>>> '|'.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']
这应该优于任何明确的成员资格测试,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
下面的列表理解将起作用
^{pr2}$您可以将黑名单合并为一个表达式:
然后过滤掉匹配的单词:
^{pr2}$模式中的单词被转义(因此
.
和其他元字符不被视为原语字符,而是作为字面字符处理),并被连接到一系列|
备选方案中:演示:
这应该优于任何明确的成员资格测试,尤其是随着黑名单中单词数量的增加:
上面测试了10个随机黑名单的短单词(2-5个字符)和1000个随机单词(3-20个字符长)的对比,regex大约快50%。在
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