K-Means算法存在的问题

收敛于局部最小值。

bisecting K-Means

首先将所有点作为一个簇,然后将该簇一分为二。
之后选择其中一个簇继续进行划分,选择哪一个簇进行划分取决于对其划分是否可以最大程度降低SSE的值。

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],:]#get the data points currently in cluster i
            centroidMat, splitClustAss = kMeans(ptsInCurrCluster, 2, distMeas)
            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: ",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