[Tutor] Rounding a Python float to the nearest half integer
Sithembewena Lloyd Dube
zebra05 at gmail.com
Fri Oct 8 15:56:28 CEST 2010
@Evert, I didn't figure out that your response was a solution, thought it
was a question. Must be coffee time :P
I tried it and, for instance, the rounded value (9) / 2 gave me 4.0 Couldn't
get it until I noticed that @Joel divided the roudned figure by a decimal
2.0. That gave 4.5, which is what I was looking for.
Thanks to all.
On Fri, Oct 8, 2010 at 3:35 PM, Joel Goldstick <joel.goldstick at gmail.com>wrote:
>
>
> On Fri, Oct 8, 2010 at 9:00 AM, Evert Rol <evert.rol at gmail.com> wrote:
>
>> > I realise that one cannot have a half integer :) I meant how would one
>> round off to the first decimal nearest to either 0.5, or a whole number.
>> >
>> > Ugh...does anyone get what I'm trying to articulate? :)
>>
>> Multiply by 2, round(), divide by 2?
>>
>
> That sounds like a good idea:
>
>>
>> >>> n = [1.0 + x/10.0 for x in range(10)] # get some numbers to test
>> >>> n
>> [1.0, 1.1000000000000001, 1.2, 1.3, 1.3999999999999999, 1.5,
>> 1.6000000000000001, 1.7, 1.8, 1.8999999999999999]
>> >>> r = [round(2*x)/2.0 for x in n] # double, round, then divide by 2.0
>> >>> r
>> [1.0, 1.0, 1.0, 1.5, 1.5, 1.5, 1.5, 1.5, 2.0, 2.0]
>> >>>
>>
>>
> --
> Joel Goldstick
>
>
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--
Regards,
Sithembewena Lloyd Dube
http://www.lloyddube.com
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