3

我想将 Apriori 算法应用于零售数据集（来自零售店的购物篮数据）。它具有以下形式的数据：-

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 
30 31 32 
33 34 35 
36 37 38 39 40 41 42 43 44 45 46 
38 39 47 48 
38 39 48 49 50 51 52 53 54 55 56 57 58 
32 41 59 60 61 62 
3 39 48

因此，为了使用 Apriori 算法，我需要将 Python 列表形式的数据放入 Numpy 数组中：-

Column Names as 0 1 2 3 4 5 6 7 8 9 10........

数据集为：

0 1 2 3 4 5 6 7 8 9 10 .........30 31 32 33 34 35....
1 1 1 1 1 1 1 1 1 1 1...........0  0  0  0  0  0...
0 0 0 0 0 0 0 0 0 0 0...........1  1  1  0  0  0..
and so on..

为此，我正在尝试使用事务编码器：-

dataset = pd.read_csv('retail.dat', header=None)
from mlxtend.preprocessing import TransactionEncoder

transactionEncoder = TransactionEncoder()
dataset = transactionEncoder.fit(dataset).transform(dataset)
dataset.astype('int')
print(dataset)

但我得到了错误： -

TypeError: 'int' object is not iterable

我还想将列名作为 0 1 2.... 附加到新形成的数据集，但print(transactionEncoder.columns_)不提供有效列。请告诉可能是什么问题以及在此数据集上应用事务编码器的正确方法是什么...

4

2 回答 2

3

IIUC，您可以堆叠数据框并尝试crosstab：

df =  pd.read_csv('retail.dat', sep=' ', header=None)
new_df = df.stack().astype(int).reset_index(name='value')
pd.crosstab(new_df['level_0'], new_df['value'])

输出：

value    0   1   2   3   4   5   6   7   8   9   ...  53  54  55  56  57  58  ...
level_0                                          ...                           
0         1   1   1   1   1   1   1   1   1   1  ...   0   0   0   0   0   0   
1         0   0   0   0   0   0   0   0   0   0  ...   0   0   0   0   0   0   
2         0   0   0   0   0   0   0   0   0   0  ...   0   0   0   0   0   0   
3         0   0   0   0   0   0   0   0   0   0  ...   0   0   0   0   0   0   
4         0   0   0   0   0   0   0   0   0   0  ...   0   0   0   0   0   0   
5         0   0   0   0   0   0   0   0   0   0  ...   1   1   1   1   1   1   
6         0   0   0   0   0   0   0   0   0   0  ...   0   0   0   0   0   0   
7         0   0   0   1   0   0   0   0   0   0  ...   0   0   0   0   0   0
于 2019-10-17T15:19:33.197 回答
1

你可以试试这个：

import pandas as pd
import numpy as np
from io import StringIO
from mlxtend.preprocessing import TransactionEncoder

inputstr = StringIO("""0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 
30 31 32 
33 34 35 
36 37 38 39 40 41 42 43 44 45 46 
38 39 47 48 
38 39 48 49 50 51 52 53 54 55 56 57 58 
32 41 59 60 61 62 
3 39 48 """)

df = pd.read_csv(inputstr, header=None,sep='\s+')
df_out = df.apply(lambda x: list(x.dropna().values), axis=1).tolist()

transactionEncoder = TransactionEncoder()
dataset = transactionEncoder.fit(df_out).transform(df_out)
dataset = dataset.astype('int')
print(dataset)

输出：

[[1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0
  0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0
  0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1
  0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  0 0 1 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  0 0 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
  0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1]
 [0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0]]

并转换为数据框：

dataset_df = pd.DataFrame(dataset)

输出：

    0   1   2   3   4   5   6   7   8   9   ...  53  54  55  56  57  58  59 
0   1   1   1   1   1   1   1   1   1   1  ...   0   0   0   0   0   0   0   
1   0   0   0   0   0   0   0   0   0   0  ...   0   0   0   0   0   0   0   
2   0   0   0   0   0   0   0   0   0   0  ...   0   0   0   0   0   0   0   
3   0   0   0   0   0   0   0   0   0   0  ...   0   0   0   0   0   0   0   
4   0   0   0   0   0   0   0   0   0   0  ...   0   0   0   0   0   0   0   
5   0   0   0   0   0   0   0   0   0   0  ...   1   1   1   1   1   1   0   
6   0   0   0   0   0   0   0   0   0   0  ...   0   0   0   0   0   0   1   
7   0   0   0   1   0   0   0   0   0   0  ...   0   0   0   0   0   0   0
于 2019-10-17T15:39:08.457 回答