我正在尝试用Python计算Davies-Bouldin Index。
以下是以下代码尝试重现的步骤。
5个步骤：
然后，
最后，
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))
参数：
X是数据labels，是由聚类算法计算的标签（即：kmeans）centroids是每个群集质心的坐标（即：cluster_centers_）另外，请注意我使用的是Python 3
问题1 ：每对质心之间的欧氏距离的计算是否正确（步骤3）？
问题2 ：我的第4步实施是否正确？
问题3 ：我是否需要规范化内部和群集间距离？
有关第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DB索引。
在第一次迭代时：
distances_means的索引0）和簇2的距离内平均值（distances_means的索引1）ctrds_distance的索引0）在第二次迭代：
distances_means的索引0）和簇3的距离内平均值（distances_means的索引2）ctrds_distance的索引1）依旧......
以10个集群为例，完整的迭代过程应如下所示：
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。
换句话说，我不确定计算出的第一个簇间距离是否对应于簇1和簇2之间的距离。计算出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，按照上述模式。
简而言之：我担心我将群内距离分成一簇不相应的群集间距离。
欢迎任何帮助！
答案 0 :(得分：2)
以上是Davies-Bouldin指数天真实施的更短，更快的修正版本。
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：
请注意，您可以找到尝试改进此索引的创新方法，特别是用“Cylindrical”距离替换欧几里德距离的“New Version of Davies-Bouldin Index”。
答案 1 :(得分：1)
感谢您的代码和修订 - 真的帮助我开始了。更短，更快的版本并不完全正确。我对其进行了修正，以正确平均每个群集中最相似群集的色散分数。
有关原始算法和解释，请参阅https://www.researchgate.net/publication/224377470_A_Cluster_Separation_Measure：
DBI是每个群集的相似性度量的平均值 最相似的集群。
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]
# calculate cluster dispersion
S = [np.mean([euclidean(p, centroids[i]) for p in k]) for i, k in enumerate(cluster_k)]
Ri = []
for i in range(n_cluster):
Rij = []
# establish similarity between each cluster and all other clusters
for j in range(n_cluster):
if j != i:
r = (S[i] + S[j]) / euclidean(centroids[i], centroids[j])
Rij.append(r)
# select Ri value of most similar cluster
Ri.append(max(Rij))
# get mean of all Ri values
dbi = np.mean(Ri)
return dbi
答案 2 :(得分：0)
感谢您的实施。我只有一个问题：最后一行是否遗漏了一个分区。在最后一步中，max（db）的值应除以实现的簇数。
def DaviesBouldin(Daten, DatenLabels):
n_cluster = len(np.bincount(DatenLabels))
cluster_k = [Daten[DatenLabels == 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)] # mittlere Entfernung zum jeweiligen Clusterzentrum
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]) / n_cluster)
return(np.max(db))
也许我监督那个部门，因为我是Python的新手。但是在我的图形中（我在一系列聚类上进行迭代），DB.max的值在开始时非常低，之后又增加。在按照簇的数量进行缩放后，图形看起来更好（开始时的DB.max值很高，并且随着簇的数量不断增加而不断下降）。
祝你好运
答案 3 :(得分：0)
这比下面的代码快20倍，所有计算都以numpy完成。
import numpy as np
from scipy.spatial.distance import euclidean, cdist, pdist, squareform
def db_index(X, y):
"""
Davies-Bouldin index is an internal evaluation method for
clustering algorithms. Lower values indicate tighter clusters that
are better separated.
"""
# get unique labels
if y.ndim == 2:
y = np.argmax(axis=1)
uniqlbls = np.unique(y)
n = len(uniqlbls)
# pre-calculate centroid and sigma
centroid_arr = np.empty((n, X.shape[1]))
sigma_arr = np.empty((n,1))
dbi_arr = np.empty((n,n))
mask_arr = np.invert(np.eye(n, dtype='bool'))
for i,k in enumerate(uniqlbls):
Xk = X[np.where(y==k)[0],...]
Ak = np.mean(Xk, axis=0)
centroid_arr[i,...] = Ak
sigma_arr[i,...] = np.mean(cdist(Xk, Ak.reshape(1,-1)))
# compute pairwise centroid distances, make diagonal elements non-zero
centroid_pdist_arr = squareform(pdist(centroid_arr)) + np.eye(n)
# compute pairwise sigma sums
sigma_psum_arr = squareform(pdist(sigma_arr, lambda u,v: u+v))
# divide
dbi_arr = np.divide(sigma_psum_arr, centroid_pdist_arr)
# get mean of max of off-diagonal elements
dbi_arr = np.where(mask_arr, dbi_arr, 0)
dbi = np.mean(np.max(dbi_arr, axis=1))
return dbi
这是使用numpy 1.14，scipy 1.1.0和python 3的实现。计算速度没有太大提高，但内存占用空间应略小。
import numpy as np
from scipy.spatial.distance import euclidean, cdist, pdist, squareform
def db_index(X, y):
"""
Davies-Bouldin index is an internal evaluation method for
clustering algorithms. Lower values indicate tighter clusters that
are better separated.
Arguments
----------
X : 2D array (n_samples, embed_dim)
Vector for each example.
y : 1D array (n_samples,) or 2D binary array (n_samples, n_classes)
True labels for each example.
Returns
----------
dbi : float
Calculated Davies-Bouldin index.
"""
# get unique labels
if y.ndim == 2:
y = np.argmax(axis=1)
uniqlbls = np.unique(y)
n = len(uniqlbls)
# pre-calculate centroid and sigma
centroid_arr = np.empty((n, X.shape[1]))
sigma_arr = np.empty(n)
for i,k in enumerate(uniqlbls):
Xk = X[np.where(y==k)[0],...]
Ak = np.mean(Xk, axis=0)
centroid_arr[i,...] = Ak
sigma_arr[i,...] = np.mean(cdist(Xk, Ak.reshape(1,-1)))
# loop over non-duplicate cluster pairs
dbi = 0
for i in range(n):
max_Rij = 0
for j in range(n):
if j != i:
Rij = np.divide(sigma_arr[i] + sigma_arr[j],
euclidean(centroid_arr[i,...], centroid_arr[j,...]))
if Rij > max_Rij:
max_Rij = Rij
dbi += max_Rij
return dbi/n