'Delaunay' object has no attribute 'circumcenters'

我很抱歉，我之前给出的示例代码中有一个错误。scipy.spatial.Delaunay对象确实没有circumcenters属性。我向你道歉。

实际上，我们可以通过计算三角形的外接圆心来估计外接圆半径。下面是修正后的示例代码：

import numpy as np
import matplotlib.pyplot as plt
from scipy.spatial import Delaunay

def plot_alpha_shape(points, alpha):
    # 计算Delaunay三角剖分
    tri = Delaunay(points)
    
    # 计算每个三角形的外接圆心和半径
    circumcenters = np.zeros((tri.simplices.shape[0], 2))
    circumradii = np.zeros(tri.simplices.shape[0])
    
    for i, simplex in enumerate(tri.simplices):
        x = points[simplex][:, 0]
        y = points[simplex][:, 1]
        A = np.column_stack((x - np.mean(x), y - np.mean(y)))
        _, _, V = np.linalg.svd(A)
        circumcenter = np.mean(points[simplex], axis=0)
        circumradius = np.linalg.norm(points[simplex[0]] - circumcenter)
        
        circumcenters[i] = circumcenter
        circumradii[i] = circumradius
    
    # 选择满足alpha条件的三角形
    alpha_triangles = tri.simplices[circumradii < alpha]
    
    # 提取边界点
    boundary_points = np.unique(alpha_triangles.flatten())
    
    # 绘制结果
    plt.figure(figsize=(8, 6))
    plt.scatter(points[:, 0], points[:, 1], s=10)
    plt.triplot(points[:, 0], points[:, 1], tri.simplices)
    plt.plot(points[boundary_points, 0], points[boundary_points, 1], 'r-', linewidth=2)
    plt.axis('equal')
    plt.show()

# 示例用法
points = np.random.rand(100, 2)  # 替换为你的点云数据
alpha = 0.1  # 替换为你想要的alpha值
plot_alpha_shape(points, alpha)

这段修正后的代码会计算每个三角形的外接圆心和半径，并根据alpha值选择满足条件的三角形。然后，绘制原始点云及其边界。

再次对之前的错误表示歉意，并希望这次能够帮到你。