I was so fixed in the string format, that I was ignoring the basic math!
=/
Thanks to everyone for the tips!!
This solve very well:
"%i.%02i" % divmod(348,100)
Em quinta-feira, 29 de novembro de 2012 09h12min39s UTC-2, Joe Barnhart
escreveu:
>
> Bzzzt! Thanks for playing!
>
> You get 3.0 for your result. (I'm not a python genius but I keep an
> interpreter handy just to type in stuff and see what I get.)
>
>
> >>> round(int("348")/100,3)
> 3.0
>
>
> In your example, you rounded the result AFTER the integer division of
> 348/100. But the result of the integer division was 3, so your rounded
> version is 3.0.
>
> On Thursday, November 29, 2012 2:02:29 AM UTC-8, Manuele wrote:
>>
>> Il 29/11/12 10:49, Joe Barnhart ha scritto:
>>
>> Hmmm...
>>
>> That would give "3" since Python 2.x keeps the calculation as integer
>> (it won't coerce it to a float).
>>
>> He could do a=int(a)/100.0 But that puts it into a floating point
>> value, and rounding sometimes gives surprising and unexpected results.
>>
>> my 2 cent...
>>
>> a = round(int(a)/100, 3)
>>
>> M.
>>
>
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