带 scipy Delaunay 的环

问题描述

我尝试绘制一个表示环的 3d 实体.我使用了 scipy 模块和 Delaunay 来进行计算.不幸的是，该图显示的是 3d 圆柱体而不是环面.有人知道如何修改代码吗?scipy 是正确的模块吗?我可以使用具有矩形形状的 Delaunay 吗?提前致谢！

将 numpy 导入为 np导入 matplotlib.pyplot 作为 plt从 mpl_toolkits.mplot3d 导入 Axes3D从 scipy.spatial 进口德劳内点数 = 50theta = np.linspace(0,2*np.pi,points)半径_中 = 7.5半径内 = 7半径_外 = 8x_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))#德劳内CH = Delaunay(M).convex_hullx,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()

解决方案

如评论中所述，凸包是凸形状，因此不能表示环.但是，凹壳(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接半径大于某个值(由 alpha 参数定义)的三角形(在 3D 情况下为四面体).

i try to draw a 3d solid that represents an annulus. I have used the scipy module and Delaunay to do the calculation.Unfortunately the plot shows a 3d cylinder and not an annulus. Has somebody an idea how to modify the code? Is scipy the right module? Can i use Delaunay with retangular shapes?thanks in advance!

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()

解决方案

As noted in the comments the convex hull is a convex shape and therefore cannot represent an annulus. However, the concept of the concave hull (also known as the alpha-shape) is probably appropriate for your needs. Basically, the alpha-shape removes from the Delaunay triangulation the triangles (tetrahedra in your 3D case) that have a circumradius greater than some value (defined by the alpha parameter).

This answer provides an implementation of the alpha-shape surface (i.e., the outer boundary) for 3D points. Using the alpha_shape_3D function from that answer, with an alpha value of 3, resulted in the figure below.

The following two lines in the code (replacing the assignment to CH and the plot function) do the job.

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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