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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 branch: master  datamining / apriori / apriori.py

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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

--------------------------------------------------------------------------------