關(guān)聯(lián)規(guī)則挖掘（Association rule mining）是數(shù)據(jù)挖掘中最活躍的研究方法之一，可以用來發(fā)現(xiàn)事情之間的聯(lián)系，最早是為了發(fā)現(xiàn)超市交易數(shù)據(jù)庫(kù)中不同的商品之間的關(guān)系。(啤酒與尿布)
基本概念
1、支持度的定義：support(X-->Y) = |X交Y|/N=集合X與集合Y中的項(xiàng)在一條記錄中同時(shí)出現(xiàn)的次數(shù)/數(shù)據(jù)記錄的個(gè)數(shù)。例如：support({啤酒}-->{尿布}) = 啤酒和尿布同時(shí)出現(xiàn)的次數(shù)/數(shù)據(jù)記錄數(shù) = 3/5=60%。
2、自信度的定義：confidence(X-->Y) = |X交Y|/|X| = 集合X與集合Y中的項(xiàng)在一條記錄中同時(shí)出現(xiàn)的次數(shù)/集合X出現(xiàn)的個(gè)數(shù) 。例如：confidence({啤酒}-->{尿布}) = 啤酒和尿布同時(shí)出現(xiàn)的次數(shù)/啤酒出現(xiàn)的次數(shù)=3/3=100%;confidence({尿布}-->{啤酒}) = 啤酒和尿布同時(shí)出現(xiàn)的次數(shù)/尿布出現(xiàn)的次數(shù) = 3/4 = 75%
同時(shí)滿足最小支持度閾值(min_sup)和最小置信度閾值(min_conf)的規(guī)則稱作強(qiáng)規(guī)則 ,如果項(xiàng)集滿足最小支持度,則稱它為頻繁項(xiàng)集
“如何由大型數(shù)據(jù)庫(kù)挖掘關(guān)聯(lián)規(guī)則?”關(guān)聯(lián)規(guī)則的挖掘是一個(gè)兩步的過程:
1、找出所有頻繁項(xiàng)集:根據(jù)定義,這些項(xiàng)集出現(xiàn)的頻繁性至少和預(yù)定義的最小支持計(jì)數(shù)一樣。
2、由頻繁項(xiàng)集產(chǎn)生強(qiáng)關(guān)聯(lián)規(guī)則:根據(jù)定義,這些規(guī)則必須滿足最小支持度和最小置信度。
Apriori定律
為了減少頻繁項(xiàng)集的生成時(shí)間，我們應(yīng)該盡早的消除一些完全不可能是頻繁項(xiàng)集的集合，Apriori的兩條定律就是干這事的。
Apriori定律1：如果一個(gè)集合是頻繁項(xiàng)集，則它的所有子集都是頻繁項(xiàng)集。舉例：假設(shè)一個(gè)集合{A,B}是頻繁項(xiàng)集，即A、B同時(shí)出現(xiàn)在一條記錄的次數(shù)大于等于最小支持度min_support，則它的子集{A},{B}出現(xiàn)次數(shù)必定大于等于min_support，即它的子集都是頻繁項(xiàng)集。
Apriori定律2：如果一個(gè)集合不是頻繁項(xiàng)集，則它的所有超集都不是頻繁項(xiàng)集。舉例：假設(shè)集合{A}不是頻繁項(xiàng)集，即A出現(xiàn)的次數(shù)小于min_support，則它的任何超集如{A,B}出現(xiàn)的次數(shù)必定小于min_support，因此其超集必定也不是頻繁項(xiàng)集。
上面的圖演示了Apriori算法的過程，注意看由二級(jí)頻繁項(xiàng)集生成三級(jí)候選項(xiàng)集時(shí)，沒有{牛奶,面包,啤酒}，那是因?yàn)閧面包,啤酒}不是二級(jí)頻繁項(xiàng)集，這里利用了Apriori定理。最后生成三級(jí)頻繁項(xiàng)集后，沒有更高一級(jí)的候選項(xiàng)集，因此整個(gè)算法結(jié)束，{牛奶,面包,尿布}是最大頻繁子集。
Python實(shí)現(xiàn)代碼:
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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的行數(shù)
??????? 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])))# 記錄item
??????? self.pre_support = [] # 保存前一個(gè)support,location,num
??????? self.pre_location = []
??????? self.pre_num = []
??????? self.item_name = [] # 項(xiàng)目名
??????? self.find_item_name()
??????? self.loop()
??????? self.confidence_sup()
??? def deal_line(self, line):
??????? "提取出需要的項(xiàng)"
??????? return [i.strip() for i in line.split(' ') if i][self.item_start - 1:self.item_end]
??? def find_item_name(self):
??????? "根據(jù)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):
??????? """
??????? 輸入[[1,2,3],[2,3,4],[1,3,5]...]
??????? 輸出每個(gè)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):
??????? "清除不滿足條件的候選集"
??????? # 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第二條規(guī)則,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
??????? # 刪除沒用的位置
??????? 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級(jí)頻繁項(xiàng)級(jí)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
??????????? # 生成下一級(jí)候選集
??????????? 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):
??????? "計(jì)算confidence"
??????? if sum(self.pre_support) == 0:
??????????? print 'min_support error' # 第一次迭代即失敗
??????? 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))] # 生成上一級(jí)頻繁項(xiàng)級(jí)
??????????????? del_num = [i for i in del_num if i in self.pre_location] # 刪除不存在上一級(jí)頻繁項(xiàng)級(jí)子集
??????????????? del_support = [self.pre_support[self.pre_location.index(i)] for i in del_num if i in self.pre_location] # 從上一級(jí)支持度查找
??????????????? # print del_num
??????????????? # print self.support[index_location]
??????????????? # print del_support
??????????????? for index,i in enumerate(del_num): # 計(jì)算每個(gè)關(guān)聯(lián)規(guī)則支持度和自信度
??????????????????? 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)
參數(shù)說明
filename:(路徑)文件名
min_support:最小支持度
item_start:item起始位置
item_end:item結(jié)束位置
使用例子：
import apriori
c = apriori.Apriori('basket.txt', 11, 3, 13)
輸出：
--------------------------------------------------------------------------------
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
--------------------------------------------------------------------------------
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