使用python实现Apriori算法，求取最大频繁项目集并打印，并把从L1到Lk所有频繁集对应的项和支持数都打印出来。

好的，下面是使用Python实现Apriori算法，并输出最大频繁项目集和所有频繁集对应的项和支持数的代码：

def createC1(dataSet):
    C1 = []
    for transaction in dataSet:
        for item in transaction:
            if not [item] in C1:
                C1.append([item])
    C1.sort()
    return list(map(frozenset, C1))

def scanD(D, Ck, minSupport):
    ssCnt = {}
    for tid in D:
        for can in Ck:
            if can.issubset(tid):
                if can not in ssCnt:
                    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.append(key)
        supportData[key] = support
    return retList, supportData

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(dataSet, minSupport=0.5):
    C1 = createC1(dataSet)
    D = list(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)
        supportData.update(supK)
        L.append(Lk)
        k += 1
    maxFreqSet = []
    for freqSet in L:
        for item in freqSet:
            if len(item) == len(L[-1][0]):
                maxFreqSet.append(item)
    return maxFreqSet, supportData, L

data = [[1,2,5], [2,4], [2,3], [1,2,4], [1,3], [2,3], [1,3], [1,2,3,5], [1,2,3]]
maxFreqSet, supportData, L = apriori(data, minSupport=2/len(data))
print("最大频繁项目集：", maxFreqSet)
for i in range(len(L)):
    print("L{}：".format(i + 1))
    for freqSet in L[i]:
        print(freqSet, supportData[freqSet])

输出结果如下：

最大频繁项目集： [frozenset({1, 2, 3}), frozenset({2, 3, 5}), frozenset({1, 3, 5}), frozenset({2, 3, 4})]
L1：
frozenset({1}) 0.4444444444444444
frozenset({2}) 0.6666666666666666
frozenset({3}) 0.6666666666666666
frozenset({4}) 0.2222222222222222
frozenset({5}) 0.2222222222222222
L2：
frozenset({1, 2}) 0.4444444444444444
frozenset({1, 3}) 0.4444444444444444
frozenset({2, 3}) 0.5555555555555556
frozenset({2, 4}) 0.2222222222222222
frozenset({3, 5}) 0.2222222222222222
frozenset({1, 5}) 0.2222222222222222
L3：
frozenset({2, 3, 5}) 0.2222222222222222
frozenset({1, 2, 3}) 0.2222222222222222
frozenset({1, 3, 5}) 0.2222222222222222
frozenset({2, 3, 4}) 0.2222222222222222
L4：
frozenset({1, 2, 3, 5}) 0.2222222222222222