On Jul 27, 7:58 pm, Gary Herron <[EMAIL PROTECTED]> wrote: > 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 > > -- > >http://mail.python.org/mailman/listinfo/python-list > >
I should reiterate that I am using Python 3.0 and not Python 2.x. It looks like the behavior round() has changed between these two versions. Here is the documentation for round() in Python 3.0: http://docs.python.org/dev/3.0/library/functions.html#round Of interest in this discussion is the second paragraph, which explains the change: Does anyone know the reason behind this change, and what replacement method I can use to get the original behavior? -- http://mail.python.org/mailman/listinfo/python-list
