John Henry wrote:
> Or more precisely:
>
> round(0.014999,2)
No, that *won't* solve the problem. Using a slightly
different example,
>>> x = 1.5 * 0.1
>>> x
0.15002
>>> round(x, 2)
0.14999
The problem is that floats are stored internally in
binary, not de
John Henry wrote:
> On Mar 9, 5:45 am, hg <[EMAIL PROTECTED]> wrote:
>> hg wrote:
>> > Hi,
>>
>> > Here is my issue:
>>
>> > f = 1.5 * 0.01
>> > f
>> >>> 0.014999
>> > '%f' % f
>> >>>'0.015000'
>>
>> > But I really want to get 0.02 as a result ... is there a way out ?
>>
>> > Thanks,
>
On Mar 9, 5:45 am, hg <[EMAIL PROTECTED]> wrote:
> hg wrote:
> > Hi,
>
> > Here is my issue:
>
> > f = 1.5 * 0.01
> > f
> >>> 0.014999
> > '%f' % f
> >>>'0.015000'
>
> > But I really want to get 0.02 as a result ... is there a way out ?
>
> > Thanks,
>
> > hg
>
> round
Or more precise
hg wrote:
> Hi,
>
> Here is my issue:
>
> f = 1.5 * 0.01
> f
>>> 0.014999
> '%f' % f
>>>'0.015000'
>
>
> But I really want to get 0.02 as a result ... is there a way out ?
>
> Thanks,
>
> hg
round
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Hi,
Here is my issue:
f = 1.5 * 0.01
f
>> 0.014999
'%f' % f
>>'0.015000'
But I really want to get 0.02 as a result ... is there a way out ?
Thanks,
hg
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