Getting spline equation from UnivariateSpline object

Question

I'm using UnivariateSpline to construct piecewise polynomials for some data that I have. I would then like to use these splines in other programs (either in C or FORTRAN) and so I would like to understand the equation behind the generated spline.

Here is my code:

import numpy as np
import scipy as sp
from  scipy.interpolate import UnivariateSpline
import matplotlib.pyplot as plt
import bisect

data = np.loadtxt('test_C12H26.dat')
Tmid = 800.0
print "Tmid", Tmid
nmid = bisect.bisect(data[:,0],Tmid)
fig = plt.figure()
plt.plot(data[:,0], data[:,7],ls='',marker='o',markevery=20)
npts = len(data[:,0])
#print "npts", npts
w = np.ones(npts)
w[0] = 100
w[nmid] = 100
w[npts-1] = 100
spline1 = UnivariateSpline(data[:nmid,0],data[:nmid,7],s=1,w=w[:nmid])
coeffs = spline1.get_coeffs()
print coeffs
print spline1.get_knots()
print spline1.get_residual()
print coeffs[0] + coeffs[1] * (data[0,0] - data[0,0]) \
                + coeffs[2] * (data[0,0] - data[0,0])**2 \
                + coeffs[3] * (data[0,0] - data[0,0])**3, \
      data[0,7]
print coeffs[0] + coeffs[1] * (data[nmid,0] - data[0,0]) \
                + coeffs[2] * (data[nmid,0] - data[0,0])**2 \
                + coeffs[3] * (data[nmid,0] - data[0,0])**3, \
      data[nmid,7]

print Tmid,data[-1,0]
spline2 = UnivariateSpline(data[nmid-1:,0],data[nmid-1:,7],s=1,w=w[nmid-1:])
print spline2.get_coeffs()
print spline2.get_knots()
print spline2.get_residual()
plt.plot(data[:,0],spline1(data[:,0]))
plt.plot(data[:,0],spline2(data[:,0]))
plt.savefig('test.png')

And here is the resulting plot. I believe I have valid splines for each interval but it looks like my spline equation is not correct... I can't find any reference to what it is supposed to be in the scipy documentation. Anybody knows? Thanks !

Samellasameness

Answer 1

The scipy documentation does not have anything to say about how one can take the coefficients and manually generate the spline curve. However, it is possible to figure out how to do this from the existing literature on B-splines. The following function bspleval shows how to construct the B-spline basis functions (the matrix B in the code), from which one can easily generate the spline curve by multiplying the coefficients with the highest-order basis functions and summing:

def bspleval(x, knots, coeffs, order, debug=False):
    '''
    Evaluate a B-spline at a set of points.

    Parameters
    ----------
    x : list or ndarray
        The set of points at which to evaluate the spline.
    knots : list or ndarray
        The set of knots used to define the spline.
    coeffs : list of ndarray
        The set of spline coefficients.
    order : int
        The order of the spline.

    Returns
    -------
    y : ndarray
        The value of the spline at each point in x.
    '''

    k = order
    t = knots
    m = alen(t)
    npts = alen(x)
    B = zeros((m-1,k+1,npts))

    if debug:
        print('k=%i, m=%i, npts=%i' % (k, m, npts))
        print('t=', t)
        print('coeffs=', coeffs)

    ## Create the zero-order B-spline basis functions.
    for i in range(m-1):
        B[i,0,:] = float64(logical_and(x >= t[i], x < t[i+1]))

    if (k == 0):
        B[m-2,0,-1] = 1.0

    ## Next iteratively define the higher-order basis functions, working from lower order to higher.
    for j in range(1,k+1):
        for i in range(m-j-1):
            if (t[i+j] - t[i] == 0.0):
                first_term = 0.0
            else:
                first_term = ((x - t[i]) / (t[i+j] - t[i])) * B[i,j-1,:]

            if (t[i+j+1] - t[i+1] == 0.0):
                second_term = 0.0
            else:
                second_term = ((t[i+j+1] - x) / (t[i+j+1] - t[i+1])) * B[i+1,j-1,:]

            B[i,j,:] = first_term + second_term
        B[m-j-2,j,-1] = 1.0

    if debug:
        plt.figure()
        for i in range(m-1):
            plt.plot(x, B[i,k,:])
        plt.title('B-spline basis functions')

