Rounding to nearest int with numpy.rint() not consistent for .5

Question

numpy\'s round int doesn\'t seem to be consistent with how it deals with xxx.5

In [2]: np.rint(1.5)
Out[2]: 2.0

In [3]: np.rint(10.5)
Out[3]: 10.0
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Answer 1

So, this kind of behavior (as noted in comments), is a very traditional form of rounding, seen in the round half to even method. Also known (according to David Heffernan) as banker's rounding. The numpy documentation around this behavior implies that they are using this type of rounding, but also implies that there may be issues with the way in which numpy interacts with the IEEE floating point format. (shown below)

Notes
-----
For values exactly halfway between rounded decimal values, Numpy
rounds to the nearest even value. Thus 1.5 and 2.5 round to 2.0,
-0.5 and 0.5 round to 0.0, etc. Results may also be surprising due
to the inexact representation of decimal fractions in the IEEE
floating point standard [1]_ and errors introduced when scaling
by powers of ten.

Whether or not that is the case, I honestly don't know. I do know that large portions of the numpy core are still written in FORTRAN 77, which predates the IEEE standard (set in 1984), but I don't know enough FORTRAN 77 to say whether or not there's some issue with the interface here.

If you're looking to just round up regardless, the np.ceil function (ceiling function in general), will do this. If you're looking for the opposite (always rounding down), the np.floor function will achieve this.