正则表达式 - 在 Python 中第二次出现连字符后删除不需要的 substring

Question

Below are the strings out of which I need to pull out the meaningful IDs

'12345-1-abcde-aBCD'
'123-Abcdefghi abcdefghijkl'
'1234567-1-AB-ABC A/1 ABC (AB1234)'
'12345-ABC-Abcdefghijkl'
'123456-Abcdefgh'
'12345-AB1CDE'

Regex should match to all the above criteria and pass for all the cases to give below output

12345-1
123
1234567-1
12345
123456
12345

Regex should omit the part from the -hyphen if there are letters.

Answer 1

You can do this:

import re

l = ['12345-1-abcde-aBCD',
     '123-Abcdefghi abcdefghijkl', 
     '1234567-1-AB-ABC A/1 ABC (AB1234)',  
     '12345-ABC-Abcdefghijkl',
     '123456-Abcdefgh',
     '12345-AB1CDE',]

In [10]: for s in l:
    ...:     print(re.match(r'^(\d+[-]?\d+?)',s))
    ...:                 
<re.Match object; span=(0, 7), match='12345-1'>
<re.Match object; span=(0, 3), match='123'>
<re.Match object; span=(0, 9), match='1234567-1'>
<re.Match object; span=(0, 5), match='12345'>
<re.Match object; span=(0, 6), match='123456'>
<re.Match object; span=(0, 5), match='12345'>

---

If you can have multiple hyphens with subsequent digits you can do something like:

l = ['12345-1-abcde-aBCD',
     '123-Abcdefghi abcdefghijkl',
     '1234567-1-AB-ABC A/1 ABC (AB1234)',
     '12345-ABC-Abcdefghijkl',
     '123456-Abcdefgh',
     '12345-AB1CDE',
     '12345-1-1-ABC',
     '1-2-3-4-5-A-B-C-D-E-F-/-(AB12345)0',
     '12345-1A Abcd',]

In [31]: for s in l: 
    ...:     match = re.match(r'^([\d|-]*)(?![A-Za-z])',s)  
    ...:     print(match.group(0).rstrip('-')) 
    ...:                                                                                            
12345-1
123
1234567-1
12345
123456
12345
12345-1-1
1-2-3-4-5
12345