Python regex find all overlapping matches?
Solution 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]
Solution 2:
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
Solution 3:
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