Processing timer for Sparse matrix is too long?

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Nghia Nguyen Thanh 2017-8-22

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编辑： Cedric 2017-9-19

采纳的回答： Cedric

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Dear All,

I am a new user with Matlab, and I meet a problem with Sparse matrix.

Time for processing with Sparse matrix is very long.

Here is my code:

lab=randi([1 7],1,100000);
M = sparse(100000,100000);
for i = 1:100000
    for j = (i+1):100000
        if lab(i,1) == lab(j,1)
            M(j,i) = 1;
        else
            M(j,i) = 0;
        end
    end
end

Do I have any method for reduce timing process? Thanks and Best regard!

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Tim Berk 2017-9-19

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Dear Nghia,

Looping through large arrays is very slow, especially in what is called 'nested loops' as you are using. Your nested loop requires 100000*100000/2 calculations (the 1/2 because of the (i+1) as starting value for j).

Lets assume that evaluating

if lab(1,i) == lab(1,j)
  M(i,j) = 1;
  else M(j,i) = 0;
end

Takes 10 micro seconds (which I think is quite a low estimate), than your script would take 1E5*1E5*0.5*1E-5 = 50000 seconds (14 hours).

Hope this clarifies why it is slow.

Often this can be solved by vectorization as explained here: https://uk.mathworks.com/help/matlab/matlab_prog/vectorization.html

Hope this helps.

Cheers,

Tim

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Answer 1

Cedric 2017-9-19

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编辑：Cedric 2017-9-19

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You won't be able to do it, vectorizing or not.

If random integers are really in 1:7, the probability, when you pick two elements, that one is equal to the other is 1/7. On average, with your lab vector of n=1e6 elements and assuming that you are interested in the lower triangular part of M, this means n^2/2/7 non-zero elements, and hence

 >> (16 * (1e6)^2/2/7 + 8 * 1e6 + 8) / 1e12
 ans =
    1.1429

more than 1.14 TB of RAM to store (just a single version of it).

Are random integers really in 1:7 or is the pool of integers larger, e.g. 1:n ? In the second case, you will gain a lot by vectorizing the approach. Often in these cases that involve sparse matrices, the best approach consists in building indices of non-zero elements ( I and J ), and performing a unique call to SPARSE at the end:

M = sparse( I, J, 1, n, n ) ; % Values are all 1 in your case.

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Answer 2

Walter Roberson 2017-9-19

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编辑：Walter Roberson 2017-9-19

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Doing M(j,i) = 0 is a waste of time on a sparse array unless M(j,i) might already have a non-zero value. sparse() initializes to 0.

Your line,

M = sparse(100000,100000);

is equivalent to

M = sparse([], [], [], 100000, 100000, 0)

which generates a sparse array that is expected to have 0 occupied cells in it. Every time a cell is written, the sparse array needs to be recreated to allocate room for the new entry. It is a lot more efficient to use

M = spalloc(100000, 100000, N)

where N is an estimate of the number of locations that will eventually be occupied with non-zero values. In your situation you can estimate N as 100000^2/7 which is roughly 1428571429. One element in seven occupancy is not very sparse.

Your inner loop could be vectorized, which could improve performance a lot.