python实现最短距离single linkage算法

好的，下面是Python实现最短距离single linkage算法的示例代码：

import numpy as np

def single_linkage(X):
    """
    :param X: numpy array of shape (n_samples, n_features)
    :return: numpy array of shape (n_samples-1, 2)
    """
    n = X.shape[0]
    clusters = [[i] for i in range(n)]
    distances = np.zeros((n, n))

    # 计算初始距离矩阵
    for i in range(n):
        for j in range(i+1, n):
            distances[i][j] = distances[j][i] = np.linalg.norm(X[i] - X[j])

    # 合并最近的两个簇，直到所有点都在同一个簇中
    while len(clusters) > 1:
        min_dist = np.inf
        for i in range(len(clusters)):
            for j in range(i+1, len(clusters)):
                for k in clusters[i]:
                    for l in clusters[j]:
                        if distances[k][l] < min_dist:
                            min_dist = distances[k][l]
                            merge_pair = (i, j)
        i, j = merge_pair
        new_cluster = clusters[i] + clusters[j]
        clusters.pop(j)
        clusters[i] = new_cluster

        # 更新距离矩阵
        for k in range(n):
            if k != i and k != j:
                distances[i][k] = distances[k][i] = min(distances[k][clusters[i][0]], distances[k][clusters[i][-1]])

    return np.array([[clusters[i][0], clusters[i][1]] for i in range(len(clusters)-1)])

这里使用了numpy库进行数组操作和计算欧几里得距离。函数接受一个形状为(n_samples, n_features)的numpy数组X, 返回一个形状为(n_samples-1, 2)的numpy数组，其中每行是合并两个簇的点的下标。