python - 如何找出一个列表中的元素是否在另一个列表中？

Question

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Answer 1

您可以使用 set.issubset 和列表理解，使用 A[:] 将更改原始/列表对象 A:

A = [[2, 5, 13, 14], [4, 5, 10, 12], [2, 9, 10, 11], [2, 5, 12, 13], [4, 5, 6, 12]]
B = [12, 5]
st = set(B)

A [:] = [sub for sub in A if st.issubset(sub)]

print(A)
[[4, 5, 10, 12], [2, 5, 12, 13], [4, 5, 6, 12]]

对于 B = [13] 也是如此

A = [[2, 5, 13, 14], [4, 5, 10, 12], [2, 9, 10, 11], [2, 5, 12, 13], [4, 5, 6, 12]]
B = [13]
st = set(B)

A [:] = [sub for sub in A if st.issubset(sub)]

print(A)
[[2, 5, 13, 14], [2, 5, 12, 13]]

set objects

s.issubset(t) s <= t test whether every element in s is in t

对于非常大的 A 或者如果您有内存限制，您可以使用生成器表达式:

A [:] = (sub for sub in A if st.issubset(sub))

如果顺序无关紧要并且可以设置，我建议您从一开始就使用它们。在集合上进行查找会更有效率。

稍微大一点的 A 上的一些时间:

In [23]: A = [[2, 5, 13, 14], [4, 5, 10, 12], [2, 9, 10, 11], [2, 5, 12, 13], [4, 5, 6, 12],[2, 5, 13, 14], [4, 5, 10, 12], [2, 9, 10, 11], [2, 5, 12, 13], [4, 5, 6, 12],[2, 5, 13, 14], [4, 5, 10, 12], [2, 9, 10, 11], [2, 5, 12, 13], [4, 5, 6, 12]]

In [24]: B = [12, 5]                                 
In [25]: timeit  filter(lambda x: all(y in x for y in B), A)
100000 loops, best of 3: 9.45 µs per loop

In [26]: %%timeit                                    
st = set(B)
[sub for sub in A if st.issubset(sub)]
   ....: 
100000 loops, best of 3: 3.88 µs per loop
 map(lambda x: not B_set-set(x), A)
In [27]: %%timeit
....: B_set = set(B)
....: map(lambda x: not B_set-set(x), A)
....: 
100000 loops, best of 3: 6.95 µs per loop

如果您已经将元素作为集合存储在 A 中:

In [33]: %%timeit                             
st = set(B)
[sub for sub in A if sub >= st]
....: 
1000000 loops, best of 3: 1.12 µs per loop