[Tutor] Count for loops

[Alan Gauld]

On 04/04/17 00:37, D.V.N.Sarma డి.వి.ఎన్.శర్మ wrote:
> I will go for this modification of the original code.

> count = 0
> b= '3'+b[2:]
> n = len(b)
> for i in range(n-4):
>     if b[i:i+4] == get_year:
>         count += 1

While I think this works OK, I would probably suggest
that this is one of the rare valid cases for using
a regex(*) with a simple string. The findall() method
with a suitable pattern and flags should catch
overlaps. And is probably faster being written in C.

(*)Assuming you want to include overlapping cases,
otherwise just use b.count()...

-- 
Alan G
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