前往小程序,Get更优阅读体验!
立即前往
首页
学习
活动
专区
工具
TVP
发布
社区首页 >专栏 >K-均值算法(二)

K-均值算法(二)

作者头像
用户6021899
发布2019-08-14 17:22:29
5700
发布2019-08-14 17:22:29
举报

一种度量聚类效果的指标是SSE(Sum of Squared Error,误差平方和),即各个点到各自簇心的距离的平方之和。一种肯定可以降低SSE的方法是增加簇的数目,但这违背了聚类的目标。聚类的目标是在保持簇数目不变的前提下提高分类的质量,使得SSE最小。

为了克服K-means算法收敛于局部最小值问题,有人提出了二分K-means算法。该算法首先将所有的点作为一个簇,随后将该簇一分为二。之后选择其中一个簇继续划分,具体选择哪个簇划分取决于是否可以最大程度降低SSE。上述基于SSE的划分过程不断重复,知道得到用户指定的簇数目为止。

此外还涉及到K-均值算法的一个具体应用,将地图上已知经度纬度信息的点根据相互距离进行聚类。

代码如下:

代码语言:javascript
复制
from numpy import *
def loadDataSet(fileName):      #general function to parse tab -delimited floats
    '''从文件加载数据集'''
    dataMat = []                #assume last column is target value
    fr = open(fileName)
    for line in fr.readlines():
        curLine = line.strip().split('\t')
        fltLine = map(float,curLine) #map all elements to float()
        dataMat.append(fltLine)
    return dataMat
def distEclud(vecA, vecB):
    '''计算欧氏距离'''
    return sqrt(sum(power(vecA - vecB, 2))) #la.norm(vecA-vecB)
def randCent(dataSet, k):
    '''随机创建簇中心,特征数n x k 大小的矩阵 '''
    n = shape(dataSet)[1]
    centroids = mat(zeros((k,n)))#create centroid mat
    for j in range(n):#create random cluster centers, within bounds of each dimension
        minJ = min(dataSet[:,j])
        rangeJ = float(max(dataSet[:,j]) - minJ)
        centroids[:,j] = mat(minJ + rangeJ * random.rand(k,1))
    return centroids
   
def kMeans(dataSet, k, distMeas=distEclud, createCent=randCent):
    m = shape(dataSet)[0]
    clusterAssment = mat(zeros((m,2)))#create mat to assign data points
                                      #to a centroid, also holds SE of each point
    centroids = createCent(dataSet, k)
    clusterChanged = True
    while clusterChanged:
        clusterChanged = False
        for i in range(m):#for each data point assign it to the closest centroid
            minDist = inf; minIndex = -1
            for j in range(k):
                distJI = distMeas(centroids[j,:],dataSet[i,:])
                if distJI < minDist:
                    minDist = distJI; minIndex = j
            if clusterAssment[i,0] != minIndex: clusterChanged = True
            clusterAssment[i,:] = minIndex,minDist**2
        print (centroids)
        for cent in range(k):#recalculate centroids
            ptsInClust = dataSet[nonzero(clusterAssment[:,0].A==cent)[0]]#get all the point in this cluster
            centroids[cent,:] = mean(ptsInClust, axis=0) #assign centroid to mean
    return centroids, clusterAssment
def biKmeans(dataSet, k, distMeas=distEclud):
    m = shape(dataSet)[0]
    clusterAssment = mat(zeros((m,2)))
    centroid0 = mean(dataSet, axis=0).tolist()[0]
    centList =[centroid0] #create a list with one centroid
    for j in range(m):#calc initial Error
        clusterAssment[j,1] = distMeas(mat(centroid0), dataSet[j,:])**2
    while (len(centList) < k):
        lowestSSE = inf
        for i in range(len(centList)): #尝试划分每一簇
            ptsInCurrCluster = dataSet[nonzero(clusterAssment[:,0].A==i)[0],:]#get the data points currently in cluster i
            centroidMat, splitClustAss = kMeans(ptsInCurrCluster, 2, distMeas)#调用kMeans(),分成两簇
            sseSplit = sum(splitClustAss[:,1])#compare the SSE to the currrent minimum
            sseNotSplit = sum(clusterAssment[nonzero(clusterAssment[:,0].A!=i)[0],1])
            print("sseSplit, and notSplit: ")
            print(sseSplit,sseNotSplit)
            if (sseSplit + sseNotSplit) < lowestSSE:
                #更新簇的分配结果
                bestCentToSplit = i
                bestNewCents = centroidMat
                bestClustAss = splitClustAss.copy()
                lowestSSE = sseSplit + sseNotSplit
        bestClustAss[nonzero(bestClustAss[:,0].A == 1)[0],0] = len(centList) #change 1 to 3,4, or whatever
        bestClustAss[nonzero(bestClustAss[:,0].A == 0)[0],0] = bestCentToSplit
        print( 'the bestCentToSplit is: ',bestCentToSplit)
        print ('the len of bestClustAss is: ', len(bestClustAss))
        centList[bestCentToSplit] = bestNewCents[0,:].tolist()[0]#replace a centroid with two best centroids
        centList.append(bestNewCents[1,:].tolist()[0])
        clusterAssment[nonzero(clusterAssment[:,0].A == bestCentToSplit)[0],:]= bestClustAss#reassign new clusters, and SSE
    return mat(centList), clusterAssment


