如何使用不以白色 space 结尾但可能在捕获组之后有白色 space 的捕获组来子字符串？

Question

我正在尝试将看起来像 ( self, False ) 的字符串替换为 (self, False)。我使用的正则表达式：

s = re.compile('\(\s*(.*)\s*\)')
s.sub(r'()', '(    self, False   )')

哪个returns(self, False )

如何捕获括号内没有尾随空格的组？

Answer 1

为什么不使用字符串替换来去除空字符中的空格

str = '(    self, False   )'
print(str.replace(' ',''))
#(self,False)

Answer 2

试试这个：

#TEST 1
>>> import re
>>> str = '(    self, False   )'
>>> re.sub(r'(\()([\s]*?)((?:[\S]+?[\s]*?(?!\))+[\S]*?)|(?:[\S]+?(?=[\s]*?\))))([\s]*?)(\))', r'', str)
#OUTPUT
'(self, False)'

#TEST 2
>>> str = '''TEbh eyendd dkdkmfkf(    self, False   ) dduddnudmd (    self, False   )
(    self, False   ) fififfj m(    self, False   )kmiff ikifkifko kfmimfimfifi k
fkmfikfk kfmifm (    self, False   ) fififi,fo'''

>>> print(re.sub(r'(\()([\s]*?)((?:[\S]+?[\s]*?(?!\))+[\S]*?)|(?:[\S]+?(?=[\s]*?\))))([\s]*?)(\))', r'', str))
#OUTPUT
'TEbh eyendd dkdkmfkf(self, False) dduddnudmd (self, False)
(self, False) fififfj m(self, False)kmiff ikifkifko kfmimfimfifi k
fkmfikfk kfmifm (self, False) fififi,fo'

#TEST 3
>>> '''TEbh eyendd dkdkmfkf(    self) dduddnudmd (    self)
(    self, False   ) fififfj m(    self, False)kmiff ikifkifko kfmimfimfifi k
fkmfikfk kfmifm (    self, False   ) fififi,fo
(self   ) dndnd (self   ) fufufjiri (    self   ) (self   ) (    self)(    self)(self   )(    self   )(self   )(    self   )'''

>>>  print(re.sub(r'(\()([\s]*?)((?:[\S]+?[\s]*?(?!\))+[\S]*?)|(?:[\S]+?(?=[\s]*?\))))([\s]*?)(\))', r'', str))

#OUTPUT
TEbh eyendd dkdkmfkf(self) dduddnudmd (self)
(self, False) fififfj m(self, False)kmiff ikifkifko kfmimfimfifi k
fkmfikfk kfmifm (self, False) fififi,fo
(self) dndnd (self) fufufjiri (self) (self) (self)(self)(self)(self)(self)(self)

利用您的简单解决方案：

>>> '''TEbh eyendd dkdkmfkf(    self) dduddnudmd (    self)
(    self, False   ) fififfj m(    self, False)kmiff ikifkifko kfmimfimfifi k
fkmfikfk kfmifm (    self, False   ) fififi,fo
(self   ) dndnd (self   ) fufufjiri (    self   ) (self   ) (    self)(    self)(self   )(    self   )(self   )(    self   )'''

>>> print(re.sub(r'(\()\s*([\S\s]*?)\s*(\))', r'', str))
#OUTPUT
TEbh eyendd dkdkmfkf(self) dduddnudmd (self)
(self, False) fififfj m(self, False)kmiff ikifkifko kfmimfimfifi k
fkmfikfk kfmifm (self, False) fififi,fo
(self) dndnd (self) fufufjiri (self) (self) (self)(self)(self)(self)(self)(self)

Answer 3

找到一个简单的解决方案。

s = re.compile('\(\s*(.*?)\s*\)')
s.sub(r'()', 'hi hello ble ble ( self, False   ) ( self      ) (self , greedy    ) (    hello)')
#Output
'hi hello ble ble (self, False) (self) (self , greedy) (hello)'

根据 python 重新文档：

The '', '+', and '?' qualifiers are all greedy; they match as much text as possible. Sometimes this behaviour isn’t desired; if the RE <.> is matched against ' b ', it will match the entire string, and not just ''. Adding ? after the qualifier makes it perform the match in non-greedy or minimal fashion; as few characters as possible will be matched. Using the RE <.*?> will match only ''.

如何使用不以白色 space 结尾但可能在捕获组之后有白色 space 的捕获组来子字符串？

How to sub a string with a capturing group that does not end with white space but might have white space after the capturing group?

python

regex

python-3.x

python-re