带scipy Delaunay的环形

Question

我尝试绘制表示环形的三维实体。我已经使用scipy模块和Delaunay进行了计算。不幸的是，该图显示的是一个3d圆柱体，而不是一个环形空间。有人知道如何修改代码吗？scipy是正确的模块吗？我可以使用三角形的Delaunay吗？提前感谢！

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from scipy.spatial import Delaunay

points = 50

theta = np.linspace(0,2*np.pi,points)
radius_middle = 7.5
radius_inner = 7
radius_outer = 8

x_m_cartesian = radius_middle * np.cos(theta)
y_m_cartesian = radius_middle * np.sin(theta)
z_m_cartesian = np.zeros(points)
M_m = np.c_[x_m_cartesian,y_m_cartesian,z_m_cartesian]

x_i_cartesian = radius_inner * np.cos(theta)
y_i_cartesian = radius_inner * np.sin(theta)
z_i_cartesian = np.zeros(points)
M_i = np.c_[x_i_cartesian,y_i_cartesian,z_i_cartesian]

x1_m_cartesian = radius_middle * np.cos(theta)
y1_m_cartesian = radius_middle * np.sin(theta)
z1_m_cartesian = np.ones(points)
M1_m = np.c_[x1_m_cartesian,y1_m_cartesian,z1_m_cartesian]

x2_i_cartesian = radius_inner * np.cos(theta)
y2_i_cartesian = radius_inner * np.sin(theta)
z2_i_cartesian = np.ones(points)
M2_i = np.c_[x2_i_cartesian,y2_i_cartesian,z2_i_cartesian]

M = np.vstack((M_m,M_i,M1_m,M2_i))

# Delaunay
CH = Delaunay(M).convex_hull

x,y,z = M[:,0],M[:,1],M[:,2]

fig = plt.figure(figsize=(12,8))
ax = fig.add_subplot(111,projection='3d')
#ax.scatter(x[:,0],y[:,1],z[:,2])
ax.plot_trisurf(x,y,z,triangles=CH, shade=False, color='lightblue',lw=1, edgecolor='k')

plt.show()

Answer 1

正如评论中所指出的，凸壳是凸形的，因此不能表示环空。但是，concave hull (也称为alpha-shape)的概念可能适合您的需要。基本上，alpha形状从Delaunay三角剖分中删除了外圆半径大于某个值(由alpha参数定义)的三角形(在3D情况下为四面体)。

This answer为3D点提供了alpha形状曲面(即外部边界)的实现。使用该答案中的alpha_shape_3D函数，alpha值为3，结果如下图所示。

代码中的以下两行(替换了对CH和绘图函数的赋值)完成了这项工作。

vertices, edges, facets = alpha_shape_3D(pos=M, alpha=3.)

ax.plot_trisurf(x,y,z,triangles=facets, shade=False, color='lightblue',lw=1, edgecolor='k')

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