以下是使用Python实现的Apriori算法的数值矩阵解决方法的代码示例：

# 导入所需的库
from itertools import combinations

# 定义函数来生成候选项集
def generate_candidates(items, length):
    candidates = []
    for item in items:
        for i in range(len(item)):
            candidate = item[:i] + item[i+1:]
            if candidate not in candidates and len(candidate) == length:
                candidates.append(candidate)
    return candidates

# 定义函数来计算项集的支持度
def calculate_support(itemset, transactions):
    count = 0
    for transaction in transactions:
        if all(item in transaction for item in itemset):
            count += 1
    return count

# 定义Apriori算法函数
def apriori(transactions, min_support, min_confidence):
    # 构建初始项集
    items = []
    for transaction in transactions:
        for item in transaction:
            if [item] not in items:
                items.append([item])

    # 生成频繁项集
    frequent_itemsets = []
    k = 2
    while True:
        candidates = generate_candidates(items, k)
        supports = []
        for candidate in candidates:
            support = calculate_support(candidate, transactions)
            supports.append(support)
            if support >= min_support:
                frequent_itemsets.append(candidate)
        if not frequent_itemsets:
            break
        items = frequent_itemsets.copy()
        k += 1

    # 生成关联规则
    association_rules = []
    for itemset in frequent_itemsets:
        if len(itemset) > 1:
            subsets = list(combinations(itemset, 1))
            for subset in subsets:
                subset_support = calculate_support(list(subset), transactions)
                confidence = calculate_support(itemset, transactions) / subset_support
                if confidence >= min_confidence:
                    association_rules.append((list(subset), list(set(itemset) - set(subset))), confidence)
    return frequent_itemsets, association_rules

# 示例用法
transactions = [
    ['A', 'B', 'C'],
    ['A', 'B', 'D'],
    ['B', 'C', 'E'],
    ['A', 'B', 'C', 'D'],
    ['B', 'D', 'E']
]

min_support = 2
min_confidence = 0.5

frequent_itemsets, association_rules = apriori(transactions, min_support, min_confidence)

# 输出结果
print("频繁项集:")
for itemset in frequent_itemsets:
    print(itemset)

print("\n关联规则:")
for rule in association_rules:
    print(rule)

这个示例代码中使用的数据集是一个包含5个事务的列表，每个事务都是一个项集。在示例中，我们设置了最小支持度为2（即项集在所有事务中出现的最小次数）和最小置信度为0.5（即关联规则必须满足的最低置信度）。运行代码后，将输出频繁项集和关联规则的列表。