Regex Capture a matching group twice

Question

I have been looking at a lot of questions but none of them answered my problem.

I am trying to identify vanity/special mobile numbers consisting of 9 digits.

I have this pattern of two numbers repeating for 4 times each

5XXXXYYYY

I was able to write a regex that catches the first group 'XXXX'. However, I am not able to acpture the second one 'YYYY'.

This is the regex I have tried so far

(\d)\1{3}

[0-9]{4,5}

I tried it on the below number, it returns '1111' and ignores '2222'

511112222

Is there a way to repeat the capture to include both groups

Thank you

Are you using `re.finditer` or `re.search`? Use `re.finditer` to get all matches. — Wiktor Stribiżew, Sep 20 '22 at 10:01
Maybe you can reach with this way `result.group(1)` or `result.group(2)`. 1111 takes in place first gruop and 2222 should be in the second group — uozcan12, Sep 20 '22 at 10:01

score 2 · Accepted Answer · answered Sep 20 '22 at 10:01

You may use this regex:

^5((\d)\2{3})((\d)\4{3})$

RegEx Breakup:

^: Start
5: Match 5
(: Start capture group #1
- (\d): Match a digit in capture group #2
- \2{3}: Repeat value captured in group #2 exactly 3 times
): End capture group #1
(: Start capture group #3
- (\d): Match a digit in capture group #4
- \4{3}: Repeat value captured in group #4 exactly 3 times
): End capture group #3
$: End

1 Answers1