<p>使用我为<a href="https://stackoverflow.com/questions/34803197/fast-b-spline-algorithm-with-numpy-scipy">another question i asked here.</a>编写的函数</p>
<p>在我的问题中,我正在寻找用scipy计算bspline的方法(这就是我是如何偶然发现你的问题的)。</p>
<p>在沉迷了很久之后,我想到了下面的功能。它将评估20度以内的任何曲线(远远超出我们的需要)。在速度方面,我测试了100000个样本,用了0.017s</p>
<pre><code>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
</code></pre>
<p>开放曲线和周期曲线的结果:</p>
<pre><code>cv = np.array([[ 50., 25.],
[ 59., 12.],
[ 50., 10.],
[ 57., 2.],
[ 40., 4.],
[ 40., 14.]])
</code></pre>
<p><a href="https://i.stack.imgur.com/q04Ww.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q04Ww.png" alt="Periodic (closed) curve"/></a>
<a href="https://i.stack.imgur.com/UipeZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UipeZ.png" alt="Opened curve"/></a></p>