Splines With Python (using Control Knots And Endpoints)

I'm trying to do something like the following (image extracted from wikipedia) #!/usr/bin/env python from scipy import interpolate import numpy as np import matplotlib.pyplot as p

Solution 1:

If what you want is to evaluate a bspline, you need to figure out the appropriate knot vector for your spline and then manually rebuild tck to fit your needs.

tck stands for knots t + coefficients c + curve degree k. splrep calculates tck for a cubic curve that passes through the given control points. So you can't use it for what you want.

The function below will show you my solution for a similar question I asked some time ago., adapted for what you want.

Fun fact: the code works for curves of any dimension (1D,2D,3D,...,nD)

import numpy as np
import scipy.interpolate as si


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

        cv :      Array ov control vertices
        n  :      Number of samples to return
        degree:   Curve degree
    """
    cv = np.asarray(cv)
    count = cv.shape[0]

    # Prevent degree from exceeding count-1, otherwise splev will crash
    degree = np.clip(degree,1,count-1)

    # Calculate knot vector
    kv = np.array([0]*degree + range(count-degree+1) + [count-degree]*degree,dtype='int')

    # Calculate query range
    u = np.linspace(0,(count-degree),n)

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

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(1,5):
    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()

Result:

Solution 2:

In this IPython Notebook http://nbviewer.ipython.org/github/empet/geom_modeling/blob/master/FP-Bezier-Bspline.ipynb you can find a detailed description of data involved in generating a B-spline curve, as well as the Python implementation of the de Boor algorithm.

Solution 3:

I just found something really interesting with the answer that I need with a bézier in this link. Then I used the code to try on my own. It's working fine apparently. This is my implementation:

#! /usr/bin/python# -*- coding: utf-8 -*-import numpy as np
import matplotlib.pyplot as plt
from scipy.special import binom

defBernstein(n, k):
    """Bernstein polynomial.

    """
    coeff = binom(n, k)

    def_bpoly(x):
        return coeff * x ** k * (1 - x) ** (n - k)

    return _bpoly


defBezier(points, num=200):
    """Build Bézier curve from points.

    """
    N = len(points)
    t = np.linspace(0, 1, num=num)
    curve = np.zeros((num, 2))
    for ii inrange(N):
        curve += np.outer(Bernstein(N - 1, ii)(t), points[ii])
    return curve
xp = np.array([2,3,4,5])
yp = np.array([2,1,4,0])
x, y = Bezier(list(zip(xp, yp))).T

plt.plot(x,y)
plt.plot(xp,yp,"ro")
plt.plot(xp,yp,"b--")

plt.show()

And an image for the example.

The red points represent the control points. That's it =)

Solution 4:

I think the problem is to do with you knots vector. It seems to cause problems if you select too many knots, it needs to have some data point between the knots. This question solves the problem Bug (?) on selecting knots on scipy.insterpolate's splrep function

#!/usr/bin/env pythonfrom scipy import interpolate
import numpy as np
import matplotlib.pyplot as plt

# sampling
x = np.linspace(0, 10, 10)
y = np.sin(x)

# spline trough all the sampled points
tck = interpolate.splrep(x, y)
print tck
x2 = np.linspace(0, 10, 200)
y2 = interpolate.splev(x2, tck)

# spline with all the middle points as knots (not working yet)
knots = np.asarray(x[1:-1])  # it should be something like this#knots = np.array([x[1]])  # not working with above line and just seeing what this line does
nknots = 5
idx_knots = (np.arange(1,len(x)-1,(len(x)-2)/np.double(nknots))).astype('int')
knots = x[idx_knots]
print knots

weights = np.concatenate(([1],np.ones(x.shape[0]-2)*.01,[1]))
tck = interpolate.splrep(x, y,  t=knots, w=weights)
x3 = np.linspace(0, 10, 200)
y3 = interpolate.splev(x2, tck)

# plot
plt.plot(x, y, 'go', x2, y2, 'b', x3, y3,'r')
plt.show()

it seems to work choosing 5 knots, 6 give wierd results, any more gives the errors.

Solution 5:

your example function is periodic and you need to add the per=True option to the interpolate.splrep method.

knots = x[1:-1]
weights = np.concatenate(([1],np.ones(x.shape[0]-2)*.01,[1]))
tck = interpolate.splrep(x, y, t=knots, w=weights, per=True)

This give me the following:

Edit: This also explains why it did work with knots = x[-2:2] which is a non-periodic subset of full range.