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关联规则挖掘（Association rule mining）是数据挖掘中最活跃的研究方法之一，可以用来发现事情之间的联系，最早是为了发现超市交易数据库中不同的商品之间的关系。(啤酒与尿布)基本概念1、支持度的定义：support(X-->Y) = |X交Y|/N=集合X与集合Y中的项在一条记录中同时出现的次数/数据记录的个数。例如：support({啤酒}-->{尿布}) = 啤酒和尿布同时出现的次数/数据记录数 = 3/5=60%。2、自信度的定义：confidence(X-->Y) = |X交Y|/|X| = 集合X与集合Y中的项在一条记录中同时出现的次数/集合X出现的个数 。例如：confidence({啤酒}-->{尿布}) = 啤酒和尿布同时出现的次数/啤酒出现的次数=3/3=100%;confidence({尿布}-->{啤酒}) = 啤酒和尿布同时出现的次数/尿布出现的次数 = 3/4 = 75%同时满足最小支持度阈值(min_sup)和最小置信度阈值(min_conf)的规则称作强规则 ,如果项集满足最小支持度,则称它为频繁项集“如何由大型数据库挖掘关联规则?”关联规则的挖掘是一个两步的过程:1、找出所有频繁项集:根据定义,这些项集出现的频繁性至少和预定义的最小支持计数一样。2、由频繁项集产生强关联规则:根据定义,这些规则必须满足最小支持度和最小置信度。Apriori定律为了减少频繁项集的生成时间，我们应该尽早的消除一些完全不可能是频繁项集的集合，Apriori的两条定律就是干这事的。Apriori定律1：如果一个集合是频繁项集，则它的所有子集都是频繁项集。举例：假设一个集合{A,B}是频繁项集，即A、B同时出现在一条记录的次数大于等于最小支持度min_support，则它的子集{A},{B}出现次数必定大于等于min_support，即它的子集都是频繁项集。Apriori定律2：如果一个集合不是频繁项集，则它的所有超集都不是频繁项集。举例：假设集合{A}不是频繁项集，即A出现的次数小于min_support，则它的任何超集如{A,B}出现的次数必定小于min_support，因此其超集必定也不是频繁项集。上面的图演示了Apriori算法的过程，注意看由二级频繁项集生成三级候选项集时，没有{牛奶,面包,啤酒}，那是因为{面包,啤酒}不是二级频繁项集，这里利用了Apriori定理。最后生成三级频繁项集后，没有更高一级的候选项集，因此整个算法结束，{牛奶,面包,尿布}是最大频繁子集。Python实现代码:复制代码 代码如下:Skip to contentSign up Sign in This repositoryExploreFeaturesEnterpriseBlog Star 0 Fork 0 taizilongxu/datamining branch: master datamining / apriori / apriori.pyhackerxutaizilongxu 20 days ago backup1 contributor156 lines (140 sloc) 6.302 kb RawBlameHistory #-*- encoding: UTF-8 -*-#---------------------------------import------------------------------------#---------------------------------------------------------------------------class Apriori(object): def __init__(self, filename, min_support, item_start, item_end): self.filename = filename self.min_support = min_support # 最小支持度 self.min_confidence = 50 self.line_num = 0 # item的行数 self.item_start = item_start # 取哪行的item self.item_end = item_end self.location = [[i] for i in range(self.item_end - self.item_start + 1)] self.support = self.sut(self.location) self.num = list(sorted(set([j for i in self.location for j in i])))# 记录item self.pre_support = [] # 保存前一个support,location,num self.pre_location = [] self.pre_num = [] self.item_name = [] # 项目名 self.find_item_name() self.loop() self.confidence_sup() def deal_line(self, line): "提取出需要的项" return [i.strip() for i in line.split(' ') if i][self.item_start - 1:self.item_end] def find_item_name(self): "根据第一行抽取item_name" with open(self.filename, 'r') as F: for index,line in enumerate(F.readlines()): if index == 0: self.item_name = self.deal_line(line) break def sut(self, location): """ 输入[[1,2,3],[2,3,4],[1,3,5]...] 输出每个位置集的support [123,435,234...] """ with open(self.filename, 'r') as F: support = [0] * len(location) for index,line in enumerate(F.readlines()): if index == 0: continue # 提取每信息 item_line = self.deal_line(line) for index_num,i in enumerate(location): flag = 0 for j in i: if item_line[j] != 'T': flag = 1 break if not flag: support[index_num] += 1 self.line_num = index # 一共多少行,出去第一行的item_name return support def select(self, c): "返回位置" stack = [] for i in self.location: for j in self.num: if j in i: if len(i) == c: stack.append(i) else: stack.append([j] + i) # 多重列表去重 import itertools s = sorted([sorted(i) for i in stack]) location = list(s for s,_ in itertools.groupby(s)) return location def del_location(self, support, location): "清除不满足条件的候选集" # 小于最小支持度的剔除 for index,i in enumerate(support): if i support[index] = 0 # apriori第二条规则,剔除 for index,j in enumerate(location): sub_location = [j[:index_loc] + j[index_loc+1:]for index_loc in range(len(j))] flag = 0 for k in sub_location: