我正在学习一个教程,其中'kmeans'算法是整个示例的主要部分。 “行”列表作为要聚集的数据传递。 Pearson函数提供第二个参数,一个关系系数,k = 3是聚类数。从kmeans函数返回的'bestmatches'是与属于每个集群的行中的元素相对应的分组/聚集索引值的列表。因为我需要制作散点图,所以我需要它们的值。我如何返回值而不是索引?Python聚类数值数据
rows=[(1,1),(3,6),(11,2),(7,19),(22,11),(32,11)]
def pearson(v1,v2):
#sums
sum1=sum(v1)
sum2=sum(v2)
print(sum1)
#sums of the sqs
sum1Sq=sum([pow(v,2) for v in v1])
sum2Sq=sum([pow(v,2) for v in v2])
#sum of products
pSum=sum([v1[i]*v2[i] for i in range(len(v1))])
#calculate pearson R
num=pSum-(sum1*sum2/len(v1))
den=sqrt((sum1Sq-pow(sum1,2)/len(v1))*(sum2Sq-pow(sum2,2)/len(v1)))
if den==0: return 0
return 1.0-num/den
def kmeans(rows,distance=pearson,k=3):
#Determine the min and max values for each point
#COunt through "rows"(data) and find min and max values
ranges=[(min([row[i] for row in rows]),max([row[i] for row in rows]))
for i in range(len(rows[0]))]
#create k randomly placed centroids within len of 'data'
clusters=[[random.random()*(ranges[i][1]-ranges[i][0])+ranges[i][0]
for i in range(len(rows[0]))] for j in range(k)]
lastmatches=None
for t in range(100):
print 'Iteration %d' % t
bestmatches=[[] for i in range(k)]
#find which centroid is the closest to each row
for j in range(len(rows)):
row=rows[j]
bestmatch=0
for i in range(k):
d=distance(clusters[i],row)
if d<distance(clusters[bestmatch],row): bestmatch=i
bestmatches[bestmatch].append(j)
if bestmatches==lastmatches: break
lastmatches=bestmatches
#move centroids to the avg of members
for i in range(k):
avgs=[0.0]*len(rows[0])
if len(bestmatches[i])>0:
#print(len(bestmatches[i]))
for rowid in bestmatches[i]:
for m in range(len(rows[rowid])):
avgs[m]+=rows[rowid][m]
for j in range(len(avgs)):
avgs[j]/=len(bestmatches[i])
clusters[i]=avgs
return bestmatches
本教程来自Toby Segaran的“Programming Collective Intelligence”。在“发现组”一章中,他涵盖了一些数据挖掘概念,并提供了这些代码。 – DomRow
这本书被报告有一些代码错误... http://www.oreilly.com/catalog/errataunconfirmed.csp?isbn=9780596529321虽然你似乎缺少基本的python体验... –