I have a function f(x) which gives f(x)=0 for two to three values of x. I want to find all the values of x for which f(x)=0 but np.interp(0, f(x), x) gives me the last value of x for which f(x)=0. How to get all the values?

I want to find the critical points of a function. I tried np.gradient(f(x)) and then interpolated it with np.interpolate(0, np.gradient(f(x)), x).

I have a function f(x) which gives f(x)=0 for two to three values of x. I want to find all the values of x for which f(x)=0 but np.interp(0, f(x), x) gives me the last value of x for which f(x)=0. How to get all the values?

I want to find the critical points of a function. I tried np.gradient(f(x)) and then interpolated it with np.interpolate(0, np.gradient(f(x)), x).

• python
• numpy
• interpolation

• 2 What does f(x) look like? – jared Commented Mar 31 at 3:29
• Your xp argument should be 'monotonically increasing'. This isn't enforced, but if the sequence xp is non-increasing, interpolation results are meaningless. – hpaulj Commented Mar 31 at 17:04

1 Answer 1

That's not what np.interp() does. From the docs: One-dimensional linear interpolation for monotonically increasing sample points. You are basically resampling, or in your case, interpolating the value of f(x) at x=0.

You seem to be giving (0,f,x) as arguments possibly(?) trying to swap xp,fp, in general this won't work. If you have exact zeros in your dataset, just use idx=np.where(f==0.0) and return the x values as shown below.

If your goal is to get the values where any function is 0, then things get kinda trickier. In a discrete case, your best bet in finding a 0 is to think of it as an intersection with the x-axis, meaning a change of sign for f (Bolzano's Theorem). Check this post for the sign change.

The next problem is that, while simply using np.sign()(like in the link above) will almost always work, for f=[1,0,-1] it will detect 2 changes of sign! One from 1 to 0 and one from 0 to -1. So we go with the solution of gathering all signs >=0, creating a boolean mask, and then with np.diff() acting on that, we should get changes of sign without 0 being counted twice.

You can try this:

import numpy as np

num = 10
x = np.linspace(0,4*np.pi,num)
f = np.sin(x)


def get_zero_crossings(f,x):
    # identify indices where adjacent element signs are opposite
    # this will return the left element
    idx = np.where(np.diff(np.sign(f)>=0))[0]
    # line equation, honestly there is probably a better way to do this
    x_zeros = x[idx] - f[idx] * (x[idx + 1] - x[idx]) / (f[idx + 1] - f[idx])
    return x_zeros

print(get_zero_crossings(f,x))

Simply get your np.gradient(f), x through the above function, and you should probably get the critical points of f.