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Update a sparse matrix efficiently

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mafoEV on 5 Mar 2020
Edited: Bruno Luong on 26 Jan 2023
Dear MATLAB community,
I have a large sparse matrix (~ 100,000 by 100,000, with about 100,000 non-zero elements), which I need to update regularly (at each solver step).
So far, I've done it by creating its elements first (rows, columns and values) as column vectors, then assembling them using the sparse function:
A = sparse(iA, jA, vA, m, n, nnzA);
However this line is now taking 13% of my total solve time (called 355 times, taking about 0.29s each time).
The matrix structure does not change from one iteration to the next, only about half of the values need updating. Is there a more efficient way to do it? I haven't been able to find any solution so far looking at the forums.
I'm using Matlab 2019b Update 3 (
Many thanks.
Bruno Luong
Bruno Luong on 25 Aug 2020
Are you updated only non-zero elements? only zero elements? Not known?
In three cases, the required time are differents.
Joel Lynch
Joel Lynch on 14 Jan 2023
Are you absolutely certain you need to regenerate A every iteration? A basic strategy in nonlinear systems is to transform the Ax=b problem to solve for perturbations in x, allowing you to recycle old A matrices (Jacobians) until they no longer converge. More importantly, you can also recycle decompositions of A as well, which is where the real time-savings are, especially if the changes in A are relatively small.

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Answers (2)

sjhstone on 25 Aug 2020
You can refer to the following blog post. It seems that you have reached the optimal efficiency.
If the sparsity pattern does not change, and there are nonzero entries on the last column and the last row, I think it might be redundant to pass arguments nnzA. Because MATLAB seems to use MEX routine, we do need to do numerical experiments to see whether not passing them into sparse can accelerate your code.

Christine Tobler
Christine Tobler on 16 Jan 2023
The fastest way to construct a sparse matrix will be when the inputs are sorted, first by columns and then by rows. You can verify that your inputs iA, jA match this by calling find on the resulting matrix A: The first two outputs of find(A) should match the order of indices in iA and jA.
I tried running sparse for just a random set of indices of the size you mentioned. As you can see, using sorted indices gives a bit of speed-up:
iA = randi(1e5, 1e5, 1);
jA = randi(1e5, 1e5, 1);
vA = randi(1e5, 1e5, 1);
tic; A = sparse(iA, jA, vA, 1e5, 1e5); toc
Elapsed time is 0.006209 seconds.
[siA, sjA, svA] = find(A);
tic; sA = sparse(siA, sjA, svA, 1e5, 1e5); toc
Elapsed time is 0.002676 seconds.
isequal(A, sA)
ans = logical
Both my calls here are faster than what you mentioned (0.006 seconds instead of 0.29 seconds), which might come down to the machine used.
Joel Lynch
Joel Lynch on 25 Jan 2023
A very minor comment, but sparse does sum co-located points in iA/jA, so the second snippet using find(A) may represent a smaller reduced dataset which skews the timing. This is not noticable for the sizes used here, but if you were to try this with a different set of I/J/V data, you may get inaccurate timing results.
Bruno Luong
Bruno Luong on 26 Jan 2023
Edited: Bruno Luong on 26 Jan 2023
@Joel Lynch Christine code returns sIA, sjA without duplication and svA accumulated of vA.
The purpose to is show that calling sparse with sorted indexes is faster.
This code might be better (fairer) since it is also accumulate with eventually duplication for both cases:
iA = randi(1e5, 1e5, 1);
jA = randi(1e5, 1e5, 1);
vA = randi(1e5, 1e5, 1);
tic; A = sparse(iA, jA, vA, 1e5, 1e5); toc
Elapsed time is 0.005231 seconds.
[~, is] = sortrows([jA iA]);
siA = iA(is);
sjA = jA(is);
svA = vA(is);
tic; sA = sparse(siA, sjA, svA, 1e5, 1e5); toc
Elapsed time is 0.002319 seconds.

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