我喜欢使用这样的函数：

def scrub(abc):
    while abc[-1] is in list(string.punctuation):
        abc=abc[:-1]
    while abc[0] is in list(string.punctuation):
        abc=abc[1:]
    return abc

在不太严格的情况下，单行线可能会有所帮助：

''.join([c for c in s if c.isalnum() or c.isspace()])

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'

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'

为什么你们都不用这个？

 ''.join(filter(str.isalnum, s)) 

太慢了？

我在寻找一个非常简单的解决方案。这是我得到的：

import re 

s = "string. With. Punctuation?" 
s = re.sub(r'[\W\s]', ' ', s)

print(s)
'string  With  Punctuation '