Java及python正则表达式详解

>>> import re
>>> m = re.match(r'(\w+) (\w+)','Isaac Newton,physicist')
>>> m.group(0)
'Isaac Newton'
>>> m.group(1)
'Isaac'
>>> m.group(2)
'Newton'
>>> m.group(1,2)
('Isaac', 'Newton')
>>>m=re.match(r'(?P<first_name>\w+)(?P<last_name>\w+)','Malcolm Reynolds')
>>> m.group('first_name')
'Malcolm'
>>> m.group('last_name')
'Reynolds'
>>> m.groups()
('Malcolm', 'Reynolds')
>>> m.groupdict()
{'first_name': 'Malcolm', 'last_name': 'Reynolds'}

>>> import re
>>> exampleString = '''There should be one--and preferably only one--obvious way to do it.
Although that way may not be obvious at first unless you're Dutch.
Now is better than never. 
Although never never is often better than right now.'''
>>> pattern = re.compile(r'(?<=\w\s)never(?=\s\w)')
>>> matchResult = pattern.search(exampleString)
>>> matchResult.span()
(171, 176)
>>> pattern = re.compile(r'(?<=\w\s)never')
>>> matchResult = pattern.search(exampleString)
>>> matchResult.span()
(154, 159)
>>> pattern = re.compile(r'(?:is\s)better(\sthan)')
>>> matchResult = pattern.search(exampleString)
>>> matchResult.span()
(139, 153)
>>> matchResult.group(0)
'is better than'
>>> matchResult.group(1)
' than'
>>> pattern = re.compile(r'\b(?i)n\w+\b')
>>> index = 0
>>> while True:
 matchResult = pattern.search(exampleString,index)
 if not matchResult:
 break
 print(matchResult.group(0),':',matchResult.span(0))
 index = matchResult.end(0)

not : (90, 93)
Now : (135, 138)
never : (154, 159)
never : (171, 176)
never : (177, 182)
now : (210, 213)
>>> pattern = re.compile(r'(?<!not\s)be\b')
>>> index = 0
>>> while True:
 matchResult = pattern.search(exampleString,index)
 if not matchResult:
 break
 print(matchResult.group(0),':',matchResult.span(0))
 index = matchResult.end(0)

be : (13, 15)
>>> exampleString[13:20]
'be one-'
>>> pattern = re.compile(r'(\b\w*(?P<f>\w+)(?P=f)\w*\b)')
>>> index = 0
>>> while True:
 matchResult = pattern.search(exampleString,index)
 if not matchResult:
 break
 print(matchResult.group(0),':',matchResult.group(2))
 index = matchResult.end(0)+1

unless : s
better : t
better : t
>>> s = 'aabc abbcd abccd abbcd abcdd'
>>> p = re.compile(r'(\b\w*(?P<f>\w+)(?P=f)\w*\b)')
>>> p.findall(s)
[('aabc', 'a'), ('abbcd', 'b'), ('abccd', 'c'), ('abbcd', 'b'), ('abcdd', 'd')]