Python regex find all overlapping matches?

Question

I'm trying to find every 10 digit series of numbers within a larger series of numbers using re in Python 2.6.

I'm easily able to grab no overlapping matches, but I want every match in the number series. Eg.

in "123456789123456789"

I should get the following list:

[1234567891,2345678912,3456789123,4567891234,5678912345,6789123456,7891234567,8912345678,9123456789]

I've found references to a "lookahead", but the examples I've seen only show pairs of numbers rather than larger groupings and I haven't been able to convert them beyond the two digits.

question from:https://stackoverflow.com/questions/65890480/how-to-find-the-longest-consecutive-repeats-of-pattern-in-text-using-regular-exp

Answer 1

Use a capturing group inside a lookahead. The lookahead captures the text you're interested in, but the actual match is technically the zero-width substring before the lookahead, so the matches are technically non-overlapping:

import re 
s = "123456789123456789"
matches = re.finditer(r'(?=(d{10}))',s)
results = [int(match.group(1)) for match in matches]
# results: 
# [1234567891,
#  2345678912,
#  3456789123,
#  4567891234,
#  5678912345,
#  6789123456,
#  7891234567,
#  8912345678,
#  9123456789]