我对 Davies-Bouldin 索引的 Python 实现是否正确?

Is my python implementation of the Davies-Bouldin Index correct?(我对 Davies-Bouldin 索引的 Python 实现是否正确?)

问题描述

我正在尝试计算 Davies-Bouldin 指数Python.

I'm trying to calculate the Davies-Bouldin Index in Python.

以下是代码尝试重现的步骤.

Here are the steps the code below tries to reproduce.

5 个步骤:

1. 对于每个集群，计算每个点到质心的欧几里德距离
2. 对于每个集群，计算这些距离的平均值
3. 对于每对集群，计算它们的质心之间的欧几里德距离

那么，

1. 对于每对聚类，求到它们各自质心的平均距离之和(在第 2 步计算)，然后除以它们之间的距离(在第 3 步计算).

最后，

1. 计算所有这些划分(= 所有索引)的平均值以获得整个聚类的 Davies-Bouldin 索引

代码

def daviesbouldin(X, labels, centroids):

    import numpy as np
    from scipy.spatial.distance import pdist, euclidean

    nbre_of_clusters = len(centroids) #Get the number of clusters
    distances = [[] for e in range(nbre_of_clusters)] #Store intra-cluster distances by cluster
    distances_means = [] #Store the mean of these distances
    DB_indexes = [] #Store Davies_Boulin index of each pair of cluster
    second_cluster_idx = [] #Store index of the second cluster of each pair
    first_cluster_idx = 0 #Set index of first cluster of each pair to 0

    # Step 1: Compute euclidean distances between each point of a cluster to their centroid
    for cluster in range(nbre_of_clusters):
        for point in range(X[labels == cluster].shape[0]):
            distances[cluster].append(euclidean(X[labels == cluster][point], centroids[cluster]))

    # Step 2: Compute the mean of these distances
    for e in distances:
        distances_means.append(np.mean(e))

    # Step 3: Compute euclidean distances between each pair of centroid
    ctrds_distance = pdist(centroids) 

    # Tricky step 4: Compute Davies-Bouldin index of each pair of cluster   
    for i, e in enumerate(e for start in range(1, nbre_of_clusters) for e in range(start, nbre_of_clusters)):
        second_cluster_idx.append(e)
        if second_cluster_idx[i-1] == nbre_of_clusters - 1:
            first_cluster_idx += 1
        DB_indexes.append((distances_means[first_cluster_idx] + distances_means[e]) / ctrds_distance[i])

    # Step 5: Compute the mean of all DB_indexes   
    print("DAVIES-BOULDIN Index: %.5f" % np.mean(DB_indexes)) 

在参数中:

• X 是数据
• labels，是由聚类算法计算的标签(即:kmeans)
• centroids 是每个集群的质心坐标(即:cluster_centers_)

• X is the data
• labels, are the labels computed by a clustering algorithm (i.e: kmeans)
• centroids are the coordinates of each cluster's centroid (i.e: cluster_centers_)

另外，请注意我使用的是 Python 3

Also, note that I'm using Python 3

问题 1:计算每对质心之间的欧氏距离是否正确(步骤 3)?

QUESTION1: Is the computation of euclidean distances between each pair of centroid correct (step 3)?

问题 2:我对第 4 步的实施是否正确?

QUESTION2: Is my implementation of step 4 correct?

问题 3:我是否需要归一化簇内和簇间距离?

QUESTION3: Do I need to normalise intra and inter cluster distances ?

关于第 4 步的进一步说明

假设我们有 10 个集群.循环应该计算每对集群的数据库索引.

Let's say we have 10 clusters. The loop should compute the DB index of each pair of cluster.

