由于需要海量的进行聚类，所以将 k-means 算法自我封装成一个方便利用的库，可以直接调用得到最优的 k值 和 中心点：
#!/usr/bin/python3.4
# -*- coding: utf-8 -*-
# k-means算法
import numpy as np
from sklearn.cluster import KMeans
from sklearn import metrics
# sklearn官方文档
# http://scikit-learn.org/stable/modules/generated/sklearn.cluster.KMeans.html#sklearn.cluster.KMeans
def calckmean(array, karr):
# array是一个二维数组
# X = X = [[1, 1], [2, 3], [3, 2], [1, 2], [5, 8], [6, 6], [5, 7], [5, 6], [6, 7], [7, 1], [8, 2], [9, 1], [7, 1], [9, 3]]
# k是待选取K值的数组
# karr = [2, 3, 4, 5, 8,...]
# 将原始数据由数组变成矩阵
x = np.array(array)
# 用来储存轮廓系数的数组
score = []
# 用来储存中心坐标点的数组
point = []
# 用来储存各个簇的坐标
coordinates = []
# 用来储存各个簇点的与中心的距离
distances = []
for k in karr:
# n_clusters为聚类的个数
# max_iter为迭代的次数，这里设置最大迭代次数为300
# n_init=10使用不同质心种子运行k-means算法的次数
kmeans_model = KMeans(n_clusters=k, max_iter=300,n_init=10).fit(x)
# title = 'K = %s, 轮廓系数 = %.03f' % (k, metrics.silhouette_score(X, kmeans_model.labels))
# print(title)
# 获取中心点的坐标
counter_point = kmeans_model.cluster_centers_
# print("k=" str(k) "时的中心点为" "n" str(counter_point))
# 记录分数
# print(metrics.silhouette_score(x, kmeans_model.labels_,metric='euclidean'))
score.append("%.03f" % (metrics.silhouette_score(x, kmeans_model.labels_)))
# 记录中心坐标
point.append(counter_point)
# 将坐标属于哪个簇的标签储存到数组
# k = 3 : [0 0 0 0 2 2 2 2 2 1 1 1 1 1]
# k = 4 : [1 1 1 1 0 0 0 0 0 3 2 2 3 2]
coordinates.append(kmeans_model.labels_)
# 每个点和中心点的距离
distances.append(KMeans(n_clusters=k, max_iter=300).fit_transform(x))
# 返回轮廓系数最大的k值中心坐标分簇坐标
maxscore = max(score, default=0)
for i in range(0, len(score)):
if maxscore == score[i]:
# 储存分簇坐标的数组
coordinate = []
# 储存簇点与中心点的距离数组
distance = []
for j in range(0, len(point[i])):
# 这里是得到分簇坐标
tempcoor = []
for item in zip(coordinates[i], array):
if item[0] == j:
tempcoor.append(item[1])
coordinate.append(tempcoor)
# 得到的样式为k=3，每个簇点的坐标群
# [[[7, 1], [8, 2], [9, 1], [7, 1], [9, 3]],
# [[5, 8], [6, 6], [5, 7], [5, 6], [6, 7]],
# [[1, 1], [2, 3], [3, 2], [1, 2]]]
# 这里是得到分簇与中心点的距离
tempdis = []
for item in zip(coordinates[i], distances[i]):
if item[0] == j:
tempdis.append(min(item[1]))
distance.append(tempdis)
# 得到k=3的各个簇点对中心的距离
# [[1.1661903789690597, 0.39999999999999575, 1.166190378969066, 1.1661903789690597, 1.7204650534085277],
# [1.2649110640673495, 0.9999999999999858, 0.4472135954999452, 0.8944271909999063, 0.6324555320336579],
# [1.25, 1.0307764064044151, 1.25, 0.75]]
# 得到k=3的中心点
# [[8.0, 1.6],
# [5.4, 6.8],
# [1.75, 2.0]]
return karr[i], point[i], coordinate, distance调用的时候直接可以：
from kmeans import *
测试数据：
代码语言：javascript复制#!/usr/bin/python3.4
# -*- coding: utf-8 -*-
from kmeans import *
x1 = np.array([1, 2, 3, 1, 5, 6, 5])
x2 = np.array([1, 3, 2, 2, 8, 6, 7])
# a = [[1, 2, 3, 1, 5, 6, 5], [1, 3, 2, 2, 8, 6, 7], [3, 5, 9, 4, 7, 6, 1], [1, 5, 3, 4, 8, 6, 7], [5, 1, 2, 3, 6, 9, 4],[8, 4, 6, 2, 1, 6, 3]]
a = [[1, 1], [2, 3], [3, 2], [1, 2], [5, 8], [6, 6], [5, 7], [5, 6], [6, 7], [7, 1], [8, 2], [9, 1], [7, 1], [9, 3]]
karr = [2, 3, 4, 5, 8]
# print(np.array(a))
# print(list(zip(x1, x2)))
K, point, coordinate, distance = calckmean(X, tests)
print("------------------------")
print("k=" str(K) "时的中心点为" "n" str(point) "n" "各个簇点为" "n" str(coordinate))
print(distance)