import urllib
import json
#用Yahoo! PlaeceFinder API 将已知的地址转化为对应的纬度与经度
def geoGrab(stAddress, city):
    apiStem = 'http://where.yahooapis.com/geocode?'  #create a dict and constants for the goecoder
    params = {}
    params['flags'] = 'J'#JSON return type
    params['appid'] = 'aaa0VN6k'
    params['location'] = '%s %s' % (stAddress, city)
    url_params = urllib.urlencode(params)
    yahooApi = apiStem + url_params      #print url_params
    print (yahooApi)
    c=urllib.urlopen(yahooApi)
    return json.loads(c.read())
from time import sleep
def massPlaceFind(fileName):
    fw = open('places.txt', 'w')
    for line in open(fileName).readlines():
        line = line.strip()
        lineArr = line.split('\t')
        retDict = geoGrab(lineArr[1], lineArr[2])
        if retDict['ResultSet']['Error'] == 0:
            lat = float(retDict['ResultSet']['Results'][0]['latitude'])
            lng = float(retDict['ResultSet']['Results'][0]['longitude'])
            print ("%s\t%f\t%f" % (lineArr[0], lat, lng))
            fw.write('%s\t%f\t%f\n' % (line, lat, lng))
        else: print( "error fetching")
        sleep(1)
    fw.close()
   
def distSLC(vecA, vecB):#Spherical Law of Cosines
    #根据球面坐标系将经度纬度信息转化为距离
    a = sin(vecA[0,1]*pi/180) * sin(vecB[0,1]*pi/180)
    b = cos(vecA[0,1]*pi/180) * cos(vecB[0,1]*pi/180) * \
                      cos(pi * (vecB[0,0]-vecA[0,0]) /180)
    return arccos(a + b)*6371.0 #pi is imported with numpy

import matplotlib
import matplotlib.pyplot as plt
def clusterClubs(numClust=5):
    datList = []
    for line in open('places.txt').readlines():
        lineArr = line.split('\t')
        datList.append([float(lineArr[4]), float(lineArr[3])])
    datMat = mat(datList)
    #对地理坐标进行2分K-均值 聚类
    myCentroids, clustAssing = biKmeans(datMat, numClust, distMeas=distSLC)
   
    #在地图上显示聚类结果
    fig = plt.figure()
    rect=[0.1,0.1,0.8,0.8]
    scatterMarkers=['s', 'o', '^', '8', 'p', \
                    'd', 'v', 'h', '>', '<']
    axprops = dict(xticks=[], yticks=[])
    ax0=fig.add_axes(rect, label='ax0', **axprops)
    imgP = plt.imread('Portland.png')
    ax0.imshow(imgP)
    ax1=fig.add_axes(rect, label='ax1', frameon=False)
    for i in range(numClust):
        ptsInCurrCluster = datMat[nonzero(clustAssing[:,0].A==i)[0],:]
        markerStyle = scatterMarkers[i % len(scatterMarkers)]
        ax1.scatter(ptsInCurrCluster[:,0].flatten().A[0], ptsInCurrCluster[:,1].flatten().A[0], marker=markerStyle, s=90)
    ax1.scatter(myCentroids[:,0].flatten().A[0], myCentroids[:,1].flatten().A[0], marker='+', s=300)
    ax1.set_xlabel("经度", fontsize =13)
    ax1.set_ylabel("纬度" ,fontsize =13)
    ax1.set_title("对地理坐标进行2分K-均值 聚类" ,fontsize =16)
    plt.show()
    
    
clusterClubs(5)
本文参与 腾讯云自媒体分享计划,分享自微信公众号。
原始发表:2019-07-23,如有侵权请联系 cloudcommunity@tencent.com 删除

本文分享自 Python可视化编程机器学习OpenCV 微信公众号,前往查看

如有侵权,请联系 cloudcommunity@tencent.com 删除。

本文参与 腾讯云自媒体分享计划  ,欢迎热爱写作的你一起参与!

评论
登录后参与评论
0 条评论
热度
最新
推荐阅读
领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档