Implementing K-means clustering with SciPy by splitting random data in 3 clusters?

Yes, we can also implement a K-means clustering algorithm by splitting the random data in 3 clusters. Let us understand with the example below −

Example

#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()

Output

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]