机器学习-层级聚类算法(Hierarchy Cluster)

Section I: Brief Introduction on Hierarchy Cluster

The two standard algorithms for agglomerative hierarchical clustering are single linkage and complete linkage. Using single linkage, the distances between the most similar members for each pair of clusters and merge the two clusters for which the distance between the most similar members is the smallest. With respect to complete linkage, the approach is similar to single linkage but, instead of comparing the most similar members in each pair of clusters, it compare the most dissimilar members to perform the merge.

Hierarchical complete linkage clustering is an iterative procedure that can be summarized by the following steps:

• Step 1: Compute the distance matrix of all samples （Euclidean Distance）
• Step 2: Represent each data point as a singleton cluster
• Step 3: Merge the two closest clusters based on the distance between the most similar/dissimilar (distant) members
• Step 4: Update similarity matrix
• Step 5: Repeat steps 2-4 until one single cluster remains

FROM
Sebastian Raschka, Vahid Mirjalili. Python机器学习第二版. 南京：东南大学出版社，2018.

第一部分: 数据初始化
代码

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import warnings
warnings.filterwarnings("ignore")

plt.rcParams['figure.dpi']=200
plt.rcParams['savefig.dpi']=200
font = {'weight': 'light'}
plt.rc("font", **font)

np.random.seed(123)

#Section 1: Generate random data
variables=['X','Y','Z']
labels=['ID_1','ID_2','ID_3','ID_4','ID_5']

X=np.random.random_sample([5,3])*10
df=pd.DataFrame(X,columns=variables,index=labels)
print("Original DataFrame:\n",df)

结果

Original DataFrame:
              X         Y         Z
ID_1  6.964692  2.861393  2.268515
ID_2  5.513148  7.194690  4.231065
ID_3  9.807642  6.848297  4.809319
ID_4  3.921175  3.431780  7.290497
ID_5  4.385722  0.596779  3.980443

第二部分：Euclidean距离计算

方法一：通过scipy包的pdist和square函数

代码

#Section 2: Perform hierarchical clustering on a distance matrix
#Section 2.1: Via pdist and squareform methods
from scipy.spatial.distance import pdist,squareform

row_dist=pd.DataFrame(squareform(pdist(df,metric='euclidean')),
              columns=labels,index=labels)
print("\nData Distance via pdist and squareform: \n",row_dist)

结果

Data Distance via pdist and squareform: 
           ID_1      ID_2      ID_3      ID_4      ID_5
ID_1  0.000000  4.973534  5.516653  5.899885  3.835396
ID_2  4.973534  0.000000  4.347073  5.104311  6.698233
ID_3  5.516653  4.347073  0.000000  7.244262  8.316594
ID_4  5.899885  5.104311  7.244262  0.000000  4.382864
ID_5  3.835396  6.698233  8.316594  4.382864  0.000000

方法二：通过linkage函数

代码

#Section 2.2: Via linkage method
from scipy.cluster.hierarchy import linkage

row_cluster=linkage(df.values,method='complete',metric='euclidean')
row_dist_linkage=pd.DataFrame(row_cluster,
                              columns=['Row Label 1','Row Label 2','Distance','Item Number in Cluster'],
                              index=['Cluster %d' % (i+1) for i in range(row_cluster.shape[0])])
print("\nData Distance via Linkage: \n",row_dist_linkage)

结果

Data Distance via Linkage: 
            Row Label 1  Row Label 2  Distance  Item Number in Cluster
Cluster 1          0.0          4.0  3.835396                     2.0
Cluster 2          1.0          2.0  4.347073                     2.0
Cluster 3          3.0          5.0  5.899885                     3.0
Cluster 4          6.0          7.0  8.316594                     5.0

参考文献
Sebastian Raschka, Vahid Mirjalili. Python机器学习第二版. 南京：东南大学出版社，2018.

转载：https://blog.csdn.net/Santorinisu/article/details/104465699