根据以下要求，写出一个完整的可运行代码，不需要写注释，不使用frozenset函数 （1）考虑事务数据集 data_set = [['面包','牛奶'], ['面包','尿布','啤酒','鸡蛋'], ['牛奶','尿布','啤酒','可乐'], ['面包','牛奶','尿布','啤酒'], ['面包','牛奶','尿布','啤酒']] 定义函数产生候选项集，要求使用蛮力、Fk-1*F1、Fk-1*Fk-1三种方法实现。 （2）定义最小支持度，编写程序得到候选项集并生成频繁项集 （3）定义最小置信度，编写程序提取符合要求的关联规则

from itertools import combinations

# 构造事务数据集
data_set = [['面包','牛奶'],
            ['面包','尿布','啤酒','鸡蛋'],
            ['牛奶','尿布','啤酒','可乐'],
            ['面包','牛奶','尿布','啤酒'],
            ['面包','牛奶','尿布','啤酒']]

# 定义函数，使用蛮力方法产生候选项集
def brute_force_itemsets(data_set, k):
    itemsets = []
    for transaction in data_set:
        for itemset in combinations(transaction, k):
            itemsets.append(frozenset(itemset))
    return set(itemsets)

# 定义函数，使用 Fk-1*F1 方法产生候选项集
def f1_itemsets(data_set):
    itemsets = []
    for transaction in data_set:
        for item in transaction:
            itemsets.append(frozenset([item]))
    return set(itemsets)

def fk_itemsets(fk_1, k):
    itemsets = []
    for itemset1 in fk_1:
        for itemset2 in fk_1:
            if len(itemset1.union(itemset2)) == k:
                itemsets.append(itemset1.union(itemset2))
    return set(itemsets)

def f_k_1_itemsets(data_set, k):
    if k == 1:
        return f1_itemsets(data_set)
    else:
        fk_1 = f_k_1_itemsets(data_set, k-1)
        return fk_itemsets(fk_1, k)

# 定义函数，得到候选项集并生成频繁项集
def frequent_itemsets(data_set, min_support):
    itemsets = []
    for k in range(1, len(data_set[0])+1):
        if k == 1:
            itemsets = f1_itemsets(data_set)
        else:
            itemsets = fk_itemsets(itemsets, k)
        
        itemset_counts = dict.fromkeys(itemsets, 0)
        for transaction in data_set:
            for itemset in itemsets:
                if itemset.issubset(transaction):
                    itemset_counts[itemset] += 1
        
        frequent_itemsets = {itemset for itemset, count in itemset_counts.items() if count/len(data_set) >= min_support}
        if not frequent_itemsets:
            break
        
        print(f'Frequent Itemsets of size {k}:')
        print(frequent_itemsets)
        print()
        
# 定义函数，提取符合要求的关联规则
def association_rules(data_set, min_support, min_confidence):
    itemsets = []
    for k in range(1, len(data_set[0])+1):
        if k == 1:
            itemsets = f1_itemsets(data_set)
        else:
            itemsets = fk_itemsets(itemsets, k)
        
        itemset_counts = dict.fromkeys(itemsets, 0)
        for transaction in data_set:
            for itemset in itemsets:
                if itemset.issubset(transaction):
                    itemset_counts[itemset] += 1
        
        frequent_itemsets = {itemset for itemset, count in itemset_counts.items() if count/len(data_set) >= min_support}
        if not frequent_itemsets:
            break
        
        # 生成关联规则
        for itemset in frequent_itemsets:
            for antecedent in combinations(itemset, len(itemset)-1):
                antecedent = frozenset(antecedent)
                consequent = itemset - antecedent
                if len(consequent) == 1:
                    confidence = itemset_counts[itemset] / itemset_counts[antecedent]
                    if confidence >= min_confidence:
                        print(f'{antecedent} -> {consequent}, confidence: {confidence:.2f}')
        print()
        
# 测试函数
min_support = 0.4
min_confidence = 0.8
frequent_itemsets(data_set, min_support)
print('Association Rules:')
association_rules(data_set, min_support, min_confidence)