On Jul 23, 5:12 pm, Billy Mays <[email protected]> wrote:
> On 7/23/2011 3:42 AM, Chris Angelico wrote:
>
>
>
> > int(s.rstrip('0').rstrip('.'))
>
> Also, it will (in?)correct parse strings such as:
>
> '165............00000000000000'
>
> to 165.
>
> --
> Bill
True enough.
If I really wanted to be 100% safe, how about this -
def get_int(s):
if '.' in s:
num, dec = s.split('.', 1)
if dec != '':
if int(dec) != 0:
raise ValueError('Invalid literal for int')
return int(num)
return int(s)
Frank
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