Fastest way to compute J' * J, where J is sparse

조회 수: 4 (최근 30일)
Oliver Woodford
Oliver Woodford . 2014년 11월 3일
댓글: Oliver Woodford . 2015년 11월 20일
I have a sparse rectangular matrix, J, for which I want to compute:
>> H = J' * J;
It's a bit slow (transpose is taking 5s and the matrix multiplication 9s), and given this is a special and very common case of a transpose and multiply, I was wondering if MATLAB had a faster way, e.g. one which avoids an explicit transpose.
  댓글 수: 15
Oliver Woodford
Oliver Woodford 2015년 11월 20일
For my size of matrix, qr(J,0) is vastly slower (i.e. several orders of magnitude; actually it crashed MATLAB after a minute or so) than a mex file that computes J'*J (a second or so). The mex file is a bit slower than standard MATLAB for smaller sparse matrices, but at least works for very tall matrices (which MATLAB produces an "Matrix too large" error on when transposing).

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답변 (2개)

Azzi Abdelmalek
Azzi Abdelmalek 2014년 11월 3일
편집: Azzi Abdelmalek 님. 2014년 11월 3일
  댓글 수: 2
John D'Errico
John D'Errico 2014년 11월 3일
편집: John D'Errico 님. 2014년 11월 3일
Um, J is already assumed to be in sparse form, and one would definitely not want to compute a full result when working with sparse matrices. Finally, you put the transpose on the wrong term, computing J*J', not J'*J.

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Matt J
Matt J 2014년 11월 4일
편집: Matt J 님. 2014년 11월 5일
I don't think there's anything available to accelerate an exact calculation of J'*J for general J. However, if you know in advance that J'*J happens to be banded to diagonals -k:k for small k (or if it can be approximated as such), then it might help to compute the 2*k+1 non-trivial diagonals individually. You can do so without transposition as below.
for i=1:k
Whether this is actually faster will probably depend on the specifics of J. If nothing else, it spares you the large memory consumption of holding wide sparse matrices such as J' in RAM
>> J=sparse(m,n); Jt=J'; whos J Jt
Name Size Bytes Class Attributes
J 3192027x3225 25824 double sparse
Jt 3225x3192027 25536240 double sparse
Replacing J'*J by a banded approximation is something I haven't tried myself with Gauss-Newton specifically, but the role of J'*J is already as an approximation there, so I think it could work. Other minimization algorithms tend to be robust to small errors in the derivatives.


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