Large Sparse Matrix Summation
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Hi, Everyone:
Suppose I have a very large M*N sparse matrix A, where M=K*N, I need to equally split it into K N*N matrices and sum it up, I can't use loop, so I tried to use:
sum(reshape(A',N,N,K),3);
However, this command can't reshape sparse matrix into a 3-D array, is there any other way to do it? (without loop or converge to dense matrix)
Many Thanks
1 个评论
James Tursa
2013-4-22
What are the actual sizes you are working with?
采纳的回答
Teja Muppirala
2013-4-22
This seems to work well:
% Making some random data...
N = 1000;
K = 100;
A = sprand(N*K,N,0.0001);
[r,c,val] = find(A);
rnew = mod(r-1,N)+1; % Shift all the row values to [1-->N]
Asum = accumarray([rnew c],val,[N N],[],[],true); % The answer
2 个评论
Teja Muppirala
2013-4-22
I guess that last line would be simpler as this:
Asum = sparse(rnew,c,val,N,N);
Ming
2013-4-22
更多回答（1 个）
Why can't you use a loop? Is K that large?
If you are looking for efficiency, I'd say that you could directly build A in a way that sums up these N*N blocks, by working on indices using a modulus for row indices. As SPARSE works like ACCUMARRAY when multiple indices are similar, you would have the summation. Example:
N = 4 ;
I = randi(3*N, 5*N, 1) ;
J = randi(N, 5*N, 1) ;
V = 50 + randi(10, 5*N, 1) ;
% - First method: assumes that A already built.
A = sparse(I, J, V, 3*N, N) ;
full(A)
[r,c,v] = find(A) ;
r = 1 + mod(r-1, N) ;
A_sum = sparse(r, c, v, N, N) ;
% - Second method: build directly the sum from indices.
I = 1 + mod(I-1, N) ;
B = sparse(I, J, V, N, N) ;
% - Check.
full(A_sum)
full(B)
full(all(A_sum(:) == B(:)))
Running this, you get (RANDI will generate something else if you test it):
ans =
0 55 0 0
60 0 51 0
0 54 0 112
0 0 0 0
52 0 0 114
0 59 115 0
57 51 0 0
0 56 0 56
0 55 0 0
0 0 0 0
54 0 56 0
0 51 0 0
ans =
52 110 0 114
60 59 166 0
111 105 56 112
0 107 0 56
ans =
52 110 0 114
60 59 166 0
111 105 56 112
0 107 0 56
ans =
1
3 个评论
Cedric
2013-4-22
Well, Teja was faster ;-)
Ming
2013-4-22
James Tursa
2013-4-23
How big? What are the actual sizes you are using?
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