How to produce a "Callable function"

Question

I am currently writing a python definition called f_from_data which uses interpolation find point on a line so far I have written this:

def f_from_data(xs, ys, x):
    xfine = np.linspace(min(xs), max(xs), 10000)
    y0 = inter.interp1d(xs, ys, kind = 'linear')
    ans = (y0(xfine))[numpy.searchsorted(xfine, x)]
    ans =  round(ans,2)
    return ans

This is giving me what I want to I need to make it so I can enter:

f = f_from_data([3, 4, 6], [0, 1, 2])
print f(3)
>>>0.0

How do I go about doing this? I've looked around but can't seem to find anything as I think its really trivial but I'm just missing somthing.

Answer 1

Using functools.partial:

from functools import partial

f = partial(f_from_data, [3, 4, 6], [0, 1, 2])

partial will create a callable object with the first 2 arguments already set.

Answer 2

interpolate.interp1d returns a callable:

import scipy.interpolate as interpolate

f_from_data = interpolate.interp1d
f = f_from_data([3, 4, 6], [0, 1, 2])
print(f(3))

yields

0.0

Since f_from_data can be assigned to interpolate.interp1d, you may not need f_from_data at all. Now, it is true that this does not chop the x-range into 10000 grid points, and use searchsorted to snap the x value to a nearby grid point, but in general you wouldn't want to do that anyway since interp1d gives you a better linear interpolation without it.

Answer 3

Perhaps something like this?

def f_from_data(xs, ys):
    def interpolate(x):
       xfine = np.linspace(min(xs), max(xs), 10000)
       y0 = inter.interp1d(xs, ys, kind = 'linear')
       ans = (y0(xfine))[numpy.searchsorted(xfine, x)]
       ans =  round(ans,2)
       return ans
    return interpolate

Warning - I don't know matplotlib well enough to say whether the code is correct.