用sklearn封装的kmeans库

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

def calckmean(array, karr):
    # array是一个二维数组
    # X = [[1, 2, 3, 4], [5, 6, 7, 8], [3, 4, 5, 6]]

    # 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 = []

    for k in karr:
        kmeans_model = KMeans(n_clusters=k).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 1 1 1 1 1 2 2 2 2 2]
        # k = 4 : [1 1 1 1 0 0 0 0 0 3 2 2 3 2]
        coordinates.append(kmeans_model.labels_)

    # 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 = []
            for j in range(0, len(point[i])):
                temp = []
                for item in zip(coordinates[i], array):
                    if item[0] == j:
                        temp.append(item[1])
                coordinate.append(temp)
                # 得到的样式为k=3，每个簇点的坐标群
                # coordinate = [[[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]]]

            return karr[i], point[i], coordinate

调用的时候直接可以：

from kmeans import *

测试数据：

#!/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 = calckmean(a, karr)
print("最好的可以分成" + str(k) + "个簇，中心点为" + "\n" + str(point))