机器学习实战--apriori

前面主要学习了机器学习的两大块：分类，回归，接下来的两节进入到频繁项集和关联规则的分析。
关联分析中最著名的例子当属啤酒和尿布了。http://www.wtoutiao.com/a/904866.html
为了定义上述的频繁和关联我们引入两个定义：
1、支持度：数据集中包含该集项的记录所占的比例
2、置信度：对于关联规则P–>M，该规则的置信度为：support(P U M) /support( P )
然而，可以通过简单的计算发现，如果要简单的重举各个数据项之间的组合关系是一个非常庞大的工程，为了解决这一计算的问题，我们引入apriori算法，即：如果某项集是非频繁的，那么该项集的子集也是非频繁的。

生成候选项集：（支持度）
算法原理：

占位符

算法实现：

def loadDataSet():    return [[1, 3, 4], [2, 3, 5], [1, 2, 3, 5], [2, 5]]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    #扫描记录，返回大于最小支持度的数据项集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

apriori实现：
算法原理：

算法实现：

#从多个子集中构造不重复的超集#这里有一个算法上的trick,结合下面的函数apriori，思考一下，如果由子集｛1，2｝，｛1，3｝，｛2，3｝构造超集｛1，2，3｝.你会怎样实现，使得所用时间最小呢？def aprioriGen(Lk, k): #creates Ck    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]            L1.sort(); L2.sort()            if L1==L2: #if first k-2 elements are equal                retList.append(Lk[i] | Lk[j]) #set union    return retListdef 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

从候选集中发现关联规则（置信度）

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        for freqSet in L[i]:            H1 = [frozenset([item]) for item in freqSet]            if (i > 1):                rulesFromConseq(freqSet, H1, supportData, bigRuleList, minConf)            else:                calcConf(freqSet, H1, supportData, bigRuleList, minConf)    return bigRuleList#计算每个规则的置信度并返回置信度>=最小置信度的规则#H中包含的是所有可能的预测项，(弄清楚P,M)def calcConf(freqSet, H, supportData, brl, minConf=0.7):    prunedH = [] #create new list to return    for conseq in H:        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)    return prunedH#这一个函数要多理解一下，这是为了从更进一步的找出关联规则，可以通过将上面的数据带进去，梳理一下程序的流程。def rulesFromConseq(freqSet, H, supportData, brl, minConf=0.7):    m = len(H[0])    if (len(freqSet) > (m + 1)): #try further merging        Hmp1 = aprioriGen(H, m+1)#create Hm+1 new candidates        Hmp1 = calcConf(freqSet, Hmp1, supportData, brl, minConf)        if (len(Hmp1) > 1):    #need at least two sets to merge            rulesFromConseq(freqSet, Hmp1, supportData, brl, minConf)