Ip parsing using regexp in python

Question

I am trying to parse IP address from a string:

>>> import re
>>> input_str = '''
kjhdkjfh shfkjdsh shfk 1.1.1.1 kaseroi 1.1.1.1 jsoiu 1.1.1.1 
1
1
11
123
132132.23213.213213.123213
23.23.23.23 2321321.33.3.3.3 3.3..3.3.3.3.3. 
3.3.3.3.3.3

3.3.3.3
34.5.6.7
agdi 123213.44.4.5 12.12.12.12
'''
>>> 
>>> 
>>> pattern = r"\b(?:(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[0-9]?[0-9])\.){3}(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[0-9]?[0-9])\b"
>>> re.findall(pattern, input_str)
['1.1.1.1', '1.1.1.1', '1.1.1.1', '23.23.23.23', '33.3.3.3', '3.3.3.3', '3.3.3.3', '3.3.3.3', '34.5.6.7', '12.12.12.12']
>>>

But the valid IP list is:

['1.1.1.1', '1.1.1.1', '1.1.1.1', '23.23.23.23', '3.3.3.3', '34.5.6.7', '12.12.12.12']

Is there anything wrong with regex?

Avinash Raj · Accepted Answer · 2014-11-20 07:08:45Z

3

You just need to add negative lookahead and lookbehind in your pattern.

(?<!\.)\b(?:(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[0-9]?[0-9])\.){3}(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[0-9]?[0-9])\b(?!\.\d?)

DEMO

OR

(?<!\S)(?:(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[0-9]?[0-9])\.){3}(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[0-9]?[0-9])(?!\S)

DEMO

(?<!\S) Negative lookbehind asserts that (what or character) precedes the match would be any but not a non-space character.
(?!\S) Negative lookahead asserts that what follows the match would be any but not a non-space character.

Code:

>>> re.findall(r'(?<!\S)(?:(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[0-9]?[0-9])\.){3}(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[0-9]?[0-9])(?!\S)', input_str)
['1.1.1.1', '1.1.1.1', '1.1.1.1', '23.23.23.23', '3.3.3.3', '34.5.6.7', '12.12.12.12']

edited Nov 20, 2014 at 7:08

answered Nov 20, 2014 at 6:54

Avinash Raj

175k32 gold badges247 silver badges289 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

nu11p01n73R · Accepted Answer · 2014-11-20 06:54:52Z

1

You cannot use \b to limit the regex because . is included within the \b. From the input string we can notice that the ips are delimited using space hence \s is a much better option.

Changing the regex with a lookarounds for \s would serve the pupose

>>> attern = r"(?<=\s)(?:(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[0-9]?[0-9])\.){3}(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[0-9]?[0-9])(?=\s)"
>>> re.findall(attern, input_str)
['1.1.1.1', '1.1.1.1', '1.1.1.1', '23.23.23.23', '3.3.3.3', '34.5.6.7', '12.12.12.12']

answered Nov 20, 2014 at 6:54

nu11p01n73R

26.8k3 gold badges42 silver badges52 bronze badges

Collectives™ on Stack Overflow

Ip parsing using regexp in python

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related