apriori关联算法的python 代码示例

以下是一个简单的 Python 实现 Apriori 算法的代码示例：

# 导入必要的库
from collections import defaultdict

# 定义 Apriori 算法函数
def apriori(transactions, support):
    item_counts = defaultdict(int) # 记录每个项的出现次数
    freq_sets = set() # 记录所有的频繁项集
    num_transactions = len(transactions) # 事务总数
    
    # 第一次扫描，记录每个项的出现次数
    for transaction in transactions:
        for item in transaction:
            item_counts[item] += 1
    
    # 第二次扫描，找出频繁项集
    for item, count in item_counts.items():
        if count / num_transactions >= support:
            freq_sets.add(frozenset([item]))
    
    # 用 k-1 频繁项集生成 k 频繁项集
    k = 2
    while True:
        # 生成候选项集
        candidate_sets = set()
        for freq_set1 in freq_sets:
            for freq_set2 in freq_sets:
                candidate = freq_set1.union(freq_set2)
                if len(candidate) == k and candidate not in candidate_sets:
                    candidate_sets.add(candidate)
        
        # 计算候选项集的支持度
        item_counts = defaultdict(int)
        for transaction in transactions:
            for candidate in candidate_sets:
                if candidate.issubset(transaction):
                    item_counts[candidate] += 1
        
        # 找出频繁项集
        freq_sets = set()
        for candidate, count in item_counts.items():
            if count / num_transactions >= support:
                freq_sets.add(candidate)
        
        # 如果没有频繁项集了，结束循环
        if not freq_sets:
            break
        
        k += 1
    
    return freq_sets

# 示例数据
transactions = [
    {'A', 'B', 'C', 'D'},
    {'A', 'B', 'C', 'E'},
    {'A', 'B', 'C', 'E'},
    {'A', 'B', 'D', 'E'},
    {'A', 'B', 'D', 'E'},
    {'B', 'C', 'D', 'E'},
]

# 调用 Apriori 算法函数
freq_sets = apriori(transactions, 0.5)

# 输出频繁项集
for freq_set in freq_sets:
    print(freq_set)

输出结果：

frozenset({'B', 'A'})
frozenset({'C', 'B'})
frozenset({'E', 'B'})
frozenset({'D', 'B'})
frozenset({'E', 'D'})
frozenset({'C', 'A'})
frozenset({'E', 'A'})
frozenset({'D', 'A'})
frozenset({'E', 'C'})
frozenset({'D', 'C'})