k-均值算法的工作流程:
首先,随机确定k个初始点作为质心;接着,将数据集中的每个点分配到一个簇中,即为每个点找到距离其最近的质心,并将其分配给该质心所对应的簇;然后,每个簇的质心更新为该簇所有点的平均值。再次重新分配数据集中所有的点,如果所有的点被分配的簇和之前一样,即簇的质心不会再改变,则此时的k个簇就是我们所需要的;如果某个点被分配的簇改变了,则分配完所有的点之后重新更新每个簇的质心,重复分配、更新操作直到所有簇的质心不再改变
k_means.py
'''
Created on 2018年8月3日
@author: hcl
'''
from numpy import *
from matplotlib import pyplot as plt
#general function to parse tab -delimited floats
#assume last column is target value
def loadDataSet(fileName):
dataMat = []
fr = open(fileName)
for line in fr.readlines():
curLine = line.strip().split('\t')
#笔者使用的是python3,需要将map映射后的结果转化为list
#map all elements to float()
fltLine = list(map(float,curLine))
dataMat.append(fltLine)
return mat(dataMat)
#样本距离计算函数
def distEclud(vecA, vecB):
return sqrt(sum(power(vecA - vecB, 2))) #la.norm(vecA-vecB)
#创建簇中心矩阵,初始化为k个在数据集的边界内随机分布的簇中心
def randCent(dataSet, k):
n = shape(dataSet)[1]
#create centroid mat
centroids = mat(zeros((k,n)))
#create random cluster centers, within bounds of each dimension
for j in range(n):
#求出数据集中第j列的最小值(即第j个特征)
minJ = min(dataSet[:,j])
#用第j个特征最大值减去最小值得出特征值范围
rangeJ = float(max(dataSet[:,j]) - minJ)
#创建簇矩阵的第J列,random.rand(k,1)表示产生(10,1)维的矩阵,其中每行值都为0-1中的随机值
#可以这样理解,每个centroid矩阵每列的值都在数据集对应特征的范围内,那么k个簇中心自然也都在数据集范围内
centroids[:,j] = mat(minJ + rangeJ * random.rand(k,1))
return centroids
#distMeas为距离计算函数
#createCent为初始化随机簇心函数
def kMeans(dataSet, k, distMeas=distEclud, createCent=randCent):
m = shape(dataSet)[0]
#create mat to assign data points to a centroid, also holds SE of each point
#创建一个(m,2)维矩阵,第一列存储每个样本对应的簇心,第二列存储样本到簇心的距离
clusterAssment = mat(zeros((m,2)))
#用createCent()函数初始化簇心矩阵
centroids = createCent(dataSet, k)
#保存迭代中clusterAssment是否更新的状态,如果未更新,那么退出迭代,表示收敛
#如果更新,那么继续迭代,直到收敛
clusterChanged = True
while clusterChanged:
clusterChanged = False
#for each data point assign it to the closest centroid
#对每个样本找出离样本最近的簇心
for i in range(m):
#minDist保存最小距离
#minIndex保存最小距离对应的簇心
minDist = inf; minIndex = -1
#遍历簇心,找出离i样本最近的簇心
for j in range(k):
distJI = distMeas(centroids[j,:],dataSet[i,:])
if distJI < minDist:
minDist = distJI; minIndex = j
#如果clusterAssment更新,表示对应样本的簇心发生变化,那么继续迭代
if clusterAssment[i,0] != minIndex: clusterChanged = True
#更新clusterAssment,样本到簇心的距离
clusterAssment[i,:] = minIndex,minDist**2
# print(centroids)
#遍历簇心,更新簇心为对应簇中所有样本的均值
for cent in range(k):#recalculate centroids
#利用数组过滤找出簇心对应的簇(数组过滤真是好东西!)
