adding matrices inside a big matrix

2011 11 15
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0 个投票

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

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Sean de Wolski
Sean de Wolski 2011-11-15

2 个投票

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.

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Fangjun Jiang
Fangjun Jiang 2011-11-15
+1, that is fun! Hope your brain is not re-shaped!
j_solar
j_solar 2011-11-15
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
Sean de Wolski 2011-11-15
I forgot to transpose the whole matrix at the end, sorry. try it now!
j_solar
j_solar 2011-11-16
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
Sean de Wolski 2011-11-16
Good catch!
Just add this line will do the same thing.
sum4x4 = reshape(sum4x4,size(sum4x4,2),size(sum4x4,1))'
j_solar
j_solar 2011-11-17
yeah, actually,right after I posted my answer I changed it to what you now say! It is more elegant. :)

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更多回答（3 个）

Fangjun Jiang
Fangjun Jiang 2011-11-15

1 个投票

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);

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j_solar
j_solar 2011-11-15
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!!!

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Sean de Wolski
Sean de Wolski 2011-11-15

1 个投票

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?

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j_solar
j_solar 2011-11-15
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
j_solar 2011-11-15
upps,thats not right. 40=10 and 240 =60 in my loop. and division_cel is 4
Sean de Wolski
Sean de Wolski 2011-11-15
in blkproc, did you change magic(16) to 'a'? the magic(16) was for demo purposes only.
Sean de Wolski
Sean de Wolski 2011-11-15
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!

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Andrei Bobrov
Andrei Bobrov 2011-11-17

0 个投票

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(:)));

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