How to perform cubic spline interpolation in python?

Short answer:

from scipy import interpolate

def f(x):
    x_points = [ 0, 1, 2, 3, 4, 5]
    y_points = [12,14,22,39,58,77]

    tck = interpolate.splrep(x_points, y_points)
    return interpolate.splev(x, tck)

print(f(1.25))

Long answer:

scipy separates the steps involved in spline interpolation into two operations, most likely for computational efficiency.

1. The coefficients describing the spline curve are computed, using splrep(). splrep returns an array of tuples containing the coefficients.

2. These coefficients are passed into splev() to actually evaluate the spline at the desired point x (in this example 1.25). x can also be an array. Calling f([1.0, 1.25, 1.5]) returns the interpolated points at 1, 1.25, and 1,5, respectively.

This approach is admittedly inconvenient for single evaluations, but since the most common use case is to start with a handful of function evaluation points, then to repeatedly use the spline to find interpolated values, it is usually quite useful in practice.

In case, scipy is not installed:

import numpy as np
from math import sqrt

def cubic_interp1d(x0, x, y):
    """
    Interpolate a 1-D function using cubic splines.
      x0 : a float or an 1d-array
      x : (N,) array_like
          A 1-D array of real/complex values.
      y : (N,) array_like
          A 1-D array of real values. The length of y along the
          interpolation axis must be equal to the length of x.

    Implement a trick to generate at first step the cholesky matrice L of
    the tridiagonal matrice A (thus L is a bidiagonal matrice that
    can be solved in two distinct loops).

    additional ref: www.math.uh.edu/~jingqiu/math4364/spline.pdf 
    """
    x = np.asfarray(x)
    y = np.asfarray(y)

    # remove non finite values
    # indexes = np.isfinite(x)
    # x = x[indexes]
    # y = y[indexes]

    # check if sorted
    if np.any(np.diff(x) < 0):
        indexes = np.argsort(x)
        x = x[indexes]
        y = y[indexes]

    size = len(x)

    xdiff = np.diff(x)
    ydiff = np.diff(y)

    # allocate buffer matrices
    Li = np.empty(size)
    Li_1 = np.empty(size-1)
    z = np.empty(size)

    # fill diagonals Li and Li-1 and solve [L][y] = [B]
    Li[0] = sqrt(2*xdiff[0])
    Li_1[0] = 0.0
    B0 = 0.0 # natural boundary
    z[0] = B0 / Li[0]

    for i in range(1, size-1, 1):
        Li_1[i] = xdiff[i-1] / Li[i-1]
        Li[i] = sqrt(2*(xdiff[i-1]+xdiff[i]) - Li_1[i-1] * Li_1[i-1])
        Bi = 6*(ydiff[i]/xdiff[i] - ydiff[i-1]/xdiff[i-1])
        z[i] = (Bi - Li_1[i-1]*z[i-1])/Li[i]

    i = size - 1
    Li_1[i-1] = xdiff[-1] / Li[i-1]
    Li[i] = sqrt(2*xdiff[-1] - Li_1[i-1] * Li_1[i-1])
    Bi = 0.0 # natural boundary
    z[i] = (Bi - Li_1[i-1]*z[i-1])/Li[i]

    # solve [L.T][x] = [y]
    i = size-1
    z[i] = z[i] / Li[i]
    for i in range(size-2, -1, -1):
        z[i] = (z[i] - Li_1[i-1]*z[i+1])/Li[i]

    # find index
    index = x.searchsorted(x0)
    np.clip(index, 1, size-1, index)

    xi1, xi0 = x[index], x[index-1]
    yi1, yi0 = y[index], y[index-1]
    zi1, zi0 = z[index], z[index-1]
    hi1 = xi1 - xi0

    # calculate cubic
    f0 = zi0/(6*hi1)*(xi1-x0)**3 + \
         zi1/(6*hi1)*(x0-xi0)**3 + \
         (yi1/hi1 - zi1*hi1/6)*(x0-xi0) + \
         (yi0/hi1 - zi0*hi1/6)*(xi1-x0)
    return f0

if __name__ == '__main__':
    import matplotlib.pyplot as plt
    x = np.linspace(0, 10, 11)
    y = np.sin(x)
    plt.scatter(x, y)

    x_new = np.linspace(0, 10, 201)
    plt.plot(x_new, cubic_interp1d(x_new, x, y))

    plt.show()