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

我在寻找一个非常简单的解决方案。这是我得到的:

import re 

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

print(s)
'string  With  Punctuation '

其他回答

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

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'

在处理Unicode字符串时,我建议使用PyPi正则表达式模块,因为它同时支持Unicode属性类(如\p{X}/\p{X})和POSIX字符类(如[:name:])。

只需在终端中键入pipinstallregex(或pip3installregex)并按回车键即可安装软件包。

如果您需要删除任何类型的标点符号(即除字母、数字和空格之外的任何其他符号),您可以使用

regex.sub(r'[\p{P}\p{S}]', '', text)  # to remove one by one
regex.sub(r'[\p{P}\p{S}]+', '', text) # to remove all consecutive punctuation/symbols with one go
regex.sub(r'[[:punct:]]+', '', text)  # Same with a POSIX character class

在线观看Python演示:

import regex

text = 'भारत India <><>^$.,,! 002'
new_text = regex.sub(r'[\p{P}\p{S}\s]+', ' ', text).lower().strip()
# OR
# new_text = regex.sub(r'[[:punct:]\s]+', ' ', text).lower().strip()

print(new_text)
# => भारत india 002

在这里,我向字符类添加了空白模式

myString.translate(None, string.punctuation)