B-spline interpolation with Python

Question

I am trying to reproduce a Mathematica example for a B-spline with Python.

The code of the mathematica example reads

pts = {{0, 0}, {0, 2}, {2, 3}, {4, 0         


        

Answer 1

I was able to recreate the Mathematica example I asked about in the previous post using Python/scipy. Here's the result:

B-Spline, Aperiodic

The trick was to either intercept the coefficients, i.e. element 1 of the tuple returned by scipy.interpolate.splrep, and to replace them with the control point values before handing them to scipy.interpolate.splev, or, if you are fine with creating the knots yourself, you can also do without splrep and create the entire tuple yourself.

What is strange about this all, though, is that, according to the manual, splrep returns (and splev expects) a tuple containing, among others, a spline coefficients vector with one coefficient per knot. However, according to all sources I found, a spline is defined as the weighted sum of the N_control_points basis splines, so I would expect the coefficients vector to have as many elements as control points, not knot positions.

In fact, when supplying splrep's result tuple with the coefficients vector modified as described above to scipy.interpolate.splev, it turns out that the first N_control_points of that vector actually are the expected coefficients for the N_control_points basis splines. The last degree + 1 elements of that vector seem to have no effect. I'm stumped as to why it's done this way. If anyone can clarify that, that would be great. Here's the source that generates the above plots:

import numpy as np
import matplotlib.pyplot as plt
import scipy.interpolate as si

points = [[0, 0], [0, 2], [2, 3], [4, 0], [6, 3], [8, 2], [8, 0]];
points = np.array(points)
x = points[:,0]
y = points[:,1]

t = range(len(points))
ipl_t = np.linspace(0.0, len(points) - 1, 100)

x_tup = si.splrep(t, x, k=3)
y_tup = si.splrep(t, y, k=3)

x_list = list(x_tup)
xl = x.tolist()
x_list[1] = xl + [0.0, 0.0, 0.0, 0.0]

y_list = list(y_tup)
yl = y.tolist()
y_list[1] = yl + [0.0, 0.0, 0.0, 0.0]

x_i = si.splev(ipl_t, x_list)
y_i = si.splev(ipl_t, y_list)

#==============================================================================
# Plot
#==============================================================================

fig = plt.figure()

ax = fig.add_subplot(231)
plt.plot(t, x, '-og')
plt.plot(ipl_t, x_i, 'r')
plt.xlim([0.0, max(t)])
plt.title('Splined x(t)')

ax = fig.add_subplot(232)
plt.plot(t, y, '-og')
plt.plot(ipl_t, y_i, 'r')
plt.xlim([0.0, max(t)])
plt.title('Splined y(t)')

ax = fig.add_subplot(233)
plt.plot(x, y, '-og')
plt.plot(x_i, y_i, 'r')
plt.xlim([min(x) - 0.3, max(x) + 0.3])
plt.ylim([min(y) - 0.3, max(y) + 0.3])
plt.title('Splined f(x(t), y(t))')

ax = fig.add_subplot(234)
for i in range(7):
    vec = np.zeros(11)
    vec[i] = 1.0
    x_list = list(x_tup)
    x_list[1] = vec.tolist()
    x_i = si.splev(ipl_t, x_list)
    plt.plot(ipl_t, x_i)
plt.xlim([0.0, max(t)])
plt.title('Basis splines')
plt.show()

B-Spline, Periodic

Now in order to create a closed curve like the following, which is another Mathematica example that can be found on the web,

it is necessary to set the per parameter in the splrep call, if you use that. After padding the list of control points with degree+1 values at the end, this seems to work well enough, as the images show.

The next peculiarity here, however, is that the first and the last degree elements in the coefficients vector have no effect, meaning that the control points must be put in the vector starting at the second position, i.e. position 1. Only then are the results ok. For degrees k=4 and k=5, that position even changes to position 2.