在第一次迭代时:

• 求和集群 1 的内距离平均值(distances_means 的索引 0)和集群 2 的内距离平均值(distances_means 的索引 1)
• 将这个总和除以 2 个集群之间的距离(ctrds_distance 的索引 0)

• sums intra-distances mean of cluster 1 (index 0 of distances_means) and intra-distances mean of cluster 2 (index 1 of distances_means)
• divides this sum by the distance between the 2 clusters (index 0 of ctrds_distance)

在第二次迭代中:

• 求和集群 1 的内距离平均值(distances_means 的索引 0)和集群 3 的内距离平均值(distances_means 的索引 2)
• 将这个总和除以 2 个集群之间的距离(ctrds_distance 的索引 1)

• sums intra-distances mean of cluster 1 (index 0 of distances_means) and intra-distances mean of cluster 3 (index 2 of distances_means)
• divides this sum by the distance between the 2 clusters (index 1 of ctrds_distance)

等等...

以 10 个集群为例，完整的迭代过程应该是这样的:

With the example of 10 clusters, the full iteration process should look like this:

intra-cluster distance intra-cluster distance       distance between their
      of cluster:             of cluster:           centroids(storage num):
         0           +             1            /             0
         0           +             2            /             1
         0           +             3            /             2
         0           +             4            /             3
         0           +             5            /             4
         0           +             6            /             5
         0           +             7            /             6
         0           +             8            /             7
         0           +             9            /             8
         1           +             2            /             9
         1           +             3            /             10
         1           +             4            /             11
         1           +             5            /             12
         1           +             6            /             13
         1           +             7            /             14
         1           +             8            /             15
         1           +             9            /             16
         2           +             3            /             17
         2           +             4            /             18
         2           +             5            /             19
         2           +             6            /             20
         2           +             7            /             21
         2           +             8            /             22
         2           +             9            /             23
         3           +             4            /             24
         3           +             5            /             25
         3           +             6            /             26
         3           +             7            /             27
         3           +             8            /             28
         3           +             9            /             29
         4           +             5            /             30
         4           +             6            /             31
         4           +             7            /             32
         4           +             8            /             33
         4           +             9            /             34
         5           +             6            /             35
         5           +             7            /             36
         5           +             8            /             37
         5           +             9            /             38
         6           +             7            /             39
         6           +             8            /             40
         6           +             9            /             41
         7           +             8            /             42
         7           +             9            /             43
         8           +             9            /             44

这里的问题是我不太确定distances_means 的索引是否与ctrds_distance 的索引匹配.

The problem here is I'm not quite sure that the index of distances_means matches the index of ctrds_distance.

换句话说，我不确定计算的第一个簇间距离对应于簇 1 和簇 2 之间的距离.计算的第二个簇间距离对应于簇 3 和簇 1 之间的距离... 依此类推，遵循上述模式.

In other words, I'm not sure that the first inter-cluster distance computed corresponds to the distance between cluster 1 and cluster 2. And that the second inter-cluster distance computed corresponds to the distance between cluster 3 and cluster 1... and so on, following the pattern above.

简而言之:恐怕我将簇内距离对除以不对应的簇间距离.

In short: I'm afraid I'm dividing pairs of intra-cluster distances by an inter-cluster distance that is not corresponding.

欢迎任何帮助！

推荐答案

这是上面 Davies-Bouldin 索引朴素实现的更短、更快速的更正版本.

Here is a shorter, faster corrected version of the Davies-Bouldin index naive implementation above.

def DaviesBouldin(X, labels):
    n_cluster = len(np.bincount(labels))
    cluster_k = [X[labels == k] for k in range(n_cluster)]
    centroids = [np.mean(k, axis = 0) for k in cluster_k]
    variances = [np.mean([euclidean(p, centroids[i]) for p in k]) for i, k in enumerate(cluster_k)]
    db = []

    for i in range(n_cluster):
        for j in range(n_cluster):
            if j != i:
                db.append((variances[i] + variances[j]) / euclidean(centroids[i], centroids[j]))

    return(np.max(db) / n_cluster)

回答我自己的问题:

• 初稿(第 4 步)上的计数器是正确的，但无关紧要
• 无需标准化集群内和集群间距离
• 计算欧几里得距离时出错

请注意，您可以找到尝试改进此索引的创新方法，特别是新版本Davies-Bouldin 指数"，用圆柱距离代替欧几里得距离.

Note you can find innovative approaches that try to improve this index, notably the "New Version of Davies-Bouldin Index" that replaces Euclidean distance by Cylindrical distance.

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