speed up the & operation

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Chaoyang Jiang
Chaoyang Jiang 2018 年 4 月 12 日
コメント済み: Walter Roberson 2018 年 4 月 13 日
How to speed up the following code? I did profiling, and this line takes 3267.72s for calling 245913489 times.
a=[1 0 1 1 0 1 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 0 1 1 1 0 1 1 0 1 1 1];
b=[1 0 1 1 0 1 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 0 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 0 1 1 1 0];
for i=1:1000000
aat=((a(1,1:24)==0)&(b(1,25:48)==1)&(~ismembc(1:24,5:50)))';
end
  5 件のコメント
Walter Roberson
Walter Roberson 2018 年 4 月 13 日
for i=1:1000000
part1 = a(1,1:24)==0;
part2 = b(1,25:48)==1;
part3 = ~ismembc(1:24,5:50);
part4 = part1 & part2 & part3;
aat = part4 .';
end
Now when you profile you can see exactly which part of the expression is taking the most time.
We suspect that it is the ismembc() that is taking all of the time and that the & is relatively fast.

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