Here's the source for generating the closed curve:

import numpy as np
import matplotlib.pyplot as plt
import scipy.interpolate as si

points = [[-2, 2], [0, 1], [-2, 0], [0, -1], [-2, -2], [-4, -4], [2, -4], [4, 0], [2, 4], [-4, 4]]

degree = 3

points = points + points[0:degree + 1]
points = np.array(points)
n_points = len(points)
x = points[:,0]
y = points[:,1]

t = range(len(x))
ipl_t = np.linspace(1.0, len(points) - degree, 1000)

x_tup = si.splrep(t, x, k=degree, per=1)
y_tup = si.splrep(t, y, k=degree, per=1)
x_list = list(x_tup)
xl = x.tolist()
x_list[1] = [0.0] + xl + [0.0, 0.0, 0.0, 0.0]

y_list = list(y_tup)
yl = y.tolist()
y_list[1] = [0.0] + yl + [0.0, 0.0, 0.0, 0.0]

x_i = si.splev(ipl_t, x_list)
y_i = si.splev(ipl_t, y_list)

#==============================================================================
# Plot
#==============================================================================

fig = plt.figure()

ax = fig.add_subplot(231)
plt.plot(t, x, '-og')
plt.plot(ipl_t, x_i, 'r')
plt.xlim([0.0, max(t)])
plt.title('Splined x(t)')

ax = fig.add_subplot(232)
plt.plot(t, y, '-og')
plt.plot(ipl_t, y_i, 'r')
plt.xlim([0.0, max(t)])
plt.title('Splined y(t)')

ax = fig.add_subplot(233)
plt.plot(x, y, '-og')
plt.plot(x_i, y_i, 'r')
plt.xlim([min(x) - 0.3, max(x) + 0.3])
plt.ylim([min(y) - 0.3, max(y) + 0.3])
plt.title('Splined f(x(t), y(t))')

ax = fig.add_subplot(234)
for i in range(n_points - degree - 1):
    vec = np.zeros(11)
    vec[i] = 1.0
    x_list = list(x_tup)
    x_list[1] = vec.tolist()
    x_i = si.splev(ipl_t, x_list)
    plt.plot(ipl_t, x_i)
plt.xlim([0.0, 9.0])
plt.title('Periodic basis splines')

plt.show()

B-Spline, Periodic, Higher Degree

Lastly, there is an effect that I can not explain either, and this is when going to degree 5, there is a small discontinuity that appears in the splined curve, see the upper right panel, which is a close-up of that 'half-moon-with-nose-shape'. The source code that produces this is listed below.

import numpy as np
import matplotlib.pyplot as plt
import scipy.interpolate as si

points = [[-2, 2], [0, 1], [-2, 0], [0, -1], [-2, -2], [-4, -4], [2, -4], [4, 0], [2, 4], [-4, 4]]

degree = 5

points = points + points[0:degree + 1]
points = np.array(points)
n_points = len(points)
x = points[:,0]
y = points[:,1]

t = range(len(x))
ipl_t = np.linspace(1.0, len(points) - degree, 1000)

knots = np.linspace(-degree, len(points), len(points) + degree + 1).tolist()

xl = x.tolist()
coeffs_x = [0.0, 0.0] + xl + [0.0, 0.0, 0.0]

yl = y.tolist()
coeffs_y = [0.0, 0.0] + yl + [0.0, 0.0, 0.0]

x_i = si.splev(ipl_t, (knots, coeffs_x, degree))
y_i = si.splev(ipl_t, (knots, coeffs_y, degree))

#==============================================================================
# Plot
#==============================================================================

fig = plt.figure()

ax = fig.add_subplot(231)
plt.plot(t, x, '-og')
plt.plot(ipl_t, x_i, 'r')
plt.xlim([0.0, max(t)])
plt.title('Splined x(t)')

ax = fig.add_subplot(232)
plt.plot(t, y, '-og')
plt.plot(ipl_t, y_i, 'r')
plt.xlim([0.0, max(t)])
plt.title('Splined y(t)')

ax = fig.add_subplot(233)
plt.plot(x, y, '-og')
plt.plot(x_i, y_i, 'r')
plt.xlim([min(x) - 0.3, max(x) + 0.3])
plt.ylim([min(y) - 0.3, max(y) + 0.3])
plt.title('Splined f(x(t), y(t))')

ax = fig.add_subplot(234)
for i in range(n_points - degree - 1):
    vec = np.zeros(11)
    vec[i] = 1.0
    x_i = si.splev(ipl_t, (knots, vec, degree))
    plt.plot(ipl_t, x_i)
plt.xlim([0.0, 9.0])
plt.title('Periodic basis splines')

plt.show()

Given that b-splines are ubiquitous in the scientific community, and that scipy is such a comprehensive toolbox, and that I have not been able to find much about what I'm asking here on the web, leads me to believe I'm on the wrong track or overlooking something. Any help would be appreciated.

