使用 SciPy 分割随机数据为 3 个聚类来实施 K 均值聚类?


是的,我们还可以通过将随机数据分成 3 个聚类来实现 K 均值聚类算法。让我们通过以下示例理解一下 −

示例

#importing the required Python libraries: import numpy as np from numpy import vstack,array from numpy.random import rand from scipy.cluster.vq import whiten, kmeans, vq from pylab import plot,show #Random data generation: data = vstack((rand(200,2) + array([.5,.5]),rand(150,2))) #Normalizing the data: data = whiten(data) # computing K-Means with K = 3 (3 clusters) centroids, mean_value = kmeans(data, 3) print("Code book :", centroids, "") print("Mean of Euclidean distances :", mean_value.round(4)) # mapping the centroids clusters, _ = vq(data, centroids) print("Cluster index :", clusters, "") #Plotting using numpy's logical indexing plot(data[clusters==0,0],data[clusters==0,1],'ob', data[clusters==1,0],data[clusters==1,1],'or', data[clusters==2,0],data[clusters==2,1],'og') plot(centroids[:,0],centroids[:,1],'sg',markersize=8) show()

输出

Code book :
[[2.10418081 1.73089074]
[2.69953885 3.04708713]
[0.6994524 1.06646081]]

Mean of Euclidean distances : 0.7661

Cluster index : [1 1 0 1 1 1 1 0 0 0 1 1 1 1 1 0 1 1 1 0 1 1 1 1 1 0 1 0 1
0 1 0 0 0 0 0 1
0 1 1 0 1 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 0 1 0 1 1 0 0 0 1 1 0
0 0 1 0 1 0 1 1 1 0 0 1 0 1 0 1 1 0 1 1 0 1 0 0 0 1 1 1 1 1 1 0 0 1 1 1 0
0 0 1 0 0 1 1 1 0 1 1 1 1 1 0 1 0 1 1 1 1 1 1 0 1 1 1 0 1 1 0 1 1 1 1 0 1
0 1 0 1 1 0 1 0 1 1 1 1 1 0 0 0 1 1 1 1 0 1 1 1 1 0 1 0 1 1 0 0 1 1 0 0 0
1 1 0 1 1 1 1 0 1 0 0 1 1 1 1 2 2 0 0 2 2 2 2 0 2 2 2 2 2 2 2 2 0 0 0 0 2
2 2 2 2 0 2 2 2 0 2 2 0 2 0 0 2 2 0 0 0 0 2 2 2 0 2 2 0 2 0 2 0 0 2 0 2 2
0 2 2 2 0 0 2 2 2 2 2 2 0 2 2 2 2 2 0 0 2 2 2 2 0 2 2 2 0 2 0 2 0 2 2 2 0
0 0 0 2 2 2 0 2 2 2 2 2 0 2 2 2 0 2 2 0 2 1 2 0 2 2 2 0 2 2 0 0 0 2 0 0 0
0 2 2 2 0 2 2 2 2 0 2 2 2 2 0 0 2]

更新于: 14-12-2021

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