您也可以这样做：

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

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（）

with open('one.txt','r')as myFile:

    str1=myFile.read()

    print(str1)


    punctuation = ['(', ')', '?', ':', ';', ',', '.', '!', '/', '"', "'"] 

for i in punctuation:

        str1 = str1.replace(i," ") 
        myList=[]
        myList.extend(str1.split(" "))
print (str1) 
for i in myList:

    print(i,end='\n')
    print ("____________")

正则表达式很简单，如果你知道的话。

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

我还没有看到这个答案。只需使用正则表达式；它删除了除单词字符（\w）和数字字符（\d）之外的所有字符，后跟一个空白字符（\s）：

import re
s = "string. With. Punctuation?" # Sample string 
out = re.sub(ur'[^\w\d\s]+', '', s)

作为更新，我重写了Python 3中的@Brian示例，并对其进行了更改，以将正则表达式编译步骤移到函数内部。我在这里的想法是对使功能工作所需的每一步进行计时。也许您使用的是分布式计算，无法在工作人员之间共享regex对象，需要在每个工作人员处执行re.compile步骤。此外，我还很好奇地对Python 3的maketrans的两种不同实现进行计时

table = str.maketrans({key: None for key in string.punctuation})

vs

table = str.maketrans('', '', string.punctuation)

另外，我添加了另一种使用集合的方法，在这里我利用交集函数来减少迭代次数。

这是完整的代码：

import re, string, timeit

s = "string. With. Punctuation"


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


def test_set2(s):
    _punctuation = set(string.punctuation)
    for punct in set(s).intersection(_punctuation):
        s = s.replace(punct, ' ')
    return ' '.join(s.split())


def test_re(s):  # From Vinko's solution, with fix.
    regex = re.compile('[%s]' % re.escape(string.punctuation))
    return regex.sub('', s)


def test_trans(s):
    table = str.maketrans({key: None for key in string.punctuation})
    return s.translate(table)


def test_trans2(s):
    table = str.maketrans('', '', string.punctuation)
    return(s.translate(table))


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("sets2      :",timeit.Timer('f(s)', 'from __main__ import s,test_set2 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("translate2 :",timeit.Timer('f(s)', 'from __main__ import s,test_trans2 as f').timeit(1000000))
print("replace   :",timeit.Timer('f(s)', 'from __main__ import s,test_repl as f').timeit(1000000))

这是我的结果：

sets      : 3.1830138750374317
sets2      : 2.189873124472797
regex     : 7.142953420989215
translate : 4.243278483860195
translate2 : 2.427158243022859
replace   : 4.579746678471565