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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。

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2 回答 2

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您应该能够使用以下方法计算它matplotlib.delaunay.triangulate.Triangulation：

Triangulation(x, y) x, y -- 点的坐标为一维浮点数组

. . .

属性：（所有都应该被视为只读以保持一致性）x，y - 点的坐标作为一维浮点数组。

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

改编自 matplotlib 示例之一（可能有一种更简洁的方法可以做到这一点，但它应该可以工作）：

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()
于 2011-04-08T14:37:50.383 回答
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这是一个计算它们的函数。它也可以用于其他三角剖分结构，例如scipy's Delaunay 三角剖分（见下文）。

def compute_triangle_circumcenters(xy_pts, tri_arr):
    """
    Compute the centers of the circumscribing circle of each triangle in a triangulation.
    :param np.array xy_pts : points array of shape (n, 2)
    :param np.array tri_arr : triangles array of shape (m, 3), each row is a triple of indices in the xy_pts array

    :return: circumcenter points array of shape (m, 2)
    """
    tri_pts = xy_pts[tri_arr]  # (m, 3, 2) - triangles as points (not indices)

    # finding the circumcenter (x, y) of a triangle defined by three points:
    # (x-x0)**2 + (y-y0)**2 = (x-x1)**2 + (y-y1)**2
    # (x-x0)**2 + (y-y0)**2 = (x-x2)**2 + (y-y2)**2
    #
    # becomes two linear equations (squares are canceled):
    # 2(x1-x0)*x + 2(y1-y0)*y = (x1**2 + y1**2) - (x0**2 + y0**2)
    # 2(x2-x0)*x + 2(y2-y0)*y = (x2**2 + y2**2) - (x0**2 + y0**2)
    a = 2 * (tri_pts[:, 1, 0] - tri_pts[:, 0, 0])
    b = 2 * (tri_pts[:, 1, 1] - tri_pts[:, 0, 1])
    c = 2 * (tri_pts[:, 2, 0] - tri_pts[:, 0, 0])
    d = 2 * (tri_pts[:, 2, 1] - tri_pts[:, 0, 1])

    v1 = (tri_pts[:, 1, 0] ** 2 + tri_pts[:, 1, 1] ** 2) - (tri_pts[:, 0, 0] ** 2 + tri_pts[:, 0, 1] ** 2)
    v2 = (tri_pts[:, 2, 0] ** 2 + tri_pts[:, 2, 1] ** 2) - (tri_pts[:, 0, 0] ** 2 + tri_pts[:, 0, 1] ** 2)

    # solve 2x2 system (see https://en.wikipedia.org/wiki/Invertible_matrix#Inversion_of_2_%C3%97_2_matrices)
    det = (a * d - b * c)
    detx = (v1 * d - v2 * b)
    dety = (a * v2 - c * v1)

    x = detx / det
    y = dety / det

    return (np.vstack((x, y))).T

在来自@JoshAdel 上述答案的数据中，添加以下代码：

cc = compute_triangle_circumcenters(np.vstack([tt.x, tt.y]).T, tt.triangle_nodes)
plt.plot(cc[:, 0], cc[:, 1], ".k")

我得到下图：

在此处输入图像描述

它也可以scipy.spatial.Delaunay像这样使用：

from scipy.spatial import Delaunay
xy_pts = np.vstack([x, y]).T
dt = Delaunay(xy_pts)
cc = compute_triangle_circumcenters(dt.points, dt.simplices)
于 2020-12-02T09:48:47.543 回答