Replacing a LOT of data by zero (in a matrix)

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Herwig Peters
Herwig Peters 2011-8-31
I have a matrix M1, below you see the whos information for that matrix:
Name Size Bytes Class Attributes M1 29624x29624 108387640 double sparse
I want to replace quite a bit of data by zeros, using logical indexing: M1(abs(M1)<1e-1) = 0;
This needs more than 16GB memory, there must be a more efficient way. I also tried the other way round, that is M1 = M1(abs(M1)>1e-1);
Same problem. What's the trick?

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采纳的回答

Herwig Peters
Herwig Peters 2011-8-31
4 way.
Find the indices of the desired matrix elements and the desired matrix elements themselves. With these 3 vectors, rebuild the sparse Matrix M1.
[i1,~,s1] = find(M1(abs(M1)>1e-1));
[i1,j1] = ind2sub(size(M1),i1);
M1 = sparse(i1,j1,s1,m,n);

更多回答（1 个）

Dmitry Borovoy
Dmitry Borovoy 2011-8-31
1 way. Use vectorization (your way). fast but requires a lot of RAM 2 way. Use loops. Very slow but no need a lot of RAM 3 way. For perverts. Write your code on C/C++/Fortran with loop and call it from Matlab. If you have restriction on RAM or on calculation speed, I suppose, it's your way.
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Herwig Peters
Herwig Peters 2011-8-31
Yes, that sums it up pretty well. 1 needs too much RAM, 2 needs too much time (CPU) and 3 needs too much time as well, my time.
See my own answer for another possibility.

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