python – 返回3D scipy.spatial.Delaunay的曲面三角形

我有这个问题.我尝试通过scipy.spatial.Delaunay对点云进行三角测量.我用了：

tri = Delaunay(points) # points: np.array() of 3d points 
indices = tri.simplices
vertices = points[indices]

但是,这段代码返回了四面体.怎么可能只返回表面三角形？

谢谢

解决方法:

要使其像代码形式一样工作,您必须将曲面参数化为2D.例如,在球(r,theta,psi)的情况下,半径是恒定的(将其丢弃),并且点由(θ,psi)给出,即2D.

Scipy Delaunay是N维三角剖分,因此如果你给3D点,它会返回3D对象.给它2D点并返回2D对象.

下面是我用于为openSCAD创建多面体的脚本. U和V是我的参数化(x和y),这些是我给Delaunay的坐标.请注意,现在“Delaunay三角剖分属性”仅适用于u,v坐标(角度在uv-spatial中最大化而不是xyz -space等).

这个例子是从http://matplotlib.org/1.3.1/mpl_toolkits/mplot3d/tutorial.html开始的修改过的副本,最初使用Triangulation函数(最终映射到Delaunay？)

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

# u, v are parameterisation variables
u = np.array([0,0,0.5,1,1]) 
v = np.array([0,1,0.5,0,1]) 

x = u
y = v
z = np.array([0,0,1,0,0])

# Triangulate parameter space to determine the triangles
#tri = mtri.Triangulation(u, v)
tri = Delaunay(np.array([u,v]).T)

print 'polyhedron(faces = ['
#for vert in tri.triangles:
for vert in tri.simplices:
    print '[%d,%d,%d],' % (vert[0],vert[1],vert[2]),
print '], points = ['
for i in range(x.shape[0]):
    print '[%f,%f,%f],' % (x[i], y[i], z[i]),
print ']);'


fig = plt.figure()
ax = fig.add_subplot(1, 1, 1, projection='3d')

# The triangles in parameter space determine which x, y, z points are
# connected by an edge
#ax.plot_trisurf(x, y, z, triangles=tri.triangles, cmap=plt.cm.Spectral)
ax.plot_trisurf(x, y, z, triangles=tri.simplices, cmap=plt.cm.Spectral)


plt.show()

下面是(稍微更结构化)文本输出：

polyhedron(
    faces = [[2,1,0], [3,2,0], [4,2,3], [2,4,1], ], 

    points = [[0.000000,0.000000,0.000000], 
              [0.000000,1.000000,0.000000], 
              [0.500000,0.500000,1.000000], 
              [1.000000,0.000000,0.000000], 
              [1.000000,1.000000,0.000000], ]);

标签：points,python,scipy,cloud,delaunay
来源： https://codeday.me/bug/20191004/1854264.html