Best way to implement numpy.sin(x) / x where x might contain 0

Question

What I am doing now is:

import numpy as np

eps = np.finfo(float).eps

def sindiv(x):
    x = np.abs(x)
    return np.maximum(eps, np.sin(x)) / np.maximum(eps, x)

But there is quite a lot of additional array operation. Is there a better way?

Answer 1

You could use numpy.sinc, which computes sin(pi x)/(pi x):

In [20]: x = 2.4

In [21]: np.sin(x)/x
Out[21]: 0.28144299189631289

In [22]: x_over_pi = x / np.pi

In [23]: np.sinc(x_over_pi)
Out[23]: 0.28144299189631289

In [24]: np.sinc(0)
Out[24]: 1.0