.

The fun answer is much faster, over 100 times, however something doesn't work right. It seems like the values are the same, however they are not in the right position.

[Sean de Wolski]

I forgot to transpose the whole matrix at the end, sorry. try it now!

[j_solar]

Almost, because when I transpose I get a 60x10 matrix instead of 10x60 which is what i want.

But what i did to solve it is:

[M,N] = size(c);

d = zeros(N,M);

d(:) = c;

e = d'

c is the solution to your answer(without tranposing at the end) and e is exactly what I want

Thank you all for your help!!!!!!!!!!1

[Sean de Wolski]

Good catch!

Just add this line will do the same thing.

sum4x4 = reshape(sum4x4,size(sum4x4,2),size(sum4x4,1))'

[j_solar]

yeah, actually,right after I posted my answer I changed it to what you now say! It is more elegant. :)

Answer 2

[Fangjun Jiang]

Because the size of your matrix, you might want to re-use the matrix so replace the variable "e" and "f" with "d".

d=rand(40,24);
[M,N]=size(d);
e=mat2cell(d,4*ones(M/4,1),4*ones(N/4,1));
f=cellfun(@(x) sum(x(:)),e);

[j_solar]

Yeah!!!! That works really well. Finally used a smaller matrix, 40x240 and its 2.2 times faster than with a loop. For bigger matrices this difference will probably increase BIG TIME!!!

Answer 3

[Sean de Wolski]

How about blkproc if you have the Image Processing Toolbox (or blockproc in newer versions).

blkproc(magic(16),[4 4], @(x)(sum(x(:))))

I'm surprised conversion to cell is faster than a well done for-loop. Perhaps you could post what you had for the nested for-loops?

[j_solar]

This is what I had before, maybe not the best. With the change offered by fangjun the time decreased by 2.2

a = zeros(40, 240);

for j = 1:240

for k= 1:40

b(k,j) = sum(sum(a((k-1)*division_cel+1:k*division_cel,(j-1)*division_cel+1:j*division_cel)));

end

end

I have tried blkproc but that returns a 4x4 and I need to add the 4x4 inside returning in my case a 10x60

[j_solar]

upps,thats not right. 40=10 and 240 =60 in my loop. and division_cel is 4

[Sean de Wolski]

in blkproc, did you change magic(16) to 'a'? the magic(16) was for demo purposes only.

[Sean de Wolski]

Your for-loop could be sped up a little by building the j-index vectors outside of the k for-loop. Overall it's good though!

Answer 4

[Andrei Bobrov]

e.g.

A = randi(25,12,16);
m = 4;
n = 4;
[M,N] = size(A);
out = reshape(sum(sum(reshape(A.',n,N/n,m,[]),3)),n,[])';

or

out2 = reshape(sum(sum(reshape(A,m,M/m,n,[]),3)),[],n);

or2

out3 = blockproc(A,[m n],@(block_struct)sum(block_struct.data(:)));