似乎应该有一种比以下更简单的方法：
import string
s = "string. With. Punctuation?" # Sample string
out = s.translate(string.maketrans("",""), string.punctuation)
有？
当前回答
这是我写的一个函数。它不是很有效，但很简单，您可以添加或删除任何您想要的标点符号：
def stripPunc(wordList):
"""Strips punctuation from list of words"""
puncList = [".",";",":","!","?","/","\\",",","#","@","$","&",")","(","\""]
for punc in puncList:
for word in wordList:
wordList=[word.replace(punc,'') for word in wordList]
return wordList
其他回答
这是我写的一个函数。它不是很有效，但很简单，您可以添加或删除任何您想要的标点符号：
def stripPunc(wordList):
"""Strips punctuation from list of words"""
puncList = [".",";",":","!","?","/","\\",",","#","@","$","&",")","(","\""]
for punc in puncList:
for word in wordList:
wordList=[word.replace(punc,'') for word in wordList]
return wordList
myString.translate(None, string.punctuation)
不一定更简单，但如果你更熟悉re家族的话，就另辟蹊径。
import re, string
s = "string. With. Punctuation?" # Sample string
out = re.sub('[%s]' % re.escape(string.punctuation), '', s)
这里有一个没有正则表达式的解决方案。
import string
input_text = "!where??and!!or$$then:)"
punctuation_replacer = string.maketrans(string.punctuation, ' '*len(string.punctuation))
print ' '.join(input_text.translate(punctuation_replacer).split()).strip()
Output>> where and or then
用空格替换标点用单个空格替换单词之间的多个空格删除尾随空格（如果有）条带（）
>>> s = "string. With. Punctuation?"
>>> s = re.sub(r'[^\w\s]','',s)
>>> re.split(r'\s*', s)
['string', 'With', 'Punctuation']
