【数据挖掘 | 关联性分析】万字长文详解关联性分析，详解Apriori算法为例，确定不来看看？

from mlxtend.preprocessing import TransactionEncoder
from mlxtend.frequent_patterns import apriori, association_rules
import pandas as pd

# 加载数据集(简单)
dataset = [["牛奶", "洋葱", "鸡蛋"],
           ["洋葱", "肉", "鸡蛋", "香蕉"],
           ["牛奶", "苹果", "鸡蛋"],
           ["牛奶", "洋葱", "肉", "鸡蛋"],
           ["洋葱","苹果", "鸡蛋"],
           ["鸡蛋", "苹果"],
           ["洋葱", "牛奶", "苹果", "鸡蛋"]]

# 处理数据
# 读取数据集
data = pd.read_csv('grocery_store.csv', header=None)

# 数据预处理
transactions = []
for i in range(len(data)):
    transactions.append([str(data.values[i, j]) for j in range(len(data.columns))])
# 根据同一名用户同一时间作为一行数据

   
# 转换数据集为布尔矩阵
te = TransactionEncoder()
te_ary = te.fit(dataset).transform(dataset)
df = pd.DataFrame(te_ary, columns=te.columns_)

# 使用Apriori算法找出频繁项集
frequent_itemsets = apriori(df, min_support=0.2, use_colnames=True)

 # 计算关联规则的置信度和支持度
  rules = association_rules(frequent_itemsets, metric="confidence", min_threshold=0.7)
print(rules)

#-*- 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

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#-*- coding: utf-8 -*-
#使用Apriori算法挖掘菜品订单关联规则
from __future__ import print_function
import pandas as pd
from apriori import * #导入自行编写的apriori函数

inputfile = '../data/menu_orders.xls'
outputfile = '../tmp/apriori_rules.xls' #结果文件
#data = pd.read_excel(inputfile, header = None)
# 加载数据集(简单)
data = [["牛奶", "洋葱", "鸡蛋"],
           ["洋葱", "肉", "鸡蛋", "香蕉"],
           ["牛奶", "苹果", "鸡蛋"],
           ["牛奶", "洋葱", "肉", "鸡蛋"],
           ["洋葱","苹果", "鸡蛋"],
           ["鸡蛋", "苹果"],
           ["洋葱", "牛奶", "苹果", "鸡蛋"]]

print(u'\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(u'\n转换完毕。')
del b #删除中间变量b，节省内存

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。需要保证原始表格中不含有该字符

find_rule(data, support, confidence, ms).to_excel(outputfile) #保存结果