Code optimization by way of selective computations

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Howard Wilton
Howard Wilton 2022 年 11 月 22 日
コメント済み: Howard Wilton 2022 年 11 月 22 日
I have the following code:
clc; clearvars;
Ts = 1e-1; t = 0:Ts:1-Ts
flag = mod(1:length(t),2)
s_t = exp(1i*2*pi*t)
I am looking to only execute the computation only when the corresponding value for is logical 1. I wish to do this without iterating as the final vector to process is long and is about 97% zero values in the result.

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Jiri Hajek
Jiri Hajek 2022 年 11 月 22 日
Hi, you need to convert your flags to logical array, initialize the results with zeros and then apply the function to the flagged positions like this:
flag = logical(mod(1:length(t),2));
s_t = zeros(size(t));
s_t(flag) = exp(1i*2*pi*t(flag));
  3 件のコメント
Jiri Hajek
Jiri Hajek 2022 年 11 月 22 日
Yes, the logical indexing probably does add to the execution time, as basic functions are quite optimized... But you mentioned that your non-zeros should be only about 3% of the array size, whereas in this test you have 50% nonzeros... So the test dis not really fair, I think.
Perhaps you can also look at sparse arrays, but truth is, their usefullness is also limited by the fraction of zero elements.
Howard Wilton
Howard Wilton 2022 年 11 月 22 日
Yeah. That is a good point. I did it again with just 1% of the flags set, but the overall processing was comparable to the unconditioned version. I shall have a look at sparse arrays. Thanks.

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その他の回答 (1 件)

Torsten
Torsten 2022 年 11 月 22 日
編集済み: Torsten 2022 年 11 月 22 日
Ts = 1e-7; t = 0:Ts:1-Ts;
tic
s_t = exp(1i*2*pi*t(1:2:end));
toc
Elapsed time is 0.111464 seconds.
or if your condition is more complicated:
tic
s_t = exp(1i*2*pi*t(mod(1:length(t),2)==1));
toc
Elapsed time is 0.431624 seconds.

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