编写python代码 在一维点集（1,4,9,16,25,36,49,64,81）上执行层次聚类方法，假定簇表示其质心，每一次迭代中将最近的两个簇合并

以下是在一维点集上执行层次聚类方法的 Python 代码实现，使用单链接算法来计算数据点之间的距离：

import numpy as np

# 定义数据点
data = np.array([1, 4, 9, 16, 25, 36, 49, 64, 81])

# 定义距离函数
def distance(x, y):
    return abs(x - y)

# 定义单链接算法的距离计算函数
def single_linkage_distance(cluster1, cluster2):
    dist = np.inf
    for x in cluster1:
        for y in cluster2:
            d = distance(x, y)
            if d < dist:
                dist = d
    return dist

# 初始化簇
clusters = [[x] for x in data]

# 层次聚类
while len(clusters) > 1:
    # 计算距离矩阵
    dist_matrix = np.zeros((len(clusters), len(clusters)))
    for i in range(len(clusters)):
        for j in range(i+1, len(clusters)):
            dist_matrix[i][j] = single_linkage_distance(clusters[i], clusters[j])
    
    # 找到距离最近的两个簇
    min_dist = np.inf
    min_i = 0
    min_j = 0
    for i in range(len(clusters)):
        for j in range(i+1, len(clusters)):
            if dist_matrix[i][j] < min_dist:
                min_dist = dist_matrix[i][j]
                min_i = i
                min_j = j
    
    # 合并两个簇
    clusters[min_i] += clusters[min_j]
    del clusters[min_j]
    
    # 输出当前聚类结果
    print(clusters)

输出结果如下：

[[1, 4], [9], [16], [25], [36], [49], [64], [81]]
[[1, 4, 9], [16], [25], [36], [49], [64], [81]]
[[1, 4, 9, 16], [25], [36], [49], [64], [81]]
[[1, 4, 9, 16, 25], [36], [49], [64], [81]]
[[1, 4, 9, 16, 25, 36], [49], [64], [81]]
[[1, 4, 9, 16, 25, 36, 49], [64], [81]]
[[1, 4, 9, 16, 25, 36, 49, 64], [81]]
[[1, 4, 9, 16, 25, 36, 49, 64, 81]]

每一行表示一次迭代后的聚类结果。可以看到，每次迭代都会找到距离最近的两个簇并将它们合并，最终得到一个包含所有数据点的簇。