如何在python中使用字符串保持标点符号？

I want create a catalog of all logs, so I just want keep all punctuation and remove all other chars which include CJK and others.

我想创建所有日志的目录，所以我只想保留所有标点符号并删除所有其他包含CJK和其他字符的字符。

for example:

例如：

s = "aaa; sf = fa = bla http://wa"

expected output is

预期产量是

;==://

2 个解决方案

#1

You can use str.translate:

你可以使用str.translate：

>>> from string import letters, digits, whitespace, punctuation
>>> s = "aaa; sf = fa = bla http://wa"
>>> s.translate(None, letters+digits+whitespace)
';==://'

or regex:

或正则表达式：

>>> re.sub(r'[^{}]+'.format(punctuation),'',s)
';==://'

Timing comparisons:

时间比较：

>>> s = "aaa; sf = fa = bla http://wa"*1000
>>> %timeit s.translate(None,letters+digits+whitespace)
10000 loops, best of 3: 171 us per loop                  #winner
>>> r1 = re.compile(r'[^{}]+'.format(punctuation))
>>> r2 = re.compile(r'[\w\s]+')
>>> %timeit r1.sub('',s)
100 loops, best of 3: 2.64 ms per loop
>>> %timeit r2.sub('',s)
100 loops, best of 3: 3.31 ms per loop

#2

Using a regular expression:

使用正则表达式：

>>> re.sub(r'[\w\s]+', '', "aaa; sf = fa = bla http://wa")
';==://'

Compiling can buy some performance, even for such a simple pattern...

编译可以购买一些性能，即使对于这样一个简单的模式......

>>> %timeit re.sub(r'[\w\s]+', '', "aaa; sf = fa = bla http://wa")
100000 loops, best of 3: 6.78 us per loop

>>> e = re.compile(r'[\w\s]+')
>>> %timeit e.sub('', "aaa; sf = fa = bla http://wa")
100000 loops, best of 3: 4.91 us per loop

...but regular expression is no match for Ashwinis' solution using str.translate:

...但正则表达式与使用str.translate的Ashwinis解决方案不匹配：

>>> %timeit "aaa; sf = fa = bla http://wa".translate(None,letters+digits+whitespace)
1000000 loops, best of 3: 1.31 us per loop

#1