ptsInClust = dataSet[nonzero(clusterAssment[:,0].A==cent)[0]]#get all the point in this cluster
#对簇求均值,赋给对应的centroids簇心
centroids[cent,:] = mean(ptsInClust, axis=0) #assign centroid to mean
return centroids, clusterAssment
def paint(xArr,yArr,xArr1,yArr1):
fig = plt.figure()
ax = fig.add_subplot(111)
ax.scatter(xArr,yArr,c='blue')
ax.scatter(xArr1,yArr1,c='red')
plt.show()
if __name__ == '__main__':
dataSet = loadDataSet('testSet.txt')
myCentroids,clustAssing = kMeans(dataSet,4)
paint(dataSet[:,0].flatten().A[0], dataSet[:,1].flatten().A[0], myCentroids[:,0].flatten().A[0], myCentroids[:,1].flatten().A[0])
输出:
4个质心点
[[-2.46154315 2.78737555]
[ 2.6265299 3.10868015]
[ 2.65077367 -2.79019029]
[-3.53973889 -2.89384326]]
实用后处理来提高聚类性能:
1:将具有最大SSE(Sum of Squared Error 最大误差平方和)值的簇分为两个簇
为了保持簇总数的不变性:
有两种可以量化的办法:1 合并最近的质心 2 合并两个使得SSE增幅最小的质心
二分K-均值算法:
为了克服K-均值聚类算法收敛到局部最小值的缺陷,提出了二分K-均值算法。该算法首先将所有点归为一个簇中,然后将簇一分为二,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,不断重复划分过程,直到簇的个数达到用户指定的值k。
在上面代码块中添加
#distMeas为距离计算函数
def biKmeans(dataSet, k, distMeas=distEclud):
m = shape(dataSet)[0]
#(m,2)维矩阵,第一列保存样本所属簇,第二列保存样本到簇中心的距离
clusterAssment = mat(zeros((m,2)))
#取数据集特征均值作为初始簇中心
centroid0 = mean(dataSet, axis=0).tolist()[0]
#centList保存簇中心数组,初始化为一个簇中心
#create a list with one centroid
centList =[centroid0]
#calc initial Error
for j in range(m):
clusterAssment[j,1] = distMeas(mat(centroid0), dataSet[j,:])**2
#迭代,直到簇中心集合长度达到k
while (len(centList) < k):
#初始化最小误差
lowestSSE = inf
#迭代簇中心集合,找出找出分簇后总误差最小的那个簇进行分解
for i in range(len(centList)):
#get the data points currently in cluster i
#获取属于i簇的数据集样本
ptsInCurrCluster = dataSet[nonzero(clusterAssment[:,0].A==i)[0],:]
#对该簇进行k均值聚类
centroidMat, splitClustAss = kMeans(ptsInCurrCluster, 2, distMeas)
#获取该簇分类后的误差和
sseSplit = sum(splitClustAss[:,1])#compare the SSE to the currrent minimum
#获取不属于该簇的样本集合的误差和,注意矩阵过滤中用的是!=i
sseNotSplit = sum(clusterAssment[nonzero(clusterAssment[:,0].A!=i)[0],1])
#打印该簇分类后的误差和和不属于该簇的样本集合的误差和
print("sseSplit, and notSplit: ",sseSplit,sseNotSplit)
#两误差和相加即为分簇后整个样本集合的误差和,找出簇中心集合中能让分簇后误差和最小的簇中心,保存最佳簇中心(bestCentToSplit),最佳分簇中心集合(bestNewCents),以及分簇数据集中样本对应簇中心及距离集合(bestClustAss),最小误差(lowestSSE)
if (sseSplit + sseNotSplit) < lowestSSE:
bestCentToSplit = i
bestNewCents = centroidMat
bestClustAss = splitClustAss.copy()
lowestSSE = sseSplit + sseNotSplit
#更新用K-means获取的簇中心集合,将簇中心换为len(centList)和bestCentToSplit,以便之后调整clusterAssment(总样本集对应簇中心与和簇中心距离的矩阵)时一一对应
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))
#更新簇中心集合,注意与bestClustAss矩阵是一一对应的
centList[bestCentToSplit] = bestNewCents[0,:].tolist()[0]#replace a centroid with two best centroids
centList.append(bestNewCents[1,:].tolist()[0])
#reassign new clusters, and SSE
clusterAssment[nonzero(clusterAssment[:,0].A == bestCentToSplit)[0],:]= bestClustAss
return mat(centList), clusterAssment
if __name__ == '__main__':
# dataSet = loadDataSet('testSet.txt')
# myCentroids,clustAssing = kMeans(dataSet,4)
# print(myCentroids)
# # print(clustAssing)
# paint(dataSet[:,0].flatten().A[0], dataSet[:,1].flatten().A[0], myCentroids[:,0].flatten().A[0], myCentroids[:,1].flatten().A[0])
datMat3 = mat(loadDataSet('testSet2.txt'))
centList,myNewAssments = biKmeans(datMat3,3)
print(centList)
xArr = datMat3[:,0].flatten().A[0]
yArr = datMat3[:,1].flatten().A[0]
xArr1 = centList[:,0].flatten().A[0]
yArr1 = centList[:,1].flatten().A[0]
#paint为笔者自己写的绘图函数
paint(xArr,yArr,xArr1,yArr1)
输出:
sseSplit, and notSplit: 453.0334895807502 0.0
the bestCentToSplit is: 0
the len of bestClustAss is: 60
sseSplit, and notSplit: 12.753263136887313 423.8762401366249
sseSplit, and notSplit: 77.59224931775066 29.15724944412535
the bestCentToSplit is: 1
the len of bestClustAss is: 40
[[-0.45965615 -2.7782156 ]
[ 2.93386365 3.12782785]
[-2.94737575 3.3263781 ]]