Python两步实现关联规则Apriori算法，参考机器学习实战，包括频繁项集的构建以及关联规则的挖掘

这是我学习了关联规则Apriori算法原理后参照《机器学习实战》实现的算法代码，首先分为两个部分，第一部分是频繁项集的构建，第二部分是关联规则的挖掘。特别的是我的测试数据也就是loadDataSet()函数中的数据进行了改变，这是为了能帮助理解第二部分。然后代码中我加了很多为了让自己理解的输出测试，保留在里面，应该也能帮助大家理解

一、构建频繁项集（注释都在代码中了）

from numpy import *

def loadDataSet():
    return [[1, 3, 4,6,7], [2, 3, 4,5,6,7], [1, 2, 3, 5,7], [2,4, 5,6],[3,4,5,6,7]]

def createC1(dataSet):
    C1 = []
    for transaction in dataSet:
        for item in transaction:
            if not [item] in C1:
                C1.append([item])
                
    C1.sort()
    return map(frozenset, C1)#use frozen set so we can use it as a key in a dict    

def scanD(D, Ck, minSupport):
    ssCnt = {}
    for tid in D:
        for can in Ck:
            if can.issubset(tid):
                if not ssCnt.has_key(can): ssCnt[can]=1
                else: ssCnt[can] += 1
    numItems = float(len(D))
    retList = []
    supportData = {}
    for key in ssCnt:
        support = ssCnt[key]/numItems
        if support >= minSupport:
            retList.insert(0,key)
        supportData[key] = support
    return retList, supportData

def aprioriGen(Lk, k): #creates Ck
    print'Lk:',Lk
    retList = []
    lenLk = len(Lk)
    for i in range(lenLk):
        for j in range(i+1, lenLk): 
            L1 = list(Lk[i])[:k-2]; L2 = list(Lk[j])[:k-2]#见《数据挖掘概念与技术》p161的连接步讲解
            L1.sort(); L2.sort()
            print'L1,L2',L1,L2
            if L1==L2: #if first k-2 elements are equal，另外在由一项频繁集L1构造C2时，以上[:k-2]=[:0]=[],所以会有L1==L2,构造出2项候选集C2
                retList.append(Lk[i] | Lk[j]) #set union
    return retList

def apriori(dataSet, minSupport = 0.5):
    C1 = createC1(dataSet)
    D = map(set, dataSet)
    L1, supportData = scanD(D, C1, minSupport)
    L = [L1]
    k = 2
    while (len(L[k-2]) > 0):
        Ck = aprioriGen(L[k-2], k)
        Lk, supK = scanD(D, Ck, minSupport)#scan DB to get Lk
        supportData.update(supK)
        L.append(Lk)
        k += 1
    return L, supportData

测试以上代码:

二、基于构造出的频繁项集挖掘关联规则（注释都在代码中了）

需要注意的就是对rulesFromConseq()函数的理解，可以参照我代码中添加的用于测试的输出函数的输出理解

def generateRules(L, supportData, minConf=0.7):  #supportData is a dict coming from scanD
    bigRuleList = []
    for i in range(1, len(L)):#only get the sets with two or more items
        print 'i:',i
        for freqSet in L[i]:
            print'L[i]:',L[i]
            print'freqSet:',freqSet
            H1 = [frozenset([item]) for item in freqSet]
            print 'H1:',H1
            if (i > 1):
                rulesFromConseq(freqSet, H1, supportData, bigRuleList, minConf)
            else:
                calcConf(freqSet, H1, supportData, bigRuleList, minConf)
    return bigRuleList         

def calcConf(freqSet, H, supportData, brl, minConf=0.7):
    prunedH = [] #create new list to return
    for conseq in H:
        print'conseq:',conseq
        print'freqSet-conseq:',freqSet-conseq
        conf = supportData[freqSet]/supportData[freqSet-conseq] #calc confidence
        if conf >= minConf: 
            print freqSet-conseq,'-->',conseq,'conf:',conf
            brl.append((freqSet-conseq, conseq, conf))
            prunedH.append(conseq)
#     print 'prunedH:',prunedH
    return prunedH

#rulesFromConseq目的就是对每一个频繁项集生成右边的规则，一个频繁项集可以生成很多右边的规则
#if (len(freqSet) > (m + 1))这是迭代停止的条件，也就是当右边的规则的元素数+1=该频繁项长度时停止迭代
#if (len(Hmp1) > 1)当Hmp1的长度大于1时才能合并，合并需要至少两个frozenset
def rulesFromConseq(freqSet, H, supportData, brl, minConf=0.7):
    m = len(H[0])
    print'm:',m
    print'len(freqSet):',len(freqSet)
    if (len(freqSet) > (m + 1)): #try further merging
        Hmp1 = aprioriGen(H, m+1)#create Hm+1 new candidates
        print'Hmp1:',Hmp1
        Hmp1 = calcConf(freqSet, Hmp1, supportData, brl, minConf)
        print 'Hmp2:',Hmp1
        if (len(Hmp1) > 1):    #need at least two sets to merge
            print'---------'
            rulesFromConseq(freqSet, Hmp1, supportData, brl, minConf)
            print'*********'

测试以上代码:











结果：


结束啦