Create BSpline from knots and coefficients

Question

How can a spline be created if only the points and the coefficients are known? I'm using scipy.interpolate.BSpline here, but am open to other standard packages as well. So basically I want to be able to give someone just those short arrays of coefficients for them to be able to recreate the fit to the data. See the failed red-dashed curve below.

import numpy as np
import matplotlib.pyplot as plt
from scipy.interpolate import BSpline, LSQUnivariateSpline

x = np.linspace(0, 10, 50) # x-data
y = np.exp(-(x-5)**2/4)    # y-data

# define the knot positions

t = [1, 2, 4, 5, 6, 8, 9]

# get spline fit

s1 = LSQUnivariateSpline(x, y, t)

x2 = np.linspace(0, 10, 200) # new x-grid
y2 = s1(x2) # evaluate spline on that new grid

# FAILED: try to construct BSpline using the knots and coefficients

k = s1.get_knots()
c = s1.get_coeffs()
s2 = BSpline(t,c,2)

# plotting

plt.plot(x, y, label='original')
plt.plot(t, s1(t),'o', label='knots')
plt.plot(x2, y2, '--', label='spline 1')
plt.plot(x2, s2(x2), 'r:', label='spline 2') 
plt.legend()

Answer 1

The fine print under get_knots says:

Internally, the knot vector contains 2*k additional boundary knots.

That means, to get a usable knot array from get_knots, one should add k copies of the left boundary knot at the beginning of array, and k copies of the right boundary knot at the end. Here k is the degree of the spline, which is usually 3 (you asked for LSQUnivariateSpline of default degree, so that's 3). So:

kn = s1.get_knots()
kn = 3*[kn[0]] + list(kn) + 3*[kn[-1]]
c = s1.get_coeffs()
s2 = BSpline(kn, c, 3)    # not "2" as in your sample; we are working with a cubic spline 

Now, the spline s2 is the same as s1:

Equivalently, kn = 4*[x[0]] + t + 4*[x[-1]] would work: your t list contains only interior knots, so x[0] and x[-1] are added, and then each repeated k times more.

The mathematical reason for the repetition is that B-splines need some room to get built, due to their inductive definition which requires (k-1)-degree splines to exist around every interval in which we define the kth degree spline.

Answer 2

Here is a slightly more compact way of doing it if you don't care too much about the details of the knot positions. The tk array is what you are looking for. Once tk is in hand the spline can be reproduced using the y=splev(x,tk,der=0) line.

import numpy as np
import matplotlib.pyplot as plt
from scipy.interpolate import splrep,splev
import matplotlib.pyplot as plt 

### Input data
x_arr = np.linspace(0, 10, 50)  # x-data
y_arr = np.exp(-(x_arr-5)**2/4)    # y-data
### Order of spline 
order = 3
### Make the spline 
tk = splrep(x_arr, y_arr, k=order) # Returns the knots and coefficents
### Evaluate the spline using the knots and coefficents on the domian x
x = np.linspace(0, 10, 1000) # new x-grid
y = splev(x, tk, der=0)
### Plot
f,ax=plt.subplots()
ax.scatter(x_arr, y_arr, label='original')
ax.plot(x,y,label='Spline')
ax.legend(fontsize=15)
plt.show()