返回3D scipy.spatial.Delaunay的表面三角形

Question

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

Answer 1

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，半径是常数（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）。

该示例是来自http://matplotlib.org/1.3.1/mpl_toolkits/mplot3d/tutorial.html的修改后的副本，该副本最初使用三角测量功能（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], ]);

Answer 2

看起来你想要计算点云的convex hull。我想这就是你想要做的：

from scipy.spatial import ConvexHull

hull = ConvexHull(points)
indices = hull.simplices
vertices = points[indices]

Answer 3

按照Jaime的回答，但用一个例子详细说明：

import matplotlib as mpl
import matplotlib.pyplot as plt
import mpl_toolkits.mplot3d as a3
import numpy as np
import scipy as sp
from scipy import spatial as sp_spatial


def icosahedron():
    h = 0.5*(1+np.sqrt(5))
    p1 = np.array([[0, 1, h], [0, 1, -h], [0, -1, h], [0, -1, -h]])
    p2 = p1[:, [1, 2, 0]]
    p3 = p1[:, [2, 0, 1]]
    return np.vstack((p1, p2, p3))


def cube():
    points = np.array([
        [0, 0, 0], [0, 0, 1], [0, 1, 0], [0, 1, 1],
        [1, 0, 0], [1, 0, 1], [1, 1, 0], [1, 1, 1],
    ])
    return points


points = icosahedron()
# points = cube()

hull = sp_spatial.ConvexHull(points)
indices = hull.simplices
faces = points[indices]

print('area: ', hull.area)
print('volume: ', hull.volume)


fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.dist = 30
ax.azim = -140
ax.set_xlim([0, 2])
ax.set_ylim([0, 2])
ax.set_zlim([0, 2])
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')

for f in faces:
    face = a3.art3d.Poly3DCollection([f])
    face.set_color(mpl.colors.rgb2hex(sp.rand(3)))
    face.set_edgecolor('k')
    face.set_alpha(0.5)
    ax.add_collection3d(face)

plt.show()

哪个应该描绘如下图：