I have a 900 x 900 sparse matrix 'S' which contains alot of zeros, when i perform the following operation
S./sum(S,2)
It takes too long because it is operating on the zeros too and resulting in alot of NaN values. Is there a more efficient way to do this

回答 (1 件)

Stephan
Stephan 2019 年 3 月 5 日

0 投票

Hi,
if you precalculate the sum and then calculate you can save some time:
N = sum(S,2);
S./N;
I tried using sparse matrices of 900x900 with about 10k non zero elements, which was more time consuming then this operation.
Best regards
Stephan

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2019 年 3 月 5 日

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2019 年 3 月 5 日

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