如果列表中包含“黑名单”中的子字符串，则从列表中删除该项目

您可以将黑名单加入一个表达式中：

import re

blacklist = re.compile('|'.join([re.escape(word) for word in B]))

然后过滤出匹配的单词：

C = [word for word in A if not blacklist.search(word)]

模式中的单词会被转义（这样，.其他元字符就不会被这样处理，而是视为文字字符），并加入一系列|替代项中：

>>> '|'.join([re.escape(word) for word in B])
'XXX|BBB'

演示：

>>> import re
>>> A = [ 'cat', 'doXXXg', 'monkey', 'hoBBBrse', 'fish', 'snake']
>>> B = ['XXX', 'BBB']
>>> blacklist = re.compile('|'.join([re.escape(word) for word in B]))
>>> [word for word in A if not blacklist.search(word)]
['cat', 'monkey', 'fish', 'snake']

这应该胜过任何显式的成员资格测试，尤其是随着黑名单中单词数量的增加：

>>> import string, random, timeit
>>> def regex_filter(words, blacklist):
...     [word for word in A if not blacklist.search(word)]
... 
>>> def any_filter(words, blacklist):
...     [word for word in A if not any(bad in word for bad in B)]
... 
>>> words = [''.join([random.choice(string.letters) for _ in range(random.randint(3, 20))])
...          for _ in range(1000)]
>>> blacklist = [''.join([random.choice(string.letters) for _ in range(random.randint(2, 5))])
...              for _ in range(10)]
>>> timeit.timeit('any_filter(words, blacklist)', 'from __main__ import any_filter, words, blacklist', number=100000)
0.36232495307922363
>>> timeit.timeit('regex_filter(words, blacklist)', "from __main__ import re, regex_filter, words, blacklist; blacklist = re.compile('|'.join([re.escape(word) for word in blacklist]))", number=100000)
0.2499098777770996

上面的测试针对1000个随机单词（长度为3-20个字符）列表中的10个随机列入黑名单的短单词（2-5个字符），正则表达式快50％。