Python 提取模式匹配

Python extract pattern matches

提问by Kannan Ekanath

Python 2.7.1 I am trying to use python regular expression to extract words inside of a pattern

Python 2.7.1 我正在尝试使用 python 正则表达式来提取模式中的单词

I have some string that looks like this

我有一些看起来像这样的字符串

someline abc
someother line
name my_user_name is valid
some more lines

I want to extract the word "my_user_name". I do something like

我想提取“my_user_name”这个词。我做类似的事情

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>

How do I extract my_user_name now?

我现在如何提取 my_user_name？

采纳答案by UltraInstinct

You need to capture from regex. searchfor the pattern, if found, retrieve the string using group(index). Assuming valid checks are performed:

您需要从正则表达式中捕获。search对于模式，如果找到，则使用 检索字符串group(index)。假设执行了有效检查：

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

回答by mgilson

You can use matching groups:

您可以使用匹配组：

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

e.g.

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']

Here I use re.findallrather than re.searchto get all instances of my_user_name. Using re.search, you'd need to get the data from the group on the match object:

在这里，我使用re.findall而不是re.search获取my_user_name. 使用re.search，您需要从匹配对象上的组中获取数据：

>>> 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'

---

---

As mentioned in the comments, you might want to make your regex non-greedy:

正如评论中提到的，您可能想让正则表达式变得非贪婪：

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

to only pick up the stuff between 'name 'and the next ' is valid'(rather than allowing your regex to pick up other ' is valid'in your group.

只拿起'name '和下一个之间的东西' is valid'（而不是让你的正则表达式拿起' is valid'你组中的其他人。

回答by Henry Keiter

You want a capture group.

你想要一个捕获组。

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.

回答by Apalala

You could use something like this:

你可以使用这样的东西：

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')

回答by John

Maybe that's a bit shorter and easier to understand:

也许这更短更容易理解：

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'

回答by Eugene Yarmash

You can use groups (indicated with '('and ')') to capture parts of the string. The match object's group()method then gives you the group's contents:

您可以使用组（用'('和表示')'）来捕获部分字符串。然后匹配对象的group()方法为您提供组的内容：

>>> import re
>>> s = 'name my_user_name is valid'
>>> match = re.search('name (.*) is valid', s)
>>> match.group(0)  # the entire match
'name my_user_name is valid'
>>> match.group(1)  # the first parenthesized subgroup
'my_user_name'

In Python 3.6+ you can also indexinto a match object instead of using group():

在 Python 3.6+ 中，您还可以索引匹配对象而不是使用group()：

>>> match[0]  # the entire match 
'name my_user_name is valid'
>>> match[1]  # the first parenthesized subgroup
'my_user_name'

回答by wolfovercats

Here's a way to do it without using groups (Python 3.6 or above):

这是一种不使用组（Python 3.6 或更高版本）的方法：

>>> re.search('2\d\d\d[01]\d[0-3]\d', 'report_20191207.xml')[0]
'20191207'

回答by Ryan Stefan

You can also use a capture group (?P<user>pattern)and access the group like a dictionary match['user'].

您还可以使用捕获组(?P<user>pattern)并像访问字典一样访问该组match['user']。

string = '''someline abc\n
            someother line\n
            name my_user_name is valid\n
            some more lines\n'''

pattern = r'name (?P<user>.*) is valid'
matches = re.search(pattern, str(string), re.DOTALL)
print(matches['user'])

# my_user_name

回答by chiceman

It seems like you're actually trying to extract a name vice simply find a match. If this is the case, having span indexes for your match is helpful and I'd recommend using re.finditer. As a shortcut, you know the namepart of your regex is length 5 and the is validis length 9, so you can slice the matching text to extract the name.

看起来您实际上是在尝试提取名称，而只是找到匹配项。如果是这种情况，为您的比赛设置跨度索引会很有帮助，我建议您使用re.finditer. 作为一种快捷方式，您知道name正则表达式的长度为 5，is valid长度为 9，因此您可以对匹配的文本进行切片以提取名称。

Note - In your example, it looks like sis string with line breaks, so that's what's assumed below.

注意 - 在您的示例中，它看起来像是s带有换行符的字符串，因此这就是下面的假设。

## covert s to list of strings separated by line:
s2 = s.splitlines()

## find matches by line: 
for i, j in enumerate(s2):
    matches = re.finditer("name (.*) is valid", j)
    ## ignore lines without a match
    if matches:
        ## loop through match group elements
        for k in matches:
            ## get text
            match_txt = k.group(0)
            ## get line span
            match_span = k.span(0)
            ## extract username
            my_user_name = match_txt[5:-9]
            ## compare with original text
            print(f'Extracted Username: {my_user_name} - found on line {i}')
            print('Match Text:', match_txt)

## covert s to list of strings separated by line:
s2 = s.splitlines()

## find matches by line: 
for i, j in enumerate(s2):
    matches = re.finditer("name (.*) is valid", j)
    ## ignore lines without a match
    if matches:
        ## loop through match group elements
        for k in matches:
            ## get text
            match_txt = k.group(0)
            ## get line span
            match_span = k.span(0)
            ## extract username
            my_user_name = match_txt[5:-9]
            ## compare with original text
            print(f'Extracted Username: {my_user_name} - found on line {i}')
            print('Match Text:', match_txt)