Finding all occurrences of alternating digits using regular expressions

Question

I would like to find all alternating digits in a string using regular expressions. An alternating digit is defined as two equal digits having a digit in between; for example, 1212 contains 2 alternations (121 and 212) and 1111 contains 2 alternations as well (111 and 111). I have the following regular expression code:

s = "1212"
re.findall(r'(\d)(?:\d)(\1)+', s)

This works for strings like "121656", but not "1212". This is a problem to do with overlapping matches I think. How can I deal with that?

Answer 1

(?=((\d)\d\2))

Use lookahead to get all overlapping matches. Use re.findall and get the first element from the tuple. See the demo:

https://regex101.com/r/fM9lY3/54

Answer 2

You can use a lookahead to allow for overlapping matches:

r'(\d)(?=(\d)\1)'

To reconstruct full matches from this:

matches = re.findall(r'(\d)(?=(\d)\1)', s)
[a + b + a for a, b in matches]

Also, to avoid other Unicode digits like ١ from being matched (assuming you don’t want them), you should use [0-9] instead of \d.

Answer 3

With the regex module you don't have to use a trick to get overlapped matches since there's a flag to obtain them:

import regex
res = [x.group(0) for x in regex.finditer(r'(\d)\d\1', s, overlapped=True)]

if s contains only digits, you can do this too:

res = [s[i-2:i+1] for i in range(2, len(s)) if s[i]==s[i-2]]

Answer 4

A non regex approach if you string is made up of just digits:

from itertools import islice as isl, izip

s = "121231132124123"
out = [a + b + c for a, b, c in zip(isl(s, 0, None), isl(s, 1, None), isl(s, 2, None)) if a == c]

Output:

['121', '212', '212']

It is actually a nice bit faster than a regex approach.