总结问题:您想通过多个点插值出平滑曲线。对于2D空间中的每个点,您都有坐标以及定义该点处曲线切线的角度。
一种解决方案可能是使用三阶
贝塞尔曲线。这样的曲线由4个点定义;两个端点是绘图中的连续两个点,另外两个中间点定义了曲线的方向。贝塞尔曲线经常用于图形软件中,并且也被matplotlib内部用于绘制路径。
![enter image description here](https://istack.dev59.com/TCWsQ.webp)
因此,我们可以定义一条贝塞尔曲线,其两个中间点位于沿着每个点的切线给定的角度。本质上没有提示这两个中间点应该位于该切线的何处,因此我们可以选择一些任意距离从两个端点开始。这就是下面代码中所谓的r
。选择一个好的参数r
的值对于获得平滑曲线非常关键。
import numpy as np
from scipy.special import binom
import matplotlib.pyplot as plt
bernstein = lambda n, k, t: binom(n,k)* t**k * (1.-t)**(n-k)
def bezier(points, num=200):
N = len(points)
t = np.linspace(0, 1, num=num)
curve = np.zeros((num, 2))
for i in range(N):
curve += np.outer(bernstein(N - 1, i, t), points[i])
return curve
class Segment():
def __init__(self, p1, p2, angle1, angle2, **kw):
self.p1 = p1; self.p2 = p2
self.angle1 = angle1; self.angle2 = angle2
self.numpoints = kw.get("numpoints", 100)
method = kw.get("method", "const")
if method=="const":
self.r = kw.get("r", 1.)
else:
r = kw.get("r", 0.3)
d = np.sqrt(np.sum((self.p2-self.p1)**2))
self.r = r*d
self.p = np.zeros((4,2))
self.p[0,:] = self.p1[:]
self.p[3,:] = self.p2[:]
self.calc_intermediate_points(self.r)
def calc_intermediate_points(self,r):
self.p[1,:] = self.p1 + np.array([self.r*np.cos(self.angle1),
self.r*np.sin(self.angle1)])
self.p[2,:] = self.p2 + np.array([self.r*np.cos(self.angle2+np.pi),
self.r*np.sin(self.angle2+np.pi)])
self.curve = bezier(self.p,self.numpoints)
def get_curve(points, **kw):
segments = []
for i in range(len(points)-1):
seg = Segment(points[i,:2], points[i+1,:2], points[i,2],points[i+1,2],**kw)
segments.append(seg)
curve = np.concatenate([s.curve for s in segments])
return segments, curve
def plot_point(ax, xy, angle, r=0.3):
ax.plot([xy[0]],[xy[1]], marker="o", ms=9, alpha=0.5, color="indigo")
p = xy + np.array([r*np.cos(angle),r*np.sin(angle)])
ax.plot([xy[0],p[0]], [xy[1],p[1]], color="limegreen")
if __name__ == "__main__":
points =np.array([[ 6.0, 0.5, 1.5],
[ 5.4, 1.2, 2.2],
[ 5.0, 1.7, 2.6],
[ 2.8, 2.4, 2.1],
[ 1.3, 3.2, 1.6],
[ 1.9, 3.9,-0.2],
[ 4.0, 3.0, 0.2],
[ 5.1, 3.7, 1.4]])
fig, ax = plt.subplots()
for point in points:
plot_point(ax, point[:2],point[2], r=0.1)
s1, c1 = get_curve(points, method="const", r=0.7)
ax.plot(c1[:,0], c1[:,1], color="crimson", zorder=0, label="const 0.7 units")
s2, c2 = get_curve(points, method="prop", r=0.3)
ax.plot(c2[:,0], c2[:,1], color="gold", zorder=0, label="prop 30% of distance")
plt.legend()
plt.show()
![enter image description here](https://istack.dev59.com/5P02B.webp)
在上面的图表中,比较了两种情况。一种是当
r
保持不变时为
0.7
个单位,另一种是当
r
相对于两点之间的距离增加30%时。
scipy.interpolate.KroghInterpolator
这样的函数可能会帮助你。然而,在这种情况下,它将失败。 - ImportanceOfBeingErnest