I use
re.compile(r"(.+?)\1+").findall('44442(2)2(2)44')
can get
['4','2(2)','4']
, but how can I get
['4444','2(2)2(2)','44']
by using regular expression?
Thanks
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3 Answers
No change to your pattern needed. Just need to use to right function for the job. re.findall will return a list of groups if there are capturing groups in the pattern. To get the entire match, use re.finditer instead, so that you can extract the full match from each actual match object.
pattern = re.compile(r"(.+?)\1+")
[match.group(0) for match in pattern.finditer('44442(2)2(2)44')]
1 Comment
Amadan
Ooo, even better. Learn something every day :D
With minimal change to OP's regular expression:
[m[0] for m in re.compile(r"((.+?)\2+)").findall('44442(2)2(2)44')]
findall will give you the full match if there are no groups, or groups if there are some. So given that you need groups for your regexp to work, we simply add another group to encompass the full match, and extract it afterwards.
Comments
You can do:
[i[0] for i in re.findall(r'((\d)(?:[()]*\2*[()]*)*)', s)]
Here the Regex is:
((\d)(?:[()]*\2*[()]*)*)
which will output a list of tuples containing the two captured groups, and we are only interest din the first one hence i[0].
Example:
In [15]: s
Out[15]: '44442(2)2(2)44'
In [16]: [i[0] for i in re.findall(r'((\d)(?:[()]*\2*[()]*)*)', s)]
Out[16]: ['4444', '2(2)2(2)', '44']
