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'

这可能不是最好的解决方案，但我就是这样做的。

import string
f = lambda x: ''.join([i for i in x if i not in 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

您也可以这样做：

import string
' '.join(word.strip(string.punctuation) for word in 'text'.split())

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。

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'