How to ignore characters with Python Regex

Question

I am wondering if there is a better Python Regex solution for the one that I currently have? Currently my code is:

import re

n = '+17021234567'
m = '7021234567'
match = re.search(r'(?:\+1)?(\d{10})', n)
match.group(1)

match = re.search(r'(?:\+1)?(\d{10})', m)
match.group(1)

The goal of the code is to only extract the 10 digit ph # if it has a leading +1 or not. Currently it works, but I am wondering is there a way to just call match.group() to get the 10 digit ph # without calling match.group(1)?

Avinash Raj · Accepted Answer · 2014-10-29 23:55:14Z

2

No, without the use of capturing groups, it couldn't be possible through re.match function. Since re.match tries to match the input string from the beginning. But it could be possible through re.search

>>> re.search(r'\d{10}$', n).group()
'7021234567'
>>> re.search(r'\d{10}$', m).group()
'7021234567'

edited Oct 29, 2014 at 23:55

answered Oct 29, 2014 at 23:49

Avinash Raj

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1 Comment

Totem Over a year ago

I don't think OP meant re.match but the variable match the they used

Hackaholic · Accepted Answer · 2014-10-30 00:00:38Z

1

you want to only capture digit use '\d' for digit

n = '+17021234567'
re.findall('\d{10}$',n)

answered Oct 30, 2014 at 0:00

Hackaholic

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Comments

alpha bravo · Accepted Answer · 2014-10-30 00:07:54Z

1

use this pattern

(?<=^|\+1)\d{10}$

Demo

(?<=                    look behind to see if there is:
  ^                       the beginning of the string
 |                        OR
  \+1                     '+1'
)                        end of look-behind
\d{10}                   digits (0-9) (10 times)
$                        before an optional \n, and the end of the string

answered Oct 30, 2014 at 0:07

alpha bravo

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