列表理解vs filter（）vs set difference:过滤集合时哪一种效率最高？

Question

考虑到我已：

• 1个整数数组，转换为一个集合（名为neighbors）

• 要避免的3组其他整数（命名为forbidden1、forbidden2和forbidden3）

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neighbors = {6, 12, 9}
forbidden1 = {1, 4, 7, 8}
forbidden2 = {2, 5, 0, 3}
forbidden3 = {6, 9}

_uuu

以下哪种解决方案最能有效地从第一组中过滤出禁止值？为什么

A/使用if语句和逻辑运算符and进行列表理解

[x for x in neighbors if x not in forbidden1 and x not in forbidden2 and x not in forbidden3]

B/使用if语句和联合运算符|进行列表理解

[x for x in neighbors if x not in forbidden1 | forbidden2 | forbidden3]

C/filter（）和逻辑运算符and

filter(lambda x: x not in forbidden1 and x not in forbidden2 and x not in forbidden3, neighbors)

D/filter（）和联合运算符|

filter(lambda x: x not in forbidden1 | forbidden2 | forbidden3, neighbors)

E/使用并集运算符设置差|

neighbors.difference(forbidden1 | forbidden2 | forbidden3)

Answer 1

已试用（timeit python）：

A) 294 ns ± 28.1 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
B) 297 ns ± 26.5 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
C) 300 ns ± 31.2 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
D) 294 ns ± 19.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
E) 291 ns ± 21.2 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

都差不多