Where is the correct round() method?

[Gary Herron]

josh logan wrote:
> Hello,
>
> I need a round function that _always_ rounds to the higher integer if
> the argument is equidistant between two integers. In Python 3.0, this
> is not the advertised behavior of the built-in function round() as
> seen below:
>
>   
>>>> round(0.5)
>>>>         
> 0
>   
>>>> round(1.5)
>>>>         
> 2
>   
>>>> round(2.5)
>>>>         
> 2
>
>   

Huh?

 >>> round(2.5)
3.0


Works for me on Python 2.5 on Linux running on "Intel(R) Core(TM)2 Duo 
CPU".  What system are you on?


It could be that 2.5 is really 2.49999... which would round down to 2, 
but on any modern CPU (using IEEE floating point), 2.5 should be 
representable exactly.

However, as with any floating point calculations, if you expect exact 
representation or calculations with any numbers, then you are misusing 
floating points. 

Gary Herron

> I would think this is a common need, but I cannot find a function in
> the Python library to do it. I wrote my own, but did I miss such a
> method in my search of the Python library?
>
> Thanks
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>