regex string and substring

Question

I have a character string 'aabaacaba'. Starting from left, I am trying to get substrings of all sizes >=2, which appear later in the string. For instance, aa appears again in the string and so is the case with ab.

I wrote following regex code:

re.findall(r'([a-z]{2,})(?:[a-z]*)(?:\1)', 'aabaacaba')

and I get ['aa'] as answer. Regular expression misses ab pattern. I think this is because of overlapping characters. Please suggest a solution, so that the expression could be fixed. Thank you.

@Chris Not necessarily. But it would be great if it could be done with regex. — Sumit
– Sumit, Commented May 14, 2017 at 2:44

falsetru · Accepted Answer · 2017-05-14 02:54:20Z

9

You can use look-ahead assertion which does not consume matched string:

>>> re.findall(r'(?=([a-z]{2,})(?=.*\1))', 'aabaacaba')
['aa', 'aba', 'ba']

NOTE: aba matched instead of ab. (trying to match as long as possible)

answered May 14, 2017 at 2:54

falsetru

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10 Comments

Sebastián Palma Over a year ago

Can [a-z] be replaced with \w as (?=(\w{2,})(?=.*\1)) (?)

falsetru Over a year ago

@SebastiánPalma, Yes it is. But it will also match digits, _. I'm not sure whether it's what OP wants or not; so I left it as is (as OP wrote). Maybe . is more appropriate if OP wants any character.

falsetru Over a year ago

@SebastiánPalma, I couldn't use look-behind assertion, because Python re allow only fixed-length look-behind assertion.

falsetru Over a year ago

@Sumit, Without the first look-ahead assertion, first matched part will be consumed; overlapped matches(aba in this case) will be excluded in the result.

Gurmanjot Singh Over a year ago

@falsetru great answer. I couldn't think about 1st look-ahead assertion. Learnt something new today :)

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