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我正在编写一个对一组数据执行 k-means 聚类的代码。我实际上使用的是 O'Reilly 的一本名为集体智慧的书中的代码。一切正常，但在他的代码中，他使用命令行，我想用记事本++编写所有内容。作为参考，他的台词是

>>>kclust=clusters.kcluster(data,k=10)
>>>[rownames[r] for r in k[0]]

这是我的代码：

from PIL import Image,ImageDraw
def readfile(filename):
    lines=[line for line in file(filename)]
    # First line is the column titles
    colnames=lines[0].strip( ).split('\t')[1:]
    rownames=[]
    data=[]
    for line in lines[1:]:
        p=line.strip( ).split('\t')
        # First column in each row is the rowname
        rownames.append(p[0])
        # The data for this row is the remainder of the row
        data.append([float(x) for x in p[1:]])
    return rownames,colnames,data


from math import sqrt
def pearson(v1,v2):
    # Simple sums
    sum1=sum(v1)
    sum2=sum(v2)

    # Sums of the squares
    sum1Sq=sum([pow(v,2) for v in v1])
    sum2Sq=sum([pow(v,2) for v in v2])

    # Sum of the products
    pSum=sum([v1[i]*v2[i] for i in range(len(v1))])

    # Calculate r (Pearson score)
    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

class bicluster:
    def __init__(self,vec,left=None,right=None,distance=0.0,id=None):
        self.left=left
        self.right=right
        self.vec=vec
        self.id=id
        self.distance=distance

def hcluster(rows,distance=pearson):
    distances={}
    currentclustid=-1

    # Clusters are initially just the rows
    clust=[bicluster(rows[i],id=i) for i in range(len(rows))]

    while len(clust)>1:
        lowestpair=(0,1)
        closest=distance(clust[0].vec,clust[1].vec)

        # loop through every pair looking for the smallest distance
        for i in range(len(clust)):
            for j in range(i+1,len(clust)):
                # distances is the cache of distance calculations
                if (clust[i].id,clust[j].id) not in distances:
                    distances[(clust[i].id,clust[j].id)]=distance(clust[i].vec,clust[j].vec)
                #print 'i'
                #print i    
                #print 
                #print 'j'
                #print j    
                #print

                d=distances[(clust[i].id,clust[j].id)]
                if d<closest:
                    closest=d
                    lowestpair=(i,j)

        # calculate the average of the two clusters
        mergevec=[
        (clust[lowestpair[0]].vec[i]+clust[lowestpair[1]].vec[i])/2.0
        for i in range(len(clust[0].vec))]

        # create the new cluster
        newcluster=bicluster(mergevec,left=clust[lowestpair[0]],
                            right=clust[lowestpair[1]],
                            distance=closest,id=currentclustid)

        # cluster ids that weren't in the original set are negative
        currentclustid-=1
        del clust[lowestpair[1]]
        del clust[lowestpair[0]]
        clust.append(newcluster)

    return clust[0]

def kcluster(rows,distance=pearson,k=4):
    # Determine the minimum and maximum values for each point
    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
    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 for 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 the results are the same as last time, this is complete
        if bestmatches==lastmatches: break
        lastmatches=bestmatches

        # Move the centroids to the average of their members
        for i in range(k):
            avgs=[0.0]*len(rows[0])
            if len(bestmatches[i])>0:
                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
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1 回答 1

2

我假设您想将其作为程序运行？最简单的方法是简单地在底部添加这些行：

kclust=clusters.kcluster(data,k=10)
print [rownames[r] for r in k[0]]

如果您打算扩展，您可能希望稍后添加一个主要功能。

编辑：Guido在这里对主要功能有很好的描述

于 2011-06-21T14:11:27.030 回答