python regex: get end digits from a string

Question

I am quite new to python and regex (regex newbie here), and I have the following simple string:

s=r"""99-my-name-is-John-Smith-6376827-%^-1-2-767980716""" 

I would like to extract only the last digits in the above string i.e 767980716 and I was wondering how I could achieve this using python regex.

I wanted to do something similar along the lines of:

re.compile(r"""-(.*?)""").search(str(s)).group(1) 

indicating that I want to find the stuff in between (.*?) which starts with a "-" and ends at the end of string - but this returns nothing..

I was wondering if anyone could point me in the right direction.. Thanks.

Answer 1

You can use re.match to find only the characters:

>>> import re >>> s=r"""99-my-name-is-John-Smith-6376827-%^-1-2-767980716""" >>> re.match('.*?([0-9]+)$', s).group(1) '767980716' 

Alternatively, re.finditer works just as well:

>>> next(re.finditer(r'\d+$', s)).group(0) '767980716' 

Explanation of all regexp components:

• .*? is a non-greedy match and consumes only as much as possible (a greedy match would consume everything except for the last digit).
• [0-9] and \d are two different ways of capturing digits. Note that the latter also matches digits in other writing schemes, like ୪ or ൨.
• Parentheses (()) make the content of the expression a group, which can be retrieved with group(1) (or 2 for the second group, 0 for the whole match).
• + means multiple entries (at least one number at the end).
• $ matches only the end of the input.

Answer 2

Nice and simple with findall:

import re  s=r"""99-my-name-is-John-Smith-6376827-%^-1-2-767980716"""  print re.findall('^.*-([0-9]+)$',s)  >>> ['767980716'] 

Regex Explanation:

^         # Match the start of the string .*        # Followed by anthing -         # Upto the last hyphen ([0-9]+)  # Capture the digits after the hyphen $         # Upto the end of the string 

Or more simply just match the digits followed at the end of the string '([0-9]+)$'