从字符串中去除标点的最佳方法

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

以下是Python 3.5的一行代码：

import string
"l*ots! o(f. p@u)n[c}t]u[a'ti\"on#$^?/".translate(str.maketrans({a:None for a in string.punctuation}))

这是我写的一个函数。虽然它不太高效，但它很简单，你可以添加或删除你想要的任何标点符号：

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

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

更新一下，我把 @Brian 的示例用 Python 3 重写了，并对其进行了更改，将正则表达式编译步骤移动到函数内部。我的想法是计时使函数正常工作所需的每个步骤。也许您正在使用分布式计算，不能在工作进程之间共享正则表达式对象，需要在每个工作器中进行 re.compile 步骤。此外，我还好奇地计时了 Python 3 中 maketrans 的两个不同实现。

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

vs

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

我增加了一种使用set的方法，利用交集函数减少迭代次数。

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))

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

在不是非常严格的情况下，一条简短的命令可能会有所帮助：

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

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


['string', 'With', 'Punctuation']

我正在寻找一个非常简单的解决方案。以下是我得到的：

import re 

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

print(s)
'string  With  Punctuation '

这里有一个不需要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

• 用空格替换标点符号
• 用单个空格替换单词之间的多个空格
• 使用strip()函数删除末尾的空格（如果有）

Why none of you use this?

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

速度太慢了吗？