Code optimization by way of selective computations

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Howard Wilton
Howard Wilton am 22 Nov. 2022
Kommentiert: Howard Wilton am 22 Nov. 2022
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 am 22 Nov. 2022
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 Kommentare
Jiri Hajek
Jiri Hajek am 22 Nov. 2022
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 am 22 Nov. 2022
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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Torsten
Torsten am 22 Nov. 2022
Bearbeitet: Torsten am 22 Nov. 2022
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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