Difference between numpy.rint and numpy.round

Question

What is the difference between numpy.rint and numpy.round/numpy.around? They both seem to perform the same function:

>>> a
array([-1.7, -1.5, -0.2,  0.2,  1.5,  1.7,  2. ])
>>> np.round(a)
array([-2., -2., -0.,  0.,  2.,  2.,  2.])
>>> np.rint(a)
array([-2., -2., -0.,  0.,  2.,  2.,  2.])

Answer 1

This is the difference:

A = np.array([-1.72, -1.58, -0.2, 0.2, 1.5, 1.7, 2.0])
np.round(A,1)
array([-1.7, -1.6, -0.2,  0.2,  1.5,  1.7,  2. ])
np.rint(A)
array([-2., -2., -0.,  0.,  2.,  2.,  2.])

You basically use np.round() when you want to round to any decimal place, like here I have done it for one decimal, so I got -1.7 for -1.72 instead of -2 which I got in np.rint()

One possible reason for having np.rint(x) when np.round(x,0) can get our job done is the speed of computation which the former offers us. When I ran both the snippets of code and recorded the timing for the operation this is what I got

%%timeit
np.round(A,0)
5.16 µs ± 495 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each) 

%%timeit
np.rint(A)
1.06 µs ± 28.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

We can very well see that np.rint() gets the job done approximately five times faster than np.round().

Hope this helps!