import re
s = "string. With. Punctuation?" # Sample string 
out = re.sub(r'[^a-zA-Z0-9\s]', '', s)

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)

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'

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')

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