python - 如何使用不以空格结尾但在捕获组之后可能有空格的捕获组子字符串？

Question

我正在尝试将看起来像这样的字符串替换( self, False )为(self, False). 我正在使用的正则表达式：

s = re.compile('\(\s*(.*)\s*\)')
s.sub(r'(\1)', '(    self, False   )')

哪个返回(self, False )

如何在没有尾随空格的情况下捕获括号内的组？

Answer 1

为什么不使用字符串替换来消除带有空字符的空格

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

Answer 2

尝试这个：

#TEST 1
>>> import re
>>> str = '(    self, False   )'
>>> re.sub(r'(\()([\s]*?)((?:[\S]+?[\s]*?(?!\))+[\S]*?)|(?:[\S]+?(?=[\s]*?\))))([\s]*?)(\))', r'\1\3\5', 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'\1\3\5', 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'\1\3\5', 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)

捎带您的简单解决方案：

>>> '''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'\1\2\3', 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

找到了一个简单的解决方案。

s = re.compile('\(\s*(.*?)\s*\)')
s.sub(r'(\1)', 'hi hello ble ble ( self, False   ) ( self      ) (self , greedy    ) (    hello)')
#Output
'hi hello ble ble (self, False) (self) (self , greedy) (hello)'

根据 python re 文档：

' '、'+' 和 '?' 限定符都是贪婪的；它们匹配尽可能多的文本。有时这种行为是不希望的；如果 RE <. > 匹配'b'，它将匹配整个字符串，而不仅仅是''。添加 ? 在预选赛使其以非贪婪或最低限度的方式执行比赛之后；将匹配尽可能少的字符。使用 RE <.*?> 将仅匹配 ''。