跟着Leo机器学习实战：Kmeans聚类

Paula ·

更新时间:2024-09-21

· 824 次阅读

跟着Leo机器学习实战：Kmeans聚类 Kmeans聚类

优点：容易实现
缺点：容易陷入局部最小值，在大规模数据收敛很慢。
适用数据类型：数值型数据

伪代码

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加载数据

from numpy import *
def loadDataSet(fileName):      #加载数据
    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):
    return sqrt(sum(power(vecA - vecB, 2))) #计算两个向量的欧氏距离
def randCent(dataSet, k):       #随机产生k个在数据范围内的中心点，并返回
    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

训练函数

def kMeans(dataSet, k, distMeas=distEclud, createCent=randCent):
    m = shape(dataSet)[0]   #获取样本数
    clusterAssment = mat(zeros((m,2)))#记录样本被分配到哪个中心点，第二列记录到最近中心点的欧氏距离
    centroids = createCent(dataSet, k)      #随机产生k个中心点
    clusterChanged = True
    while clusterChanged:
        clusterChanged = False
        for i in range(m):  #对每个样本寻找与其最近的中心点
            minDist = inf; minIndex = -1
            for j in range(k):      #对每个样本寻找与其最近的中心点匹配
                distJI = distMeas(centroids[j,:],dataSet[i,:])
                if distJI < minDist:
                    minDist = distJI; minIndex = j
            if clusterAssment[i,0] != minIndex: clusterChanged = True       #判断分配是否发生改变
            clusterAssment[i,:] = minIndex,minDist**2
        print(centroids)
        for cent in range(k):#recalculate centroids
            ptsInClust = dataSet[nonzero(clusterAssment[:,0].A==cent)[0]]#获取分到同一类的样本
            centroids[cent,:] = mean(ptsInClust, axis=0) #求分配到同一类样本的均值
    return centroids, clusterAssment

二分Kmeans聚类 原理

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代码

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)):
            ptsInCurrCluster = dataSet[nonzero(clusterAssment[:,0].A==i)[0],:]#获取分为i类的数据
            centroidMat, splitClustAss = kMeans(ptsInCurrCluster, 2, distMeas)      #对同一类再进行内部二分类
            sseSplit = sum(splitClustAss[:,1])#获取二分类之后的SSE误差平方差
            sseNotSplit = sum(clusterAssment[nonzero(clusterAssment[:,0].A!=i)[0],1])#获取二分类之前的SSE误差平方差
            print("sseSplit, and notSplit: ",sseSplit,sseNotSplit)
            if (sseSplit + sseNotSplit) < lowestSSE:
                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

作者：LuckyLeo26

实战学习 kmeans

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