亲和性分析根据样本个体之间的相似度,确定它们关系的亲疏。应用场景:
1.向网站用户提供多样化的服务或投放定向广告。
2.为了向用户推荐电影或商品
3.根据基因寻找有亲缘关系的人
比如:统计顾客购买了商品1,然后再购买商品2的比率,算相似度。
import numpy as np dataset_filename = "affinity_dataset.txt" x = np.loadtxt(dataset_filename) # print x[:5] # 上述代码的结果代表前5次交易中顾客购买了什么。用“1”表示购买,“0”表示没有购买。 # 这五种商品分别是:面包,牛奶,奶酪,苹果和香蕉。 # 现在我们要找出“如果顾客购买了商品x,那么他们可能愿意购买商品y”的规则(一条规则有前提条件和结论两部分组成)。衡量一个规则的优劣通常有:支持度(指数据集中规则应验的次数)和置信度(指规则准确率如何,计算方法是:规则应验次数除以满足前提条件的所有次数)。 # 举个例子计算有多少人购买了苹果。 num_apples_purchases = 0 for sample in x: if sample[3] == 1: num_apples_purchases += 1 # print "{0} people bought Apples".format(num_apples_purchases) # 接着我们计算有多少人购买了苹果,后又购买了香蕉。同时计算支持度和置信度。 num_apples_bananas_purchases = 0 for sample in x: if sample[3] == 1 and sample[4] == 1: num_apples_bananas_purchases += 1 valid_rules = num_apples_bananas_purchases num_occurances = num_apples_purchases support = valid_rules confidence = valid_rules/float(num_occurances) print "{0} people bought Apples, but {1} people also bought bananas".format(num_apples_purchases, num_apples_bananas_purchases) print "------" # 支持度 print support # 置信度 print "{0:.3f}".format(confidence) # 我们接着将所有规则下的可能性都统计出来,找出亲和性最高的几个。首先,分为两种:一种是规则应验,一种是规则无效。分别创建字典。字典的键是由条件和结论组成的元组,元组元素为特征在特征列表中的索引值,比如“如果顾客买了苹果,他们也会买香蕉”就用(3,4)表示。这里使用defaultdict,好处是如果查找的键不存在,返回一个默认值。 from collections import defaultdict features = ["bread", "milk", "cheese", "apple", "banana"] valib_rules = defaultdict(int) invalib_rules = defaultdict(int) num_occurances = defaultdict(int) # 依次对样本的每个个体及个体的每个特征值进行处理。第一个特征为规则的前提条件。 for sample in x: for premise in xrange(4): if sample[premise] == 0: continue num_occurances[premise] += 1 # 比如“顾客买了苹果,他们也买了苹果”,这样的规则是没有意义的。 for conclusion in xrange(len(features)): if premise == conclusion: continue if sample[conclusion] == 1: valib_rules[(premise, conclusion)] += 1 else: invalib_rules[(premise, conclusion)] += 1 support = valib_rules confidence = defaultdict(float) ''' for premise, conclusion in valib_rules.keys(): rule = (premise, conclusion) confidence[rule] = valib_rules[rule] / num_occurances[premise] ''' # 这样我们就得到了支持度字典和置信度字典。我们再来创建一个函数,以便更加方便查看结果。 def print_rule(premise, conclusion, support, confidence, features): premise_name = features[premise] conclusion_name = features[conclusion] confidence[(premise, conclusion)] = valib_rules[(premise, conclusion)] / float(num_occurances[premise]) print "Rule: If a person buys {0} they will also buy {1}".format(premise_name, conclusion_name) print "- Support: {0}".format(support[(premise, conclusion)]) print "- Confidence: {0:.3f}".format(confidence[(premise, conclusion)]) if __name__ == "__main__": premise = 1 conclusion = 3 # print print_rule(premise, conclusion, support, confidence, features) # 排序找出最佳的规则。对字典排序:首先字典的items()函数返回包含字典所有元素的列表,再使用itemgetter()类作为键,这样就可以对嵌套列表进行排序了。 from operator import itemgetter sorted_support = sorted(support.items(), key=itemgetter(1), reverse=True) # 提取支持度最高的5条 for index in range(5): print "Rule #{0}".format(index + 1) premise, conclusion = sorted_support[index][0] print_rule(premise, conclusion, support, confidence, features) # 总结亲和性分析,可以清楚的看出哪两种商品一起购买的几率要大些,经理就可以根据这些规则来调整商品摆放的位置,从而为商家带来更大的经济效益。
affinity_dataset.txt
0 1 0 0 0
1 1 0 0 0
0 0 1 0 1
1 1 0 0 0
0 0 1 1 1
0 1 0 0 0
0 0 1 1 1
0 0 1 1 0
0 1 0 1 0
0 1 0 0 1
0 0 0 1 0
1 0 1 0 0
1 0 0 0 1
0 1 1 0 0
0 0 1 0 1
0 1 0 1 0
1 1 0 1 1
0 0 0 1 1
0 1 0 0 1
1 1 0 1 0
0 1 1 0 0
0 1 0 0 1
0 0 1 0 0
1 0 0 0 1
0 1 0 1 0
1 0 0 1 1
0 1 1 0 0
0 1 0 0 1
0 0 0 0 1
1 0 0 0 1
0 1 0 1 1
1 0 0 0 0
0 1 0 0 0
1 0 0 0 0
0 0 1 1 1
0 0 1 1 1
0 0 1 1 1
1 0 0 1 0
0 1 0 0 1
1 1 0 0 0
0 0 0 0 1
0 1 0 1 1
0 1 0 1 0
0 1 0 0 1
1 1 1 1 0
1 0 0 0 1
0 0 0 1 1
1 1 0 0 1
0 1 0 0 0
0 1 1 0 0
0 1 0 1 1
0 1 0 0 1
0 0 1 1 1
0 0 0 1 1
0 0 1 0 0
0 0 1 1 1
1 0 0 0 0
1 1 1 0 1
0 0 1 1 1
0 1 0 0 0
0 0 1 1 0
0 1 0 0 1
0 0 1 0 0
0 1 0 0 0
1 0 0 0 1
0 1 0 0 0
0 1 1 0 1
0 0 1 0 0
0 0 1 0 0
0 0 0 1 1
0 0 1 0 0
0 0 1 1 0
0 1 0 0 0
0 1 1 1 1
1 1 0 0 1
0 0 1 1 0
0 0 1 1 0
0 0 1 1 1
0 0 1 1 1
0 1 0 0 0
0 1 0 1 0
1 1 0 0 1
0 1 0 0 1
0 0 1 1 1
0 1 0 0 1
0 1 0 1 1
0 1 0 0 1
1 0 0 0 0
1 0 0 1 1
0 1 1 1 1
1 0 0 0 1
0 0 1 0 1
0 1 1 1 0
1 1 0 1 1
1 0 1 0 1
0 0 1 1 1
1 1 1 1 0
0 1 0 0 1
0 1 0 0 1
1 1 0 1 1