Replacing a LOT of data by zero (in a matrix)
古いコメントを表示
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?
採用された回答
その他の回答 (1 件)
Dmitry Borovoy
2011 年 8 月 31 日
0 投票
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.
カテゴリ
ヘルプ センター および File Exchange で Matrix Indexing についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!