基于度量的列表中的聚类元素

Posted 2023-03-12

【中文标题】基于度量的列表中的聚类元素

【英文标题】：clustering element in list based on a metric

【发布时间】：2021-03-08 05:56:02

【问题描述】：

我有一个字典列表，其中包含关键字及其向量距离，我正在尝试应用聚类技术对它们进行分组

# data = ["key": "str1", "weight": float value, ...]
# distances = [item['weight'] for item in data]
distances = [0.004906579754566209, 0.008361678408906337, 0.010228429212122636, 0.013671005756098031, 0.013671005756098031, 0.013713535105272179]

mean_distances_differences = mean([j-i for i, j in zip(distances[:-1], distances[1:])])

我计算了列表中两个连续元素之间差异的平均值。如果两个元素之间的距离小于我想要对它们进行聚类的平均值，那么结果将是

[[0.004906579754566209], [0.008361678408906337], [0.010228429212122636], [0.013671005756098031, 0.013671005756098031, 0.013713535105272179]]

在这里我想我不能使用 knn，因为我不知道会出现多少个集群。所以我试过这样

distances = [item['weight'] for item in data]
mean_distances_differences = mean([j-i for i, j in zip(distances[:-1], distances[1:])])
distances_new = distances
required_list = []
while distances_new:
    temp = []
    if len(distances_new) == 1:
        temp = distances_new
        required_list.append(temp)
        break
    else:
        for i,j in zip(distances_new[:-1], distances_new[1:]):
            if j-1 < mean_distances_differences:
                temp.append(i)
            else:
                break
        distances_new = [_i for _i in distances_new if _i not in temp]
    required_list.append(temp)

但我得到了答案

[[0.004906579754566209, 0.008361678408906337, 0.010228429212122636, 0.013671005756098031], [0.013713535105272179]]

有什么办法吗？

【参考方案1】：

你可以使用 diff 来计算距离，我取了绝对值，因为我不确定距离是否会被排序：

import numpy as np
distance_diff = abs(np.diff(distances))

如果对距离是否大于某个值进行cumsum，它将连续的小于阈值的元素组合在一起：

np.cumsum(distance_diff > abs(np.mean(distance_diff)))]

array([1, 2, 3, 3, 3])

所以剩下的就是提供一个起始组 0：

np.hstack([0,np.cumsum(distance_diff > abs(np.mean(distance_diff)))])

array([0, 1, 2, 3, 3, 3])