使用肘部法确定k-means均值的k值

Posted 2021-03-25 starcrm

import numpy as np
from sklearn.cluster import KMeans
from scipy.spatial.distance import cdist
import matplotlib.pyplot as plt

c1x = np.random.uniform(0.5, 1.5, (1, 10))
c1y = np.random.uniform(0.5, 1.5, (1, 10))
c2x = np.random.uniform(3.5, 4.5, (1, 10))
c2y = np.random.uniform(3.5, 4.5, (1, 10))
x = np.hstack((c1x, c2x))
y = np.hstack((c1y, c2y))
X = np.vstack((x, y)).T

K = range(1, 10)
meanDispersions = []
for k in K:
    kmeans = KMeans(n_clusters=k)
    kmeans.fit(X)
    #理解为计算某个与其所属类聚中心的欧式距离
    #最终是计算所有点与对应中心的距离的平方和的均值
    meanDispersions.append(sum(np.min(cdist(X, kmeans.cluster_centers_, ‘euclidean‘), axis=1)) / X.shape[0])

plt.plot(K, meanDispersions, ‘bx-‘)
plt.xlabel(‘k‘)
plt.ylabel(‘Average Dispersion‘)
plt.title(‘Selecting k with the Elbow Method‘)
plt.show()

X为：

[[0.84223858 1.18059879]
 [0.84834276 0.84499409]
 [1.13263229 1.34316399]
 [0.95487981 0.59743761]
 [0.81646041 1.32361288]
 [0.90405171 0.54047701]
 [1.2723004  1.3461647 ]
 [0.52939142 1.03325549]
 [0.84592514 0.74344317]
 [1.07882783 1.4286598 ]
 [3.71702311 3.97510452]
 [3.95476036 3.83842502]
 [4.4297804  3.91854623]
 [4.08686159 4.15798624]
 [3.90406684 3.84413461]
 [4.32395689 4.06825926]
 [4.23112269 3.78578326]
 [3.70602931 4.08608482]
 [3.58690191 4.37072349]
 [4.38564657 4.02168693]]

随着K的增加，纵轴呈下降趋势且最终趋于稳定，那么拐点肘部处的位置所对应的k 值，不妨认为是相对最佳的类聚数量值。