Raymond Hettinger <[email protected]> added the comment:
When groups are present in the regex, findall() returns the subgroups rather
than the entire match:
>>> mo = re.search('(12)+', '121212 and 121212')
>>> mo[0] # Entire match
'121212'
>>> mo[1] # Group match
'12'
To get the result you were looking for use a non-capturing expression:
>>> re.findall('(?:12)+', '121212 and 121212')
['121212', '121212']
Also consider using finditer() which gives more fine grained control:
>>> for mo in re.finditer('(12)+', '121212 and 121212'):
print(mo.span())
print(mo[0])
print(mo[1])
print()
(0, 6)
121212
12
(11, 17)
121212
12
----------
nosy: +rhettinger
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<https://bugs.python.org/issue42448>
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