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

在不太严格的情况下,单行线可能会有所帮助:

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

其他回答

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()

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

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])

为了方便使用,我总结了Python 2和Python 3中从字符串中删除标点符号的注意事项。有关详细说明,请参阅其他答案。


Python 2

import string

s = "string. With. Punctuation?"
table = string.maketrans("","")
new_s = s.translate(table, string.punctuation)      # Output: string without punctuation

Python 3

import string

s = "string. With. Punctuation?"
table = str.maketrans(dict.fromkeys(string.punctuation))  # OR {key: None for key in string.punctuation}
new_s = s.translate(table)                          # Output: string without punctuation

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)