Finding recurring patterns in a string

Question

Ok the problem is that given a fraction eg:1/3 the answer 0.3333333 should represented as 0.(3) and 0.2325555 as 0.232(5) i have figured out a way to split the string when its a single digit repeating:

using re.findall(r'^(.+?)((.)\3+)$', '42344444' )[0][:-1] (ignoring the 0. before the number)

but i want to know how to do this if the pattern is 0.324324324.. to get 0.(324)

Are you given the numbers as a fraction or a floating point? If it's the later, this is impossible. — Antimony
– Antimony, Commented Jul 3, 2012 at 1:11
Is the aim to discover recurring patterns of arbitrary length? If so, that cannot be done with regex. — Oliver Charlesworth
– Oliver Charlesworth, Commented Jul 3, 2012 at 1:12
actually forget about the fraction part .. say you are give the string 23154545454 how would you split the string into the non recurring part and the recurring part — insane
– insane, Commented Jul 3, 2012 at 1:13
ok to make things simpler lets assume strings are of length 20 — insane
– insane, Commented Jul 3, 2012 at 1:15

Igor Chubin · Accepted Answer · 2012-07-03 01:13:03Z

2

add + after . in the recurring part:

>>> re.findall(r'^(.+?)((.+)\3+)$', '42344343434' )[0][:-1]
('42344', '343434')

answered Jul 3, 2012 at 1:13

Igor Chubin

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