Question

我有两个清单：

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

我试图找出A中哪些列表包含B中的元素（顺序无关紧要）并删除其余列表。

在这种情况下答案是：

[[4, 5, 10, 12], [2, 5, 12, 13], [4, 5, 6, 12]]

如果我们更改B并将其设为B = [13]，答案就是：

[[2, 5, 13, 14], [2, 5, 12, 13]]

Answer 1

您可以set.issubset使用列表理解，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]]

B = [13]

相同

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＆lt; = t测试s中的每个元素是否都在t

对于非常大的A或者如果你有内存限制，你可以使用生成器表达式：

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

如果订单永远不重要且可以设置我建议你从一开始就使用它们。在集合上执行查找会更有效率。

稍微大一点的时间A：

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

如果您已将元素存储为A：

中的集合

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

Answer 2

您可以在此filter结合使用all：

print filter(lambda x: all(y in x for y in B), A)

更高效的答案：

B_set = set(B)
print map(lambda x: not B_set-set(x), A)

如何确定一个列表中的元素是否在另一个列表中？

2 个答案: