Python的提取模式匹配

Question

的Python 2.7.1 我试图使用python的正则表达式来提取模式中的单词

我有一些字符串，它看起来像这样

someline abc 
someother line 
name my_user_name is valid 
some more lines 

我想提取单词“my_user_name”。我做类似

import re 
s = #that big string 
p = re.compile("name .* is valid", re.flags) 
p.match(s) #this gives me <_sre.SRE_Match object at 0x026B6838> 

现在如何提取my_user_name？

Answer 1

你需要从正则表达式捕获。 search为模式，如果找到，则使用group(index)检索字符串。假设执行有效的检查：

>>> p = re.compile("name (.*) is valid") 
>>> p.search(s) # The result of this is referenced by variable name '_' 
<_sre.SRE_Match object at 0x10555e738> 
>>> _.group(1)  # group(1) will return the 1st capture. 
'my_user_name' 

Answer 2

您可以使用匹配的组：

p = re.compile('name (.*) is valid') 

例如

>>> import re 
>>> p = re.compile('name (.*) is valid') 
>>> s = """ 
... someline abc 
... someother line 
... name my_user_name is valid 
... some more lines""" 
>>> p.findall(s) 
['my_user_name'] 

这里我用re.findall而不是re.search得到的my_user_name所有实例。使用re.search，你需要拿到赛对象从该组数据：

>>> p.search(s) #gives a match object or None if no match is found 
<_sre.SRE_Match object at 0xf5c60> 
>>> p.search(s).group() #entire string that matched 
'name my_user_name is valid' 
>>> p.search(s).group(1) #first group that match in the string that matched 
'my_user_name' 

---

正如在评论中提到，你可能想使你的正则表达式非贪婪：

p = re.compile('name (.*?) is valid') 

只能拿起'name '和下' is valid'（之间的东西，而不是让你的正则表达式到论坛中拿起其他' is valid'。

Answer 3

你想要一个capture group。

p = re.compile("name (.*) is valid", re.flags) # parentheses for capture groups 
print p.match(s).groups() # This gives you a tuple of your matches. 

Answer 4

你可以使用这样的事情：

import re 
s = #that big string 
# the parenthesis create a group with what was matched 
# and '\w' matches only alphanumeric charactes 
p = re.compile("name +(\w+) +is valid", re.flags) 
# use search(), so the match doesn't have to happen 
# at the beginning of "big string" 
m = p.search(s) 
# search() returns a Match object with information about what was matched 
if m: 
    name = m.group(1) 
else: 
    raise Exception('name not found') 

Answer 5

也许这是一个有点短，更容易理解：

import re 
text = '... someline abc... someother line... name my_user_name is valid.. some more lines' 
>>> re.search('name (.*) is valid', text).group(1) 
'my_user_name'