或更简洁地

if set(L) & set(M):
    # there is an intersection
else:
    # no intersection

如果您确实需要True或False

bool(set(L) & set(M))

经过一些计时后，这似乎也是尝试的好选择

After running some timings, this seems to be a good option to try too

m_set=set(M)
any(x in m_set  for x in L)

如果M或L中的项目不可散列，则必须使用这种效率较低的方法

If the items in M or L are not hashable you have to use a less efficient approach like this

any(x in M for x in L)

以下是100个商品列表的一些时间安排.在没有交集的情况下，使用集合的速度要快得多，在有交集的情况下，使用集的速度会慢一些.

Here are some timings for 100 item lists. Using sets is considerably faster when there is no intersection, and a bit slower when there is a considerable intersection.

M=range(100)
L=range(100,200)

timeit set(L) & set(M)
10000 loops, best of 3: 32.3 µs per loop

timeit any(x in M for x in L)
1000 loops, best of 3: 374 µs per loop

timeit m_set=frozenset(M);any(x in m_set  for x in L)
10000 loops, best of 3: 31 µs per loop

L=range(50,150)

timeit set(L) & set(M)
10000 loops, best of 3: 18 µs per loop

timeit any(x in M for x in L)
100000 loops, best of 3: 4.88 µs per loop

timeit m_set=frozenset(M);any(x in m_set  for x in L)
100000 loops, best of 3: 9.39 µs per loop


# Now for some random lists
import random
L=[random.randrange(200000) for x in xrange(1000)]
M=[random.randrange(200000) for x in xrange(1000)]

timeit set(L) & set(M)
1000 loops, best of 3: 420 µs per loop

timeit any(x in M for x in L)
10 loops, best of 3: 21.2 ms per loop

timeit m_set=set(M);any(x in m_set  for x in L)
1000 loops, best of 3: 168 µs per loop

timeit m_set=frozenset(M);any(x in m_set  for x in L)
1000 loops, best of 3: 371 µs per loop