应用:NumPy实现k均值聚类算法(K-means)

NumPy实现k均值聚类算法(K-means)

一、K-means聚类算法简介

在这里插入图片描述
其伪代码如下:

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

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。
在这里插入图片描述

二、NumPy实现

kmeans.py

from numpy import *
import time
import matplotlib.pyplot as plt

# calculate Euclidean distance
def euclDistance(vector1, vector2):
    return sqrt(sum(power(vector2 - vector1, 2)))

# init centroids with random samples
def initCentroids(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
def kmeans(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
def showCluster(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!"
        return 1

    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"
        return 1

    # 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()

测试

测试数据是二维的,共80个样本。有4个类。如下

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

test_kmeans.py

测试代码:

from numpy import *
import time
import matplotlib.pyplot as plt

## step 1: load data
print "step 1: load data..."
dataSet = []
fileIn = open('E:/Python/Machine Learning in Action/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)

运行前后结果对比:
在这里插入图片描述

三、算法分析

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:
    在这里插入图片描述
  • (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:
    在这里插入图片描述
  • (3)存在局限性,如下面这种非球状的数据分布就搞不定了:
    在这里插入图片描述
  • (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,这个算法我们以后再分析和实现。

文章出处

作者:zouxy09
来源:CSDN
原文:https://blog.csdn.net/zouxy09/article/details/17589329

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转载自blog.csdn.net/m0_37827994/article/details/86550985