机器学习实战5:k-means聚类:二分k均值聚类+地理位置聚簇实例

  k-均值聚类是非监督学习的一种,输入必须指定聚簇中心个数k。k均值是基于相似度的聚类,为没有标签的一簇实例分为一类。

  一 经典的k-均值聚类  

  思路:  

  1 随机创建k个质心(k必须指定,二维的很容易确定,可视化数据分布,直观确定即可);

  2 遍历数据集的每个实例,计算其到每个质心的相似度,这里也就是欧氏距离;把每个实例都分配到距离最近的质心的那一类,用一个二维数组数据结构保存,第一列是最近质心序号,第二列是距离;

  3 根据二维数组保存的数据,重新计算每个聚簇新的质心;

  4 迭代2 和 3,直到收敛,即质心不再变化;

from numpy import *

def loadDataSet(fileName):      #general function to parse tab -delimited floats
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, ))) #la.norm(vecA-vecB) def randCent(dataSet, k):
n = shape(dataSet)[]
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,))
return centroids def kMeans(dataSet, k, distMeas=distEclud, createCent=randCent):
m = shape(dataSet)[]
clusterAssment = mat(zeros((m,)))#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 = -
for j in range(k):
distJI = distMeas(centroids[j,:],dataSet[i,:])
if distJI < minDist:
minDist = distJI; minIndex = j
if clusterAssment[i,] != minIndex: clusterChanged = True
clusterAssment[i,:] = minIndex,minDist**
print centroids
for cent in range(k):#recalculate centroids
ptsInClust = dataSet[nonzero(clusterAssment[:,].A==cent)[]]#get all the point in this cluster
centroids[cent,:] = mean(ptsInClust, axis=) #assign centroid to mean
return centroids, clusterAssment

  经典的k均值聚类有很大的缺点就是很容易收敛到局部最优,为了避免这种局部最优,我们引入了二分k-均值算法。

  二 二分k-均值聚类算法

  二分k-均值聚类算法是基于经典k-均值算法实现的;里面调用经典k-均值(k=2),把一个聚簇分成两个,迭代到分成k个停止;

  具体思路:

  1 把整个数据集看成一个聚簇,计算质心;并用同样的数据结构二维数组保存每个实例到质心的距离;

  2 对每一个聚簇进行2-均值聚类划分;

  3 计算划分后的误差,选择所有被划分的聚簇中总误差最小的划分保存;

  4 迭代2 和 3 直到聚簇数目达到k停止;

def biKmeans(dataSet, k, distMeas=distEclud):
m = shape(dataSet)[]
clusterAssment = mat(zeros((m,)))
centroid0 = mean(dataSet, axis=).tolist()[]
centList =[centroid0] #create a list with one centroid
for j in range(m):#calc initial Error
clusterAssment[j,] = distMeas(mat(centroid0), dataSet[j,:])**
while (len(centList) < k):
lowestSSE = inf
for i in range(len(centList)):
ptsInCurrCluster = dataSet[nonzero(clusterAssment[:,].A==i)[],:]#get the data points currently in cluster i
centroidMat, splitClustAss = kMeans(ptsInCurrCluster, , distMeas)
sseSplit = sum(splitClustAss[:,])#compare the SSE to the currrent minimum
sseNotSplit = sum(clusterAssment[nonzero(clusterAssment[:,].A!=i)[],])
print "sseSplit, and notSplit: ",sseSplit,'--',sseNotSplit
if (sseSplit + sseNotSplit) < lowestSSE:
bestCentToSplit = i
bestNewCents = centroidMat
bestClustAss = splitClustAss.copy()
lowestSSE = sseSplit + sseNotSplit
bestClustAss[nonzero(bestClustAss[:,].A == )[],] = len(centList) #change to ,, or whatever
bestClustAss[nonzero(bestClustAss[:,].A == )[],] = bestCentToSplit
print 'the bestCentToSplit is: ',bestCentToSplit
print 'the len of bestClustAss is: ', len(bestClustAss)
centList[bestCentToSplit] = bestNewCents[,:].tolist()[]#replace a centroid with two best centroids
centList.append(bestNewCents[,:].tolist()[])
clusterAssment[nonzero(clusterAssment[:,].A == bestCentToSplit)[],:]= bestClustAss#reassign new clusters, and SSE
return mat(centList), clusterAssment

  三 地理位置聚簇实例

  地理位置的经纬度正好是二维的,可以可视化出来,所以很适合聚类算法确定质心个数k值;值得注意的是,球面计算距离,不能简单的用欧式距离,而需要用球面距离公式,见函数distSLC;

  代码的含义给定n个俱乐部地址名称,然后使用urllib包,调用yahoo地图的API返回经纬度,调用我们上面实现的k均值聚类算法,找到聚簇的中心,最后利用matplotlib工具可视化出来;

import urllib
import json
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[], lineArr[])
if retDict['ResultSet']['Error'] == :
lat = float(retDict['ResultSet']['Results'][]['latitude'])
lng = float(retDict['ResultSet']['Results'][]['longitude'])
print "%s\t%f\t%f" % (lineArr[], lat, lng)
fw.write('%s\t%f\t%f\n' % (line, lat, lng))
else: print "error fetching"
sleep()
fw.close() def distSLC(vecA, vecB):#Spherical Law of Cosines
a = sin(vecA[,]*pi/) * sin(vecB[,]*pi/)
b = cos(vecA[,]*pi/) * cos(vecB[,]*pi/) * \
cos(pi * (vecB[,]-vecA[,]) /)
return arccos(a + b)*6371.0 #pi is imported with numpy import matplotlib
import matplotlib.pyplot as plt
def clusterClubs(numClust=):
datList = []
for line in open('places.txt').readlines():
lineArr = line.split('\t')
datList.append([float(lineArr[]), float(lineArr[])])
datMat = mat(datList)
myCentroids, clustAssing = biKmeans(datMat, numClust, distMeas=distSLC)
fig = plt.figure()
rect=[0.1,0.1,0.8,0.8]
scatterMarkers=['s', 'o', '^', '', '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[:,].A==i)[],:]
markerStyle = scatterMarkers[i % len(scatterMarkers)]
ax1.scatter(ptsInCurrCluster[:,].flatten().A[], ptsInCurrCluster[:,].flatten().A[], marker=markerStyle, s=)
ax1.scatter(myCentroids[:,].flatten().A[], myCentroids[:,].flatten().A[], marker='+', s=)
plt.show()

  四 总结

  优点:易实现;

  缺点:可能收敛到局部最小值,在大数据集上收敛较慢;

  适用数据类型:数值型;

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