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>>> match = re.findall('a.*?a', 'a 1 a 2 a 3 a 4 a')
>>> match
['a 1 a', 'a 3 a']
How do I get it to print
['a 1 a', 'a 2 a', 'a 3 a', 'a 4 a']
Thank you!
674
findall() is probably the single most powerful function in the re module. Above we used re.search() to find the first match for a pattern. findall() finds *all* the matches and returns them as a list of strings, with each string representing one match.
findall(pattern, string) returns a list of matching strings. re. finditer(pattern, string) returns an iterator over MatchObject objects.
Search multiple words using regex Use | (pipe) operator to specify multiple patterns.
findall() module is used to search for “all” occurrences that match a given pattern. In contrast, search() module will only return the first occurrence that matches the specified pattern. findall() will iterate over all the lines of the file and will return all non-overlapping matches of pattern in a single step.
I think using a positive lookahead assertion should do the trick:
>>> re.findall('(?=(a.*?a))', 'a 1 a 2 a 3 a 4 a')
['a 1 a', 'a 2 a', 'a 3 a', 'a 4 a']
re.findall returns all the groups in the regex - including those in look-aheads. This works because the look-ahead assertion doesn't consume any of the string.
83
You may use alternative regex module which allows overlapping matches:
>>> regex.findall('a.*?a', 'a 1 a 2 a 3 a 4 a', overlapped = True)
['a 1 a', 'a 2 a', 'a 3 a', 'a 4 a']
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