k-meas非监督聚类分析

对于每一个类 j，重新计算该类的质心：

}

下图展示了对n个样本点进行K-means聚类的效果，这里k取2。

其伪代码如下：

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创建k个点作为初始的质心点（随机选择）

当任意一个点的簇分配结果发生改变时

对数据集中的每一个数据点

对每一个质心

计算质心与数据点的距离

将数据点分配到距离最近的簇

对每一个簇，计算簇中所有点的均值，并将均值作为质心

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四、实验过程描述

k-means算法比较简单，但也有几个比较大的缺点：

（1）k值的选择是用户指定的，不同的k得到的结果会有挺大的不同，如下图所示，左边是k=3的结果，这个就太稀疏了，蓝色的那个簇其实是可以再划分成两个簇的。而右图是k=5的结果，可以看到红色菱形和蓝色菱形这两个簇应该是可以合并成一个簇的：

（2）对k个初始质心的选择比较敏感，容易陷入局部最小值。例如，我们上面的算法运行的时候，有可能会得到不同的结果，如下面这两种情况。K-means也是收敛了，只是收敛到了局部最小值：

（3）存在局限性，如下面这种非球状的数据分布就搞不定了：

（4）数据库比较大的时候，收敛会比较慢。

k-means老早就出现在江湖了。所以以上的这些不足也被世人的目光敏锐的捕捉到，并融入世人的智慧进行了某种程度上的改良。例如问题（1）对k的选择可以先用一些算法分析数据的分布，如重心和密度等，然后选择合适的k。而对问题（2），有人提出了另一个成为二分k均值（bisecting k-means）算法，它对初始的k个质心的选择就不太敏感。

五、实验结果

六、附录代码

：

Kmeans.py:

from numpy import *

import time

import matplotlib.pyplot as plt

# calculate Euclidean distance

defeuclDistance(vector1, vector2):

    return sqrt(sum(power(vector2 - vector1, 2)))

# init centroids with random samples

definitCentroids(dataSet, k):

    numSamples, dim = dataSet.shape

    centroids = zeros((k, dim))

    for i in range(k):

        index = int(random.uniform(0, numSamples))

        centroids[i, :] = dataSet[index, :]

    return centroids

# k-means cluster

defkmeans(dataSet, k):

    numSamples = dataSet.shape[0]

    # first column stores which cluster this sample belongs to,

    # second column stores the error between this sample and its centroid

    clusterAssment = mat(zeros((numSamples, 2)))

    clusterChanged = True

    ## step 1: init centroids

    centroids = initCentroids(dataSet, k)

    while clusterChanged:

        clusterChanged = False

        ## for each sample

        for i in xrange(numSamples):

            minDist = 100000.0

            minIndex = 0

            ## for each centroid

            ## step 2: find the centroid who is closest

            for j in range(k):

                distance = euclDistance(centroids[j, :], dataSet[i, :])

                if distance < minDist:

                    minDist = distance

                    minIndex = j

            ## step 3: update its cluster

            if clusterAssment[i, 0] != minIndex:

                clusterChanged = True

                clusterAssment[i, :] = minIndex, minDist**2

        ## step 4: update centroids

        for j in range(k):

            pointsInCluster = dataSet[nonzero(clusterAssment[:, 0].A == j)[0]]

            centroids[j, :] = mean(pointsInCluster, axis = 0)

    print‘Congratulations, cluster complete!‘

    return centroids, clusterAssment

# show your cluster only available with 2-D data

defshowCluster(dataSet, k, centroids, clusterAssment):

    numSamples, dim = dataSet.shape

    if dim != 2:

        print"Sorry! I can not draw because the dimension of your data is not 2!"

        return1

    mark = [‘or‘, ‘ob‘, ‘og‘, ‘ok‘, ‘^r‘, ‘+r‘, ‘sr‘, ‘dr‘, ‘<r‘, ‘pr‘]

    if k > len(mark):

        print"Sorry! Your k is too large! please contact Zouxy"

        return1

    # draw all samples

    for i in xrange(numSamples):

        markIndex = int(clusterAssment[i, 0])

        plt.plot(dataSet[i, 0], dataSet[i, 1], mark[markIndex])

    mark = [‘Dr‘, ‘Db‘, ‘Dg‘, ‘Dk‘, ‘^b‘, ‘+b‘, ‘sb‘, ‘db‘, ‘<b‘, ‘pb‘]

    # draw the centroids

    for i in range(k):

        plt.plot(centroids[i, 0], centroids[i, 1], mark[i], markersize = 12)

    plt.show()

Test_kmeans.py:

from numpy import *

import time

import matplotlib.pyplot as plt

import kmeans

from kmeans import kmeans,showCluster

## step 1: load data

print"step 1: load data..."

dataSet = []

fileIn = open(‘F:\code\python\k-means-2014-5-8/testSet.txt‘)

for line in fileIn.readlines():

    lineArr = line.strip().split(‘\t‘)

    dataSet.append([float(lineArr[0]), float(lineArr[1])])

## step 2: clustering...

print"step 2: clustering..."

dataSet = mat(dataSet)

k = 4

centroids, clusterAssment = kmeans(dataSet, k)

## step 3: show the result

print"step 3: show the result..."

showCluster(dataSet, k, centroids, clusterAssment)

数据点：testSet.txt

1.658985    4.285136

-3.453687    3.424321

4.838138    -1.151539

-5.379713    -3.362104

0.972564    2.924086

-3.567919    1.531611

0.450614    -3.302219

-3.487105    -1.724432

2.668759    1.594842

-3.156485    3.191137

3.165506    -3.999838

-2.786837    -3.099354

4.208187    2.984927

-2.123337    2.943366

0.704199    -0.479481

-0.392370    -3.963704

2.831667    1.574018

-0.790153    3.343144

2.943496    -3.357075

-3.195883    -2.283926

2.336445    2.875106

-1.786345    2.554248

2.190101    -1.906020

-3.403367    -2.778288

1.778124    3.880832

-1.688346    2.230267

2.592976    -2.054368

-4.007257    -3.207066

2.257734    3.387564

-2.679011    0.785119

0.939512    -4.023563

-3.674424    -2.261084

2.046259    2.735279

-3.189470    1.780269

4.372646    -0.822248

-2.579316    -3.497576

1.889034    5.190400

-0.798747    2.185588

2.836520    -2.658556

-3.837877    -3.253815

2.096701    3.886007

-2.709034    2.923887

3.367037    -3.184789

-2.121479    -4.232586

2.329546    3.179764

-3.284816    3.273099

3.091414    -3.815232

-3.762093    -2.432191

3.542056    2.778832

-1.736822    4.241041

2.127073    -2.983680

-4.323818    -3.938116

3.792121    5.135768

-4.786473    3.358547

2.624081    -3.260715

-4.009299    -2.978115

2.493525    1.963710

-2.513661    2.642162

1.864375    -3.176309

-3.171184    -3.572452

2.894220    2.489128

-2.562539    2.884438

3.491078    -3.947487

-2.565729    -2.012114

3.332948    3.983102

-1.616805    3.573188

2.280615    -2.559444

-2.651229    -3.103198

2.321395    3.154987

-1.685703    2.939697

3.031012    -3.620252

-4.599622    -2.185829

4.196223    1.126677

-2.133863    3.093686

4.668892    -2.562705

-2.793241    -2.149706

2.884105    3.043438

-2.967647    2.848696

4.479332    -1.764772

-4.905566    -2.911070