数据挖掘作业——Apriori算法

设minsupport=0.4，利用Apriori算法求出所有的频繁项集，指出最大频繁项集。
数据集：

代码实现：

# 设minsupport=40%，利用Apriori算法求出所有的频繁项目集，指出其中的最大频繁项目集。
import pandas as pd

def get_D():
    D = pd.read_csv("work1.csv",index_col='TID')
    D = D['Itemset'].values.tolist()
    for i in range(len(D)):
        D[i] = D[i].split(',')
    return D

def get_C1(D):
    C1=[]
    for index in D:
        for item in index:
            if not [item] in C1:
                C1.append([item])
    C1.sort()
    return list(map(frozenset,C1))

def scanD(D,CK,minSupport):
    ssCnt = {}
    for index in D:   # 遍历行
        for can in CK: # 遍历Ck
            if can.issubset(index):
                if not can in ssCnt: # 计数
                    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

#频繁项集两两组合
def aprioriGen(Lk,k):
    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:
                retList.append(Lk[i]|Lk[j])
    return retList

def apriori(D,minSupport=0.4):
    C1=get_C1(D)
    D=list(map(set,D))
    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)
        supportData.update(supK)
        L.append(Lk)
        k+=1
    return L,supportData


if __name__=='__main__':
    D=get_D()
    L,supportData=apriori(D)
    L.pop()
    print("所有的频繁项目集为：")
    for i in range(len(L)):
        print("L{}".format(i+1),L[i])
    print("所有频繁项目集对应的支持度为：")
    print(supportData)
    print("最大频繁项目集为：",L[len(L)-1])
    

运行结果：