Association rules Apriori algorithm example

Association rules

The following data is calculated using association rules

import pandas as pd
#import Apriori
from apriori import *

inputfile ='../menu_orders.xls'
outputfile = 'tmp/apriori_rules.xls'
data = pd.read_excel(inputfile,header = None)

print('\n转换原始数据至0-1矩阵...')
ct=lambda x:pd.Series(1,index = x[pd.notnull(x)]) #转换0-1矩阵的过渡函数
b=map(ct,data.as_matrix()) #用map方式执行
data=pd.DataFrame(list(b)).fillna(0) #实现矩阵转换，空值用0填充
print('\n转换完毕')
del b #删除中间变量b

support = 0.2 #最小支持度
confidence=0.5 #最小置信度
ms= '---' #连接符，默认‘--’，用来区分不同元素，如A--B
find_rule(data,support,confidence,ms).to_excel(outputfile)

Calculation results

结果为：
           support  confidence
e---a          0.3    1.000000
e---c          0.3    1.000000
c---e---a      0.3    1.000000
a---e---c      0.3    1.000000
c---a          0.5    0.714286
a---c          0.5    0.714286
a---b          0.5    0.714286
c---b          0.5    0.714286
b---a          0.5    0.625000
b---c          0.5    0.625000
a---c---e      0.3    0.600000
b---c---a      0.3    0.600000
a---c---b      0.3    0.600000
a---b---c      0.3    0.600000

Form data processing and association rule calculation

Table data processing

import pandas as pd
from apriori import *

inputdata='data/caidang.xlsx'

data=pd.read_excel(inputdata)
print(data.columns)

print('\n转换数据')
df=data[['订单号','菜品','序列']]

print('*****重构数据*****')
data=df.set_index(['订单号','菜品']).unstack().replace(range(len(df)+1),1).fillna(0)

print('*****删除索引*****') 
data=pd.DataFrame(data.values,columns=data.columns.levels[1])

support = 0.2 #最小支持度
confidence=0.5 #最小置信度
ms= '---' #连接符，默认‘--’，用来区分不同元素，如A--B
find_rule(data,support,confidence).to_excel(outputfile)

Apriori algorithm

#-*- coding: utf-8 -*-
from __future__ import print_function
import pandas as pd

#自定义连接函数，用于实现L_{k-1}到C_k的连接
def connect_string(x, ms):
  x = list(map(lambda i:sorted(i.split(ms)), x))
  l = len(x[0])
  r = []
  for i in range(len(x)):
    for j in range(i,len(x)):
      if x[i][:l-1] == x[j][:l-1] and x[i][l-1] != x[j][l-1]:
        r.append(x[i][:l-1]+sorted([x[j][l-1],x[i][l-1]]))
  return r

#寻找关联规则的函数
def find_rule(d, support, confidence, ms = u'--'):
  result = pd.DataFrame(index=['support', 'confidence']) #定义输出结果
  
  support_series = 1.0*d.sum()/len(d) #支持度序列
  column = list(support_series[support_series > support].index) #初步根据支持度筛选
  k = 0
  
  while len(column) > 1:
    k = k+1
    print(u'\n正在进行第%s次搜索...' %k)
    column = connect_string(column, ms)
    print(u'数目：%s...' %len(column))
    sf = lambda i: d[i].prod(axis=1, numeric_only = True) #新一批支持度的计算函数
    
    #创建连接数据，这一步耗时、耗内存最严重。当数据集较大时，可以考虑并行运算优化。
    d_2 = pd.DataFrame(list(map(sf,column)), index = [ms.join(i) for i in column]).T
    
    support_series_2 = 1.0*d_2[[ms.join(i) for i in column]].sum()/len(d) #计算连接后的支持度
    column = list(support_series_2[support_series_2 > support].index) #新一轮支持度筛选
    support_series = support_series.append(support_series_2)
    column2 = []
    
    for i in column: #遍历可能的推理，如{A,B,C}究竟是A+B-->C还是B+C-->A还是C+A-->B？
      i = i.split(ms)
      for j in range(len(i)):
        column2.append(i[:j]+i[j+1:]+i[j:j+1])
    
    cofidence_series = pd.Series(index=[ms.join(i) for i in column2]) #定义置信度序列
 
    for i in column2: #计算置信度序列
      cofidence_series[ms.join(i)] = support_series[ms.join(sorted(i))]/support_series[ms.join(i[:len(i)-1])]
    
    for i in cofidence_series[cofidence_series > confidence].index: #置信度筛选
      result[i] = 0.0
      result[i]['confidence'] = cofidence_series[i]
      result[i]['support'] = support_series[ms.join(sorted(i.split(ms)))]
  
  result = result.T.sort_values(['confidence','support'], ascending = False) #结果整理，输出
  print(u'\n结果为：')
  print(result)
  
  return result