Extract a number to the right of a text python

Question

Lets assume I have some string in python ending with some number.

2te2xt12
text1
text143
te2xt341

How can I get the right most number sequence:

2te2xt12  >>  12
text1     >>  1
text143   >>  143
te2xt341  >>  341

Is there a simple and easy way.

I know nothing about re

Thanks in advance

Answer 1

import re
s=r"""1231ololo767980716"""
re.match('.*?([0-9]+)$', s).group(1)

>>> '767980716'

or

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

Explanation of 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.

and all components you can find here: https://www.rexegg.com/regex-quickstart.html very helpful cheat sheet

Answer 2

This code which uses regex \\d+$ should get you what you are exactly looking for,

import re

arr = ['2te2xt12',
'text1',
'text143',
'te2xt341']

for s in arr:
 m = re.search(r'\d+$', s)
 if (m):
  print(s + ' >> ' + m.group())

Prints,

2te2xt12 >> 12
text1 >> 1
text143 >> 143
te2xt341 >> 341

Answer 3

Try this? The function test should do what you need.

import re

def test(string):
    m = re.search(r'\d+$', string)
    if m is not None:
        print(m.group())

Answer 4

It is quite simple using regex in python

re.findall('\d+$',yourString)[0]

To know more about regular expressions refer the docs here