我试图用Python重现B样条的Mathematica示例.
mathematica示例的代码读取
pts = {{0, 0}, {0, 2}, {2, 3}, {4, 0}, {6, 3}, {8, 2}, {8, 0}};
Graphics[{BSplineCurve[pts, SplineKnots -> {0, 0, 0, 0, 2, 3, 4, 6, 6, 6, 6}], Green, Line[pts], Red, Point[pts]}]
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并产生我所期望的.现在我尝试用Python/scipy做同样的事情:
import numpy as np
import matplotlib.pyplot as plt
import scipy.interpolate as si
points = np.array([[0, 0], [0, 2], [2, 3], [4, 0], [6, 3], [8, 2], [8, 0]])
x = points[:,0]
y = points[:,1]
t = range(len(x))
knots = [2, 3, 4]
ipl_t …Run Code Online (Sandbox Code Playgroud)