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
>>> s = "string. With. Punctuation?"
>>> s = re.sub(r'[^\w\s]','',s)
>>> re.split(r'\s*', s)
['string', 'With', 'Punctuation']
其他回答
字符串标点符号仅为ASCII！一种更正确（但也慢得多）的方法是使用unicodedata模块：
# -*- coding: utf-8 -*-
from unicodedata import category
s = u'String — with - «punctation »...'
s = ''.join(ch for ch in s if category(ch)[0] != 'P')
print 'stripped', s
您还可以概括和剥离其他类型的字符：
''.join(ch for ch in s if category(ch)[0] not in 'SP')
它还将删除~*+§$等字符，这些字符可能是“标点符号”，也可能不是“标点符号。
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
为什么你们都不用这个？
''.join(filter(str.isalnum, s))
太慢了？
使用Python从文本文件中删除停止词
print('====THIS IS HOW TO REMOVE STOP WORS====')
with open('one.txt','r')as myFile:
str1=myFile.read()
stop_words ="not", "is", "it", "By","between","This","By","A","when","And","up","Then","was","by","It","If","can","an","he","This","or","And","a","i","it","am","at","on","in","of","to","is","so","too","my","the","and","but","are","very","here","even","from","them","then","than","this","that","though","be","But","these"
myList=[]
myList.extend(str1.split(" "))
for i in myList:
if i not in stop_words:
print ("____________")
print(i,end='\n')
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（）
