使用 matplotlib 对多边形进行三角剖分

Question

我尝试使用 matplot.Delaunay 对一个简单的二维多边形进行三角剖分...这里的问题是我需要正三角形。多边形是由numpy随机创建的，也许Delaunay不是这样。

import matplotlib.delaunay as triang
import pylab
import numpy

# 10 random points (x,y) in the plane
x,y =  numpy.array(numpy.random.standard_normal((2,10)))
cens,edg,tri,neig = triang.delaunay(x,y)

for t in tri:
 # t[0], t[1], t[2] are the points indexes of the triangle
 t_i = [t[0], t[1], t[2], t[0]]
 pylab.plot(x[t_i],y[t_i])


pylab.plot(x,y,'^')
pylab.show()

Answer 1

我想，正三角形的三角剖分对于随机多边形来说并不是一项微不足道的任务。但是，如果您只想使用正三角形，则必须手动定义点的坐标。就像这个例子：

import matplotlib.pyplot as plt
import matplotlib.tri as tri
import numpy as np
import math

# Creating a Triangulation without specifying the triangles results in the
# Delaunay triangulation of the points.

# First create the x and y coordinates of the points.
n_angles = 36
n_radii = 8
min_radius = 0.25
radii = np.linspace(min_radius, 0.95, n_radii)

angles = np.linspace(0, 2*math.pi, n_angles, endpoint=False)
angles = np.repeat(angles[..., np.newaxis], n_radii, axis=1)
angles[:, 1::2] += math.pi/n_angles

x = (radii*np.cos(angles)).flatten()
y = (radii*np.sin(angles)).flatten()

# Create the Triangulation; no triangles so Delaunay triangulation created.
triang = tri.Triangulation(x, y)

# Mask off unwanted triangles.
xmid = x[triang.triangles].mean(axis=1)
ymid = y[triang.triangles].mean(axis=1)
mask = np.where(xmid*xmid + ymid*ymid < min_radius*min_radius, 1, 0)
triang.set_mask(mask)

# Plot the triangulation.
plt.figure()
plt.gca().set_aspect('equal')
plt.triplot(triang, 'bo-')
plt.title('triplot of Delaunay triangulation')

plt.show()

使用 matplotlib 对多边形进行三角剖分

Triangulation of a polygon using matplotlib

python

2d

polygon

matplotlib

triangulation