    ## Evaluate the spline by multiplying the coefficients with the highest-order basis functions.
    y = zeros(npts)
    for i in range(m-k-1):
        y += coeffs[i] * B[i,k,:]

    if debug:
        plt.figure()
        plt.plot(x, y)
        plt.title('spline curve')
        plt.show()

    return(y)

To give an example of how this can be used with Scipy's existing univariate spline functions, the following is an example script. This takes the input data and uses Scipy's functional and also its object-oriented approach to spline fitting. Taking the coefficients and knot points from either of the two and using these as inputs to our manually-calculated routine bspleval, we reproduce the same curve that they do. Note that the difference between the manually evaluated curve and Scipy's evaluation method is so small that it is almost certainly floating-point noise.

x = array([-273.0, -176.4, -79.8, 16.9, 113.5, 210.1, 306.8, 403.4, 500.0])
y = array([2.25927498e-53, 2.56028619e-03, 8.64512988e-01, 6.27456769e+00, 1.73894734e+01,
        3.29052124e+01, 5.14612316e+01, 7.20531200e+01, 9.40718450e+01])

x_nodes = array([-273.0, -263.5, -234.8, -187.1, -120.3, -34.4, 70.6, 194.6, 337.8, 500.0])
y_nodes = array([2.25927498e-53, 3.83520726e-46, 8.46685318e-11, 6.10568083e-04, 1.82380809e-01,
                2.66344008e+00, 1.18164677e+01, 3.01811501e+01, 5.78812583e+01, 9.40718450e+01])

## Now get scipy's spline fit.
k = 3
tck = splrep(x_nodes, y_nodes, k=k, s=0)
knots = tck[0]
coeffs = tck[1]
print('knot points=', knots)
print('coefficients=', coeffs)

## Now try scipy's object-oriented version. The result is exactly the same as "tck": the knots are the
## same and the coeffs are the same, they are just queried in a different way.
uspline = UnivariateSpline(x_nodes, y_nodes, s=0)
uspline_knots = uspline.get_knots()
uspline_coeffs = uspline.get_coeffs()

## Here are scipy's native spline evaluation methods. Again, "ytck" and "y_uspline" are exactly equal.
ytck = splev(x, tck)
y_uspline = uspline(x)
y_knots = uspline(knots)

## Now let's try our manually-calculated evaluation function.
y_eval = bspleval(x, knots, coeffs, k, debug=False)

plt.plot(x, ytck, label='tck')
plt.plot(x, y_uspline, label='uspline')
plt.plot(x, y_eval, label='manual')

## Next plot the knots and nodes.
plt.plot(x_nodes, y_nodes, 'ko', markersize=7, label='input nodes')            ## nodes
plt.plot(knots, y_knots, 'mo', markersize=5, label='tck knots')    ## knots
plt.xlim((-300.0,530.0))
plt.legend(loc='best', prop={'size':14})

plt.figure()
plt.title('difference')
plt.plot(x, ytck-y_uspline, label='tck-uspl')
plt.plot(x, ytck-y_eval, label='tck-manual')
plt.legend(loc='best', prop={'size':14})

plt.show()

Byssus

Nice work, however, it seems to fail for the following example: x = np.linspace(0, 100, 1000), d = np.sin(x * 0.5) + 2 + np.cos(x * 0.1), x_sample = x[::50], d_sample = d[::50], s = UnivariateSpline(x_sample, d_sample, k=3, s=0.005), spl = bspleval(x, s.get_knots(), s.get_coeffs(), 3, debug=False). The resulting plot shows that d and spl look quite different. Any ideas? – Onym

@Cleb: I can see a couple of things going on there in your example. First, you usex_sample and d_sample as inputs to the spline fit, but you compare it with x and d. If you let x_sample = x and d_sample=d, then you get a better fit. But there is still an error. Next, if you reduce the smoothing factor from s=0.005 to something much smaller (say s=1.0e-15) then you get a good fit. – Byssus

Thanks for the reply; will check it later on. If you have a solution for this question based on the answer here, feel free to add on :) – Onym

Answer 2

The coefficients given by get_coeffs are B-spline (Basis spline) coefficients, described here: B-spline (Wikipedia)

Probably whatever other program/language you will be using has an implementation. Supply the knot locations and coefficients, and you should be all set.

Southern