Answer 2

I believe scipy's fitpack Library is doing something more complicated than what Mathematica is doing. I was confused as to what was going on as well.

There is the smoothing parameter in these functions, and the default interpolation behavior is to try to make points go through lines. That's what this fitpack software does, so I guess scipy just inherited it? (http://www.netlib.org/fitpack/all -- I'm not sure this is the right fitpack)

I took some ideas from http://research.microsoft.com/en-us/um/people/ablake/contours/ and coded up your example with the B-splines in there.

import numpy

import matplotlib.pyplot as plt

# This is the basis function described in eq 3.6 in http://research.microsoft.com/en-us/um/people/ablake/contours/
def func(x, offset):
    out = numpy.ndarray((len(x)))

    for i, v in enumerate(x):
        s = v - offset

        if s >= 0 and s < 1:
            out[i] = s * s / 2.0
        elif s >= 1 and s < 2:
            out[i] = 3.0 / 4.0 - (s - 3.0 / 2.0) * (s - 3.0 / 2.0)
        elif s >= 2 and s < 3:
            out[i] = (s - 3.0) * (s - 3.0) / 2.0
        else:
            out[i] = 0.0

    return out

# We have 7 things to fit, so let's do 7 basis functions?
y = numpy.array([0, 2, 3, 0, 3, 2, 0])

# We need enough x points for all the basis functions... That's why the weird linspace max here
x = numpy.linspace(0, len(y) + 2, 100)

B = numpy.ndarray((len(x), len(y)))

for k in range(len(y)):
    B[:, k] = func(x, k)

plt.plot(x, B.dot(y))
# The x values in the next statement are the maximums of each basis function. I'm not sure at all this is right
plt.plot(numpy.array(range(len(y))) + 1.5, y, '-o')
plt.legend('B-spline', 'Control points')
plt.show()

for k in range(len(y)):
    plt.plot(x, B[:, k])
plt.title('Basis functions')
plt.show()

Anyway I think other folks have the same problems, have a look at: Behavior of scipy's splrep

Answer 3

Use this function i wrote for another question i asked here.

In my question i was looking for ways to calculate bsplines with scipy (this is how i actually stumbled upon your question).

After much obsession, i came up with the function below. It'll evaluate any curve up to the 20th degree (way more than we need). And speed wise i tested it for 100,000 samples and it took 0.017s

import numpy as np
import scipy.interpolate as si


def bspline(cv, n=100, degree=3, periodic=False):
    """ Calculate n samples on a bspline

        cv :      Array ov control vertices
        n  :      Number of samples to return
        degree:   Curve degree
        periodic: True - Curve is closed
                  False - Curve is open
    """

    # If periodic, extend the point array by count+degree+1
    cv = np.asarray(cv)
    count = len(cv)

    if periodic:
        factor, fraction = divmod(count+degree+1, count)
        cv = np.concatenate((cv,) * factor + (cv[:fraction],))
        count = len(cv)
        degree = np.clip(degree,1,degree)

    # If opened, prevent degree from exceeding count-1
    else:
        degree = np.clip(degree,1,count-1)


    # Calculate knot vector
    kv = None
    if periodic:
        kv = np.arange(0-degree,count+degree+degree-1,dtype='int')
    else:
        kv = np.concatenate(([0]*degree, np.arange(count-degree+1), [count-degree]*degree))


    # Calculate query range
    u = np.linspace(periodic,(count-degree),n)


    # Calculate result
    return np.array(si.splev(u, (kv,cv.T,degree))).T

Results for both open and periodic curves:

cv = np.array([[ 50.,  25.],
   [ 59.,  12.],
   [ 50.,  10.],
   [ 57.,   2.],
   [ 40.,   4.],
   [ 40.,   14.]])