这里有一个使用RegEx的简单方法

import re

punct = re.compile(r'(\w+)')

sentence = 'This ! is : a # sample $ sentence.' # Text with punctuation
tokenized = [m.group() for m in punct.finditer(sentence)]
sentence = ' '.join(tokenized)
print(sentence) 
'This is a sample sentence'

myString.translate(None, string.punctuation)

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

对于严肃的自然语言处理（NLP），您应该让像SpaCy这样的库通过标记化处理标点符号，然后您可以根据需要手动调整。

例如，您希望如何处理单词中的连字符？例外情况，如缩写？开始和结束引号？URL？在NLP中，将“let’s”这样的收缩分隔为“let”和“s”以进行进一步处理通常很有用。

从效率的角度来看，你不会击败

s.translate(None, string.punctuation)

对于更高版本的Python，请使用以下代码：

s.translate(str.maketrans('', '', string.punctuation))

它使用查找表在C语言中执行原始字符串操作——除了编写自己的C代码之外，没有什么能比这更好的了。

如果速度不令人担忧，另一个选择是：

exclude = set(string.punctuation)
s = ''.join(ch for ch in s if ch not in exclude)

这比用每个字符替换s.replace更快，但不会像正则表达式或字符串转换等非纯python方法那样执行得好，正如您从下面的计时中看到的那样。对于这种类型的问题，在尽可能低的水平上解决是有回报的。

计时代码：

import re, string, timeit

s = "string. With. Punctuation"
exclude = set(string.punctuation)
table = string.maketrans("","")
regex = re.compile('[%s]' % re.escape(string.punctuation))

def test_set(s):
    return ''.join(ch for ch in s if ch not in exclude)

def test_re(s):  # From Vinko's solution, with fix.
    return regex.sub('', s)

def test_trans(s):
    return s.translate(table, string.punctuation)

def test_repl(s):  # From S.Lott's solution
    for c in string.punctuation:
        s=s.replace(c,"")
    return s

print "sets      :",timeit.Timer('f(s)', 'from __main__ import s,test_set as f').timeit(1000000)
print "regex     :",timeit.Timer('f(s)', 'from __main__ import s,test_re as f').timeit(1000000)
print "translate :",timeit.Timer('f(s)', 'from __main__ import s,test_trans as f').timeit(1000000)
print "replace   :",timeit.Timer('f(s)', 'from __main__ import s,test_repl as f').timeit(1000000)

结果如下：

sets      : 19.8566138744
regex     : 6.86155414581
translate : 2.12455511093
replace   : 28.4436721802

我通常用这样的词：

>>> s = "string. With. Punctuation?" # Sample string
>>> import string
>>> for c in string.punctuation:
...     s= s.replace(c,"")
...
>>> s
'string With Punctuation'