2020-03-05 16:45

# python数据挖掘 关联规则 apriori算法运算中的问题

-*- 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

# -*- coding: utf-8 -*-

inputfile = 'C:/course/c5_data1.xlsx'

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)

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