python - Fast b-spline algorithm with numpy/scipy

Question

I need to compute bspline curves in python. I looked into scipy.interpolate.splprep and a few other scipy modules but couldn't find anything that readily gave me what I needed. So i wrote my own module below. The code works fine, but it is slow (test function runs in 0.03s, which seems like a lot considering i'm only asking for 100 samples with 6 control vertices).

Is there a way to simplify the code below with a few scipy module calls, which presumably would speed it up? And if not, what could i do to my code to improve its performance?

import numpy as np

# cv = np.array of 3d control vertices
# n = number of samples (default: 100)
# d = curve degree (default: cubic)
# closed = is the curve closed (periodic) or open? (default: open)
def bspline(cv, n=100, d=3, closed=False):

    # Create a range of u values
    count = len(cv)
    knots = None
    u = None
    if not closed:
        u = np.arange(0,n,dtype='float')/(n-1) * (count-d)
        knots = np.array([0]*d + range(count-d+1) + [count-d]*d,dtype='int')
    else:
        u = ((np.arange(0,n,dtype='float')/(n-1) * count) - (0.5 * (d-1))) % count # keep u=0 relative to 1st cv
        knots = np.arange(0-d,count+d+d-1,dtype='int')


    # Simple Cox - DeBoor recursion
    def coxDeBoor(u, k, d):

        # Test for end conditions
        if (d == 0):
            if (knots[k] <= u and u < knots[k+1]):
                return 1
            return 0

        Den1 = knots[k+d] - knots[k]
        Den2 = knots[k+d+1] - knots[k+1]
        Eq1  = 0;
        Eq2  = 0;

        if Den1 > 0:
            Eq1 = ((u-knots[k]) / Den1) * coxDeBoor(u,k,(d-1))
        if Den2 > 0:
            Eq2 = ((knots[k+d+1]-u) / Den2) * coxDeBoor(u,(k+1),(d-1))

        return Eq1 + Eq2


    # Sample the curve at each u value
    samples = np.zeros((n,3))
    for i in xrange(n):
        if not closed:
            if u[i] == count-d:
                samples[i] = np.array(cv[-1])
            else:
                for k in xrange(count):
                    samples[i] += coxDeBoor(u[i],k,d) * cv[k]

        else:
            for k in xrange(count+d):
                samples[i] += coxDeBoor(u[i],k,d) * cv[k%count]


    return samples




if __name__ == "__main__":
    import matplotlib.pyplot as plt
    def test(closed):
        cv = np.array([[ 50.,  25.,  -0.],
               [ 59.,  12.,  -0.],
               [ 50.,  10.,   0.],
               [ 57.,   2.,   0.],
               [ 40.,   4.,   0.],
               [ 40.,   14.,  -0.]])

        p = bspline(cv,closed=closed)
        x,y,z = p.T
        cv = cv.T
        plt.plot(cv[0],cv[1], 'o-', label='Control Points')
        plt.plot(x,y,'k-',label='Curve')
        plt.minorticks_on()
        plt.legend()
        plt.xlabel('x')
        plt.ylabel('y')
        plt.xlim(35, 70)
        plt.ylim(0, 30)
        plt.gca().set_aspect('equal', adjustable='box')
        plt.show()

    test(False)

The two images below shows what my code returns with both closed conditions:

Answer 1

So after obsessing a lot about my question, and much research, i finally have my answer. Everything is available in scipy , and i'm putting my code here so hopefully someone else can find this useful.

The function takes in an array of N-d points, a curve degree, a periodic state (opened or closed) and will return n samples along that curve. There are ways to make sure the curve samples are equidistant but for the time being i'll focus on this question, as it is all about speed.

Worthy of note: I can't seem to be able to go beyond a curve of 20th degree. Granted, that's overkill already, but i figured it's worth mentioning.

Also worthy of note: on my machine the code below can calculate 100,000 samples in 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)
    else:
        kv = np.clip(np.arange(count+degree+1)-degree,0,count-degree)

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


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

To test it:

import matplotlib.pyplot as plt
colors = ('b', 'g', 'r', 'c', 'm', 'y', 'k')

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

plt.plot(cv[:,0],cv[:,1], 'o-', label='Control Points')

for d in range(1,21):
    p = bspline(cv,n=100,degree=d,periodic=True)
    x,y = p.T
    plt.plot(x,y,'k-',label='Degree %s'%d,color=colors[d%len(colors)])

plt.minorticks_on()
plt.legend()
plt.xlabel('x')
plt.ylabel('y')
plt.xlim(35, 70)
plt.ylim(0, 30)
plt.gca().set_aspect('equal', adjustable='box')
plt.show()

Results for both opened or periodic curves:

ADDENDUM

As of scipy-0.19.0 there is a new scipy.interpolate.BSpline function that can be used.

import numpy as np
import scipy.interpolate as si
def scipy_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
    """
    cv = np.asarray(cv)
    count = cv.shape[0]

    # Closed curve
    if periodic:
        kv = np.arange(-degree,count+degree+1)
        factor, fraction = divmod(count+degree+1, count)
        cv = np.roll(np.concatenate((cv,) * factor + (cv[:fraction],)),-1,axis=0)
        degree = np.clip(degree,1,degree)

    # Opened curve
    else:
        degree = np.clip(degree,1,count-1)
        kv = np.clip(np.arange(count+degree+1)-degree,0,count-degree)

    # Return samples
    max_param = count - (degree * (1-periodic))
    spl = si.BSpline(kv, cv, degree)
    return spl(np.linspace(0,max_param,n))

Testing for equivalency:

p1 = bspline(cv,n=10**6,degree=3,periodic=True) # 1 million samples: 0.0882 sec
p2 = scipy_bspline(cv,n=10**6,degree=3,periodic=True) # 1 million samples: 0.0789 sec
print np.allclose(p1,p2) # returns True