正则表达式在第二次出现字符后匹配相邻数字

Question

Stuck with the following issue:

I have a string 'ABC.123.456XX' and I want to use regex to extract the 3 numeric characters that come after the second period. Really struggling with this and would appreciate any new insights, this is the closest I got but its not really close to what I want:

'.*\.(.*?\.\d{3})'

I appreciate any help in advance - thanks.

Answer 1

If your input will always be in a similar format, like xxx.xxx.xxxxx , then one solution is string manipulation:

>>> s = 'ABC.123.456XX'
>>> '.'.join(s.split('.')[2:])[0:3]

Explanation

In the line '.'.join(s.split('.')[2:])[0:3] :

• s.split('.') splits the string into the list ['ABC', '123', '456XX']
• '.'.join(s.split('.')[2:]) joins the remainder of the list after the second element, so '456XX'
• [0:3] selects the substring from index 0 to index 2 (inclusive), so the result is 456

Answer 2

Dot, not-Dot twice then the 3 digits follow in capture group 1

[^.]*(?:\.[^.]*){2}(\d{3})

https://regex101.com/r/qWpfHx/1

Expanded

 [^.]* 
 (?: \. [^.]* ){2}
 ( \d{3} )                     # (1)

Answer 3

This expression might also work just OK:

[^\r\n.]+\.[^\r\n.]+\.([0-9]{3})

Test

import re

regex = r'[^\r\n.]+\.[^\r\n.]+\.([0-9]{3})'
string = '''
ABC.123.456XX
ABCOUOU.123123123.000871XX
ABCanything_else.123123123.111871XX
'''

print(re.findall(regex, string))

Output

['456', '000', '111']

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If you wish to simplify/modify/explore the expression, it's been explained on the top right panel of regex101.com . If you'd like, you can also watch in this link , how it would match against some sample inputs.

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