从使用 matplotlib 生成的 delaunay 三角剖分中获取外心 [英] Getting the circumcentres from a delaunay triangulation generated using matplotlib

问题描述

如果我使用 matplotlib 为一组点生成 delaunay 三角剖分，那么获取已生成三角形的外心的最合适的方法是什么?我还没有设法在 Triangulation 库中找到一个明显的方法来做到这一点.

If I use matplotlib to generate a delaunay triangulation for a group of points, what is the most appropraite way of getting the circumcentres of the triangles that have been geenrated? I haven't yet managed to find an obvious method in the Triangulation library to do this.

推荐答案

您应该可以使用 matplotlib.delaunay.triangulate.Triangulation 进行计算:

You should be able to calculate it using matplotlib.delaunay.triangulate.Triangulation:

三角剖分(x, y)x，y-点的坐标为浮点数的一维数组

Triangulation(x, y) x, y -- the coordinates of the points as 1-D arrays of floats

...

属性:(所有应视为只读以保持一致性)x, y -- 一维浮点数组形式的点坐标.

Attributes: (all should be treated as read-only to maintain consistency) x, y -- the coordinates of the points as 1-D arrays of floats.

  circumcenters -- (ntriangles, 2) array of floats giving the (x,y)
    coordinates of the circumcenters of each triangle (indexed by a triangle_id).

从其中一个matplotlib示例改编(可能有一种更简洁的方法，但是应该可以):

Adapted from one of the matplotlib examples (there is probably a cleaner way to do this, but it should work):

import matplotlib.pyplot as plt
import matplotlib.delaunay
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()

tt = matplotlib.delaunay.triangulate.Triangulation(x,y)
triang = tri.Triangulation(x, y)

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

plt.plot(tt.circumcenters[:,0],tt.circumcenters[:,1],'r.')
plt.show()

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