Python regex find all overlapping matches?

Question

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]

You can also try using the third-party regex module (not re), which supports overlapping matches.

>>> import regex as re
>>> s = "123456789123456789"
>>> matches = re.findall(r'\d{10}', s, overlapped=True)
>>> for match in matches: print(match)  # print match
...
1234567891
2345678912
3456789123
4567891234
5678912345
6789123456
7891234567
8912345678
9123456789

I'm fond of regexes, but they are not needed here.

Simply

s =  "123456789123456789"

n = 10
li = [ s[i:i+n] for i in xrange(len(s)-n+1) ]
print '\n'.join(li)

result

1234567891
2345678912
3456789123
4567891234
5678912345
6789123456
7891234567
8912345678
9123456789