if k not in self.location: flag = 1 break if flag: support[index] = 0 # 删除没用的位置 location = [i for i,j in zip(location,support) if j != 0] support = [i for i in support if i != 0] return support, location def loop(self): "s级频繁项级的迭代" s = 2 while True: print '-'*80 print 'The' ,s - 1,'loop' print 'location' , self.location print 'support' , self.support print 'num' , self.num print '-'*80 # 生成下一级候选集 location = self.select(s) support = self.sut(location) support, location = self.del_location(support, location) num = list(sorted(set([j for i in location for j in i]))) s += 1 if location and support and num: self.pre_num = self.num self.pre_location = self.location self.pre_support = self.support self.num = num self.location = location self.support = support else: break def confidence_sup(self): "计算confidence" if sum(self.pre_support) == 0: print 'min_support error' # 第一次迭代即失败 else: for index_location,each_location in enumerate(self.location): del_num = [each_location[:index] + each_location[index+1:] for index in range(len(each_location))] # 生成上一级频繁项级 del_num = [i for i in del_num if i in self.pre_location] # 删除不存在上一级频繁项级子集 del_support = [self.pre_support[self.pre_location.index(i)] for i in del_num if i in self.pre_location] # 从上一级支持度查找 # print del_num # print self.support[index_location] # print del_support for index,i in enumerate(del_num): # 计算每个关联规则支持度和自信度 index_support = 0 if len(self.support) != 1: index_support = index support = float(self.support[index_location])/self.line_num * 100 # 支持度 s = [j for index_item,j in enumerate(self.item_name) if index_item in i] if del_support[index]: confidence = float(self.support[index_location])/del_support[index] * 100 if confidence > self.min_confidence: print ','.join(s) , '->>' , self.item_name[each_location[index]] , ' min_support: ' , str(support) + '%' , ' min_confidence:' , str(confidence) + '%'def main(): c = Apriori('basket.txt', 14, 3, 13) d = Apriori('simple.txt', 50, 2, 6)if __name__ == '__main__': main()############################################################################Status API Training Shop Blog About© 2014 GitHub, Inc. Terms Privacy Security ContactApriori算法Apriori(filename, min_support, item_start, item_end)参数说明filename:(路径)文件名min_support:最小支持度item_start:item起始位置item_end:item结束位置使用例子：复制代码 代码如下:import aprioric = apriori.Apriori('basket.txt', 11, 3, 13)输出：复制代码 代码如下:--------------------------------------------------------------------------------The 1 looplocation [[0], [1], [2], [3], [4], [5], [6], [7], [8], [9], [10]]support [299, 183, 177, 303, 204, 302, 293, 287, 184, 292, 276]num [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]----------------------------------------------------------------------------------------------------------------------------------------------------------------The 2 looplocation [[0, 9], [3, 5], [3, 6], [5, 6], [7, 10]]support [145, 173, 167, 170, 144]num [0, 3, 5, 6, 7, 9, 10]----------------------------------------------------------------------------------------------------------------------------------------------------------------The 3 looplocation [[3, 5, 6]]support [146]num [3, 5, 6]--------------------------------------------------------------------------------frozenmeal,beer ->> cannedveg min_support: 14.6% min_confidence: 0.858823529412cannedveg,beer ->> frozenmeal min_support: 14.6% min_confidence: 0.874251497006cannedveg,frozenmeal ->> beer min_support: 14.6% min_confidence: 0.843930635838-------------------------------------------------------------------------------- 09-08 14:47