Regex Replace All But Pattern

If you mean by the second would yield nothing, you could match any char except a digit or newline, followed by capturing the pattern in a group.

If the sub should leave an empty string, you could match the whole line using an alternation.

[^\d\r\n]+(\d{1,2}-\d{1,2})|.+

In parts

• [^\d\r\n]+ Match 1+ times any char except a digit or a newline
• (\d{1,2}-\d{1,2}) Capture group 1, match 1-2 digits, - and 1-2 digits
• | Or
• .+ Match any char except a newline 1+ more times

Regex demo | Python demo

Example code

import re

lines = [
    'dkas;6-17',
    'dsajdl 10',
    'dsjalkdj16-20'
]

for text in lines:
    print(re.sub('[^\d\r\n]+(\d{1,2}-\d{1,2})|.+', r'\1', text))

Output

6-17

16-20

How about just looking for all the matches in the string and concatenating them together?

>>> ''.join(re.findall('[0-9]{1,2}\-[0-9]{1,2}', 'dkas;6-17abc19-10'))
'6-1719-10'

>>> ''.join(re.findall('[0-9]{1,2}\-[0-9]{1,2}', 'dsajdl 10'))
''

Consider matching

\d+-\d+|$

Demo

If the string were

dkas;6-17

the first match would be 6-17, the second would be the empty string at the end of the line.

If the string were

dsjalkdj16-20kl21-33mn

there would be three matches, 16-20, 21-33 and the empty space at the end of the line.

If the string were

dsajdl 10

the first (and only) match would be the empty string at the end of the line.

It follows that if there it one match it will be the empty string at the end of the string, which is to be returned; else, return the first, or all but the last, match(es), depending on requirements.