Python extract pattern matches

Question

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

I have some string that looks like this

someline abc
s         


        

Answer 1

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:

>>> 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 index into a match object instead of using group():

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

Answer 2

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

Answer 3

You can use matching groups:

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

e.g.

>>> 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.findall rather than re.search to get all instances of my_user_name. Using re.search, you'd need to get the data from the group on the match object:

>>> 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:

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.

Answer 4

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

>>> 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 (stuff within the brackets).
                        # group(0) will returned the entire matched text.
'my_user_name'

Answer 5

You could use something like this:

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 6

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

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