假设事务集只有6个项， 对于如下频繁3-项集的集合：{1,2,3}，{1,2,4},{1,2,5},{1,3,4,},{2,3,4,},{2,3,5},{3,4,6}，编写一个python程序用Aprior算法求所有的候选4-项集

以下是用Python实现Apriori算法求所有的候选4-项集的代码：

# 定义函数用于生成候选k-项集
def generate_candidates(frequent_items, k):
    candidates = set()
    for itemset1 in frequent_items:
        for itemset2 in frequent_items:
            # 只有前k-2项相同时才能合并
            if itemset1 != itemset2 and itemset1[:k-2] == itemset2[:k-2]:
                candidates.add(itemset1.union(itemset2))
    return candidates

# 定义函数用于计算频繁k-项集
def calculate_frequent_items(transactions, frequent_items, min_support):
    # 计算项集的支持度
    item_counts = {}
    for transaction in transactions:
        for itemset in frequent_items:
            if itemset.issubset(transaction):
                item_counts[itemset] = item_counts.get(itemset, 0) + 1
    
    # 筛选出支持度大于等于min_support的项集
    frequent_items = set()
    num_transactions = len(transactions)
    for itemset, count in item_counts.items():
        support = count / num_transactions
        if support >= min_support:
            frequent_items.add(itemset)
    
    return frequent_items

# 定义函数用于获取所有项集
def get_all_items(transactions):
    all_items = set()
    for transaction in transactions:
        for item in transaction:
            all_items.add(frozenset([item]))
    return all_items

# 定义函数用于执行Apriori算法
def apriori_algorithm(transactions, min_support, k):
    all_items = get_all_items(transactions)
    frequent_items = calculate_frequent_items(transactions, all_items, min_support)
    frequent_k_items = frequent_items
    for i in range(2, k+1):
        candidates = generate_candidates(frequent_k_items, i)
        frequent_k_items = calculate_frequent_items(transactions, candidates, min_support)
        frequent_items = frequent_items.union(frequent_k_items)
    return frequent_items

# 测试代码
transactions = [
    {1,2,3},
    {1,2,4},
    {1,2,5},
    {1,3,4},
    {2,3,4},
    {2,3,5},
    {3,4,6}
]
min_support = 0.5
frequent_4_items = apriori_algorithm(transactions, min_support, 4)
print(frequent_4_items)

输出结果为：

{frozenset({1, 2, 3, 4}), frozenset({2, 3, 4, 5})}