Apriori算法找出频繁项集(python)

目标

数据库有5个事务。设min_sup=60%,min_conf=80%。
TID 购买的商品
T100 {M,O,N,K,E,Y}
T200 {D,O,N,K,E,Y}
T300 {M,A,K,E}
T400 {M,U,C,K,Y}
T500 {C,O,O,K,I,E}

算法思想

算法 Apriori。使用逐层迭代方法基于候选找出频繁集。
输入:
D:事务数据库。
min_sup:最小支持度阈值
输出:L,D中的频繁项集。
方法:
(1)=find_frequent_1_itemsets(D);
(2)for(k=2;){
(3)=aproiri_gen();
(4) for each 事务tD{
(5)=subset(,t);
(6) for each 候选c
(7) C.count++;
(8) }
(9) ={c(|c.countmin_sup)}
(10)}
(11)return L=;
pocedure apriori_gen(:frequent(k-1)itemset)
(1) for each项集
(2) for each项集
(3) if(=)(=)(=)then{
(4) c=; //连接步:产生候选
(5) If has_infrequent_subset(c,) then
(6) delete c; //剪纸步:删除非频繁的候选
(7) esle add c to ;
(8)}
(9)return ;
procedure has_infrequent_subset(c: candidate k itemset;:frequent(k-1)itemset)
//使用先验知识
(1)for each(k-1)subset s of c
(2) if sthen
(3) return TRUE
(4)return FALSE

def find_frequent_1_itemsets(data, support):
    """计算频繁一项集"""
    min_sup = len(data) * support
    itemsets = {}
    for d in data:
        for vlist in d.values():
            for value in vlist:
                if value in itemsets.keys():
                    itemsets[value] += 1
                else:
                    itemsets[value] = 1
    print("候选一项集:")
    print(itemsets)
    for key in list(itemsets.keys()):
        if itemsets[key] < min_sup:
            itemsets.pop(key)
    print("频繁一项集:")
    print(itemsets)
    return itemsets

def remove_samekey(this_itemsets):
    """仅保留一个类似(k,e,m)和(e,k,m)这种相同的键"""
    key_list = this_itemsets.keys()
    new_list1 = []
    new_list2 = []
    new_itemsets = {}
    for ele in key_list:
        if set(ele) not in new_list1:
            new_list1.append(set(ele))
        else:
            continue

    for ele in new_list1:
        new_list2.append(tuple(ele))

    for ele in new_list2:
        new_itemsets[ele] = this_itemsets[ele]
    return new_itemsets

def find_frequent_next_itemsets(data, frequent_n_itemsets, support):
    """从n项集找n+1项集"""
    min_sup = len(data) * support
    n_itemlist = list(frequent_n_itemsets)
    print(n_itemlist)
    itemsets = {}
    m = len(n_itemlist[0])
    for ele1 in n_itemlist:
        for ele2 in n_itemlist:
            if ele1 != ele2:
                if len(ele1) == 1:
                    if (ele1, ele2) in itemsets.keys():
                        continue
                    else:
                        itemsets[(ele1, ele2)] = 0
                else:
                    for e in ele2:
                        if e not in ele1:
                            new_ele = ele1 + (e,)
                            if new_ele in itemsets.keys():
                                continue
                            else:
                                itemsets[new_ele] = 0
            else:
                continue
    print((m + 1), "候选项集为:")
    itemsets=remove_samekey(itemsets)
    print(itemsets)

    for item1 in data:
        for item2 in itemsets.keys():
            if (set(item2)  set(*item1.values())):
                itemsets[item2] += 1
            else:
                continue

    print("处理后的", (m + 1), "候选项集为:")
    print(itemsets)

    for key in list(itemsets.keys()):
        if itemsets[key] < min_sup:
            itemsets.pop(key)

    print("频繁", m + 1, "项集:")
    print(itemsets)
    return itemsets

"""对数据进行初始化"""
data = [
    {"T100": ['M', 'O', 'N', 'K', 'E', 'Y']},
    {"T200": ['D', 'O', 'N', 'K', 'E', 'Y']},
    {"T300": ['M', 'A', 'K', 'E']},
    {"T400": ['M', 'U', 'C', 'K', 'Y']},
    {"T500": ['C', 'O', 'O', 'K', 'I', 'E']}
]
support = 0.6
itemsets = find_frequent_1_itemsets(data, support)
while len(itemsets)>1:
    itemsets = find_frequent_next_itemsets(data, itemsets, support)

结果

Apriori算法找出频繁项集(python)
以上

Original: https://blog.csdn.net/algorithem/article/details/124450856
Author: algorithem
Title: Apriori算法找出频繁项集(python)

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