您可以尝试通过会员资格检查来理解列表
>>> lestring = "Text123">>> lelist = ["Text", "foo", "bar"]>>> [e for e in lelist if e in lestring]['Text']
与您的实现相比，尽管LC有一个隐式循环，但它更快，因为没有显式函数调用（如您的情况）
count
与Joe的实现相比，您的实现更快，因为filter函数需要循环调用两个函数，
lambda并且
count
>>> def joe(lelist, lestring): return ''.join(random.sample(x + 'b'*len(x), len(x)))>>> def uz(lelist, lestring): for x in lelist: if lestring.count(x): return 'Yep. "%s" contains characters from "%s" item.' % (lestring, x)>>> def ab(lelist, lestring): return [e for e in lelist if e in lestring]>>> t_ab = timeit.Timer("ab(lelist, lestring)", setup="from __main__ import lelist, lestring, ab")>>> t_uz = timeit.Timer("uz(lelist, lestring)", setup="from __main__ import lelist, lestring, uz")>>> t_joe = timeit.Timer("joe(lelist, lestring)", setup="from __main__ import lelist, lestring, joe")>>> t_ab.timeit(100000)0.09391469893125759>>> t_uz.timeit(100000)0.1528471407273173>>> t_joe.timeit(100000)1.4272649857800843对于较短的字符串，Jamie的评论解决方案比较慢。这是测试结果
>>> def jamie(lelist, lestring): return next(itertools.chain((e for e in lelist if e in lestring), (None,))) is not None>>> t_jamie = timeit.Timer("jamie(lelist, lestring)", setup="from __main__ import lelist, lestring, jamie")>>> t_jamie.timeit(100000)0.22237164127909637如果需要布尔值，对于较短的字符串，只需修改上面的LC表达式
[e in lestring for e in lelist if e in lestring]
或者对于更长的字符串，您可以执行以下操作
>>> next(e in lestring for e in lelist if e in lestring)True
要么
>>> any(e in lestring for e in lelist)
