round() returns floating point, not int, for some numpy floats when no second arg

The semantics of round() changed in Python 3:

round(number[, ndigits]) Return number rounded to ndigits precision after the decimal point. If ndigits is omitted or is None, it returns the nearest integer to its input.

This works incorrectly in the case of np.float64, which returns a float. I believe the __round__ method is calling __rint__, which should return an integer but doesn’t.

Reproducing code example:

import numpy as np
In [50]: round(1.0)
Out[50]: 1

In [51]: round(float(1.0))
Out[51]: 1

In [52]: round(np.float(1.0))
Out[52]: 1

In [53]: round(np.float64(1.0))
Out[53]: 1.0

This behavior is the same for float16, float32, and float128.

[charris]

NumPy round applied to numpy floats does not return integers. It is a feature, not a bug. The Python behavior you illustrate is new in Python 3.

[miccoli]

Another thought on this issue: since isinstance(np.float64(1), float) is true, the current implementation breaks Liskov substitution principle making the use of numpy scalars very unSOLID.

The problem is that one has a lot of paths (sometimes unexpected) in which numpy.float64 values sneaks into existing code, which makes unit testing and maintenance unnecessarily cumbersome.