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

Numpy uses bankers rounding so .5 is rounded to the nearest even number. If you always want to round .5 up but round .4 down:

np.rint(np.nextafter(a, a+1))

or if you always want to round .5 down and .4 down but .6 up:

np.rint(np.nextafter(a, a-1))

NB this also works with np.around if you want the same logic but not integers.

>>> a = np.array([1, 1.5, 2, 2.5, 3, 3.5])
>>> np.rint(a)
array([1., 2., 2., 2., 3., 4.])
>>> np.rint(np.nextafter(a, a+1))
array([1., 2., 2., 3., 3., 4.])
>>> np.rint(np.nextafter(a, a-1))
array([1., 1., 2., 2., 3., 3.])

What is happening? nextafter gives the next representable number in a direction, so this is enough to push the number off 'exactly' 2.5.

Note this is different to ceil and floor.

>>> np.ceil(a)
array([1., 2., 2., 3., 3., 4.])
>>> np.floor(a)
array([1., 1., 2., 2., 3., 3.])