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I have created an open, clamped, cubic, b-spline on the set of data points xp and yp.

The spline is parametarized by the vector u which spans the domain of xp.

My goal is to determine the "y" coordinate of the b-spline at a given "x" coordinate in the domain of xp.

As is expected behavior when generating parametric curves, when I pass the value "4" into splev after calculating tck, the value for both the x and y coordinate corresponding to the parameter 4 is returned.

I am able to use Newton's method to determine the value of the parameter u at a given "x" coordinate; This is however indirect and requires more computation time than my final application can permit.

Can anyone suggest a more direct way to determine the "y" coordinate on the b-spline for a given "x"?

import numpy as np

import matplotlib.pyplot as plt

from scipy import interpolate

xp = [0., 0.71428571, 1.42857143, 2.14285714, 2.85714286, 3.57142857, 4.28571429, 5.]

yp = [0., -0.86217009, -2.4457478, -2.19354839, -2.32844575, -0.48680352, -0.41055718, -3.]

length = len(xp)

t = np.linspace(0., xp[-1], length - 2, endpoint=True)

t = np.append([0, 0, 0], t)

t = np.append(t, [xp[-1], xp[-1], xp[-1]])

tck = [t, [xp, yp], 3]

u = np.linspace(0, 5., 1000, endpoint=True)

out = interpolate.splev(u, tck)

x_value_in_xp_domain = 4.

y_value_out = interpolate.splev(x_value_in_xp_domain, tck)

plt.plot(xp, yp, linestyle='--', marker='o', color='purple')

plt.plot(out[0], out[1], color = 'teal')

plt.plot(x_value_in_xp_domain, y_value_out[1], marker='o', color = 'orangered')

plt.plot(y_value_out[0], y_value_out[1], marker='o', color = 'black')

plt.axvline(x=x_value_in_xp_domain, color = 'orangered')

plt.show()

Plot from above script

The below image shows the guide polygon and the b-spline generated by the above code. The orange point at x=4 corresponds the point I wish to directly determine the y value of the b-spline at. The black point is the value of the b-spline when the value of 4 is passed as a parameter.

providing a few useful references:

Fast b-spline algorithm with numpy/scipy

https://github.com/kawache/Python-B-spline-examples

https://pages.mtu.edu/~shene/COURSES/cs3621/NOTES/spline/B-spline/bspline-curve.html

http://web.mit.edu/hyperbook/Patrikalakis-Maekawa-Cho/node17.html

1 Answer

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by (38.4k points)

You can use roots() method of PPoly, that is faster than Newton, and it will give you all the solutions in case if  there is more than one.

sx = interpolate.BSpline(t, xp, 3)

sy = interpolate.BSpline(t, yp, 3)

x0 = 4

u0 = interpolate.PPoly.from_spline((sx.t, sx.c - x0, 3)).roots()

sy(u0)

plt.plot(xp, yp, linestyle='--', marker='o', color='purple')

plt.plot(out[0], out[1], color = 'teal')

plt.plot(sx(u0), sy(u0), 'o')

plt.show()

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