数据挖掘之Apriori算法详解和Python实现代码分享

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:

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#-*- 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 < self.line_num * self.min_support / 100:
                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()
############################################################################
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apriori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：

import apriori
c = apriori.apriori('basket.txt', 11, 3, 13)

输出：

--------------------------------------------------------------------------------
the 1 loop
location [[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 loop
location [[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 loop
location [[3, 5, 6]]
support [146]
num [3, 5, 6]
--------------------------------------------------------------------------------
frozenmeal,beer ->> cannedveg  min_support:  14.6%  min_confidence: 0.858823529412
cannedveg,beer ->> frozenmeal  min_support:  14.6%  min_confidence: 0.874251497006
cannedveg,frozenmeal ->> beer  min_support:  14.6%  min_confidence: 0.843930635838
--------------------------------------------------------------------------------