数据挖掘-商品购买记录-商品推荐_亲和度分析

痛定思痛。 2022-07-11 23:57 218阅读 0赞

# coding: utf-8  
    import numpy as np
    dataset_filename="affinity_dataset.txt"
    #加载数据文件
    dataset=np.loadtxt(dataset_filename)
    '''
    数据结果:
    array([[ 0.,  0.,  1.,  1.,  1.],
           [ 1.,  1.,  0.,  1.,  0.]
           ])
           '''
    #一行表示一个用户对五个商品的购买情况,1表示买了，0表示没买
    n_samples,n_features=dataset.shape
    print("{0} samples in total with {1} features".format(n_samples,n_features))
    
    #定义特征名集合
    features=["bread","milk","cheese","apple","banana"]
    
    #目标一 ， 找出以下规则:要找出用户购买了apple, 那他也可能购买banana
    # 前提：计算规则的支持度和置信度
    #支持度:数据集中规则应验的次数.
    #置信度: 符合给定条件的所有规则里，跟当前规则结论一致的比例有多少. 
    
    #首先, 有多少行有这条规则的前提条件
    num_apple_purchases=0
    for sample in dataset:
        if sample[3]==1:
            num_apple_purchases+=1
    print("{0} people bought apples".format(num_apple_purchases))
    
    #下面可以计算这条规则的应验数和无效数
    rule_valid=0.0
    rule_invalid=0
    for sample in dataset:
        if sample[3]==1:
            if sample[4]==1:
                rule_valid+=1
            else:
                rule_invalid+=1
    print("{0} cases of the rule being valid".format(rule_valid))
    print("{0} cases of the rule being invalid".format(rule_invalid))
    
    #支持度
    support=rule_valid
    #置信度
    confidence=rule_valid/num_apple_purchases
    print("the support is {0} and the confidence is {1:.3f}".format(support, confidence))
    
    
    ################################################################
    #有了上面的案例，用于理解支持度和置信度
    #下面开始所有的规则的支持度和置信度的计算
    
    from collections import defaultdict
    #数据格式采用字典：{ 键(前提,结论), 值(数值)}
    valid_rules=defaultdict(int)    #每条规则的支持度
    invalid_rules=defaultdict(int)  #每条规则的无效数
    #数据格式采用字典：{ 键(前提), 值(数值)}
    num_rule_occurances=defaultdict(float)   #规则条件出现的次数
    
    for sample in dataset:
        # premise 前提条件
        for premise in range(n_features):    
            if sample[premise] ==0: continue    #这个商品没有购买，则表示这个前提不成立,跳过循环
            num_rule_occurances[premise]+=1.0     #出现了一次前提条件
            for conclusion in range(n_features):   #计算它的支持度和无效数
                if premise==conclusion:
                    continue
                if sample[conclusion]==1:
                    valid_rules[(premise,conclusion)]+=1
                else:
                    invalid_rules[(premise,conclusion)]+=1
                                  
    support=valid_rules
    confidence=defaultdict(float)
    for premise,conclusion in valid_rules.keys():
        confidence[(premise,conclusion)]=valid_rules[(premise,conclusion)]/num_rule_occurances[premise]
    #下面将结果输出一下
    for premise, conclusion in confidence:
        #求出规则前提条件商品名
        premise_name=features[premise]
        #求出规则条件的结论对应的商品名
        conclusion_name=features[conclusion]
        print("Rule: if a person buys {0} they will also buy {1}".format(premise_name,conclusion_name))
        print(" this rule's support is :{0}".format( support[(premise,conclusion)]     )  )
        print(" this rule's confidence is :{0}".format( confidence[(premise,conclusion)]     )  )
        print("")
        
    
    '''
      将输出部分定义成一个函数
      premise: 前提 的索引
      conclusion: 结论的索引
      support: 支持度字典
      confidence: 置信度字典
      features: 特征数组
    '''
    def print_rule( premise,conclusion,support,confidence,features):
         #求出规则前提条件商品名
        premise_name=features[premise]
        #求出规则条件的结论对应的商品名
        conclusion_name=features[conclusion]
        print("Rule: if a person buys {0} they will also buy {1}".format(premise_name,conclusion_name))
        print(" this rule's support is :{0}".format( support[(premise,conclusion)]     )  )
        print(" this rule's confidence is :{0}".format( confidence[(premise,conclusion)]     )  )
        print("")
        
    #调用上面的函数，输出计算好的规则的支持度和置信度,比如，买了milk后又买apple的支持度和置信度
    premise=1
    conclusion=3
    print_rule(premise,conclusion,support,confidence,features)
    
    ###################################################################################
    '''
      下面对支持度进行排序
    '''
    #先排序输出
    from pprint import pprint
    pprint(list(support.items()))
    #排好序后存到字典中
    from operator import itemgetter
    #                根据支持度的值排序 
    sorted_support=sorted(support.items(), key=itemgetter(1),reverse=True)
    #输出前三名
    for index in range(5):
        print("Rule #{0}".format(index+1))
        (premise,conclusion)=sorted_support[index][0]   #取出键
        print_rule( premise, conclusion, support,confidence,features)
    '''
      下面对信任度进行排序
    '''
    #先排序输出
    pprint(list(confidence.items()))
    #排好序后存到字典中
    #                根据信任度的值排序 
    sorted_confidence=sorted(confidence.items(), key=itemgetter(1),reverse=True)
    #输出前三名
    for index in range(5):
        print("Rule #{0}".format(index+1))
        (premise,conclusion)=sorted_confidence[index][0]   #取出键
        print_rule( premise, conclusion, support,confidence,features)

结论:

1. 支持度排名前五的规则

![Center][]

[Center]: /images/20220711/fdbe1b9d46674266a181334600b0cbb7.png