如何使用 Scikit Learn dictvectorizer 从 Python 中的密集数据帧中获取编码数据帧？

Question

我有一个如下的数据框：

   user  item  affinity
0     1    13       0.1
1     2    11       0.4
2     3    14       0.9
3     4    12       1.0

据此，我想创建一个编码数据集（用于fastFM），如下所示：

  user1 user2 user4 user4 item11 item12 item13 item14 affinity
    1     0     0     0     0      0      1      0       0.1
    0     1     0     0     1      0      0      0       0.4
    0     0     1     0     0      0      0      1       0.9
    0     0     0     1     0      1      0      0       1.0

我需要来自sklearn 的dictvectorizer 吗？如果是，那么有没有办法将原始数据帧转换为字典，可以将其提供给dictvectorizer，这又会给我编码的数据集，如图所示？

Answer 1

您可以将get_dummies 与concat 一起使用如果user 或item 列中的值是数字，则由astype 转换为string：

df = pd.DataFrame({'item': {0: 13, 1: 11, 2: 14, 3: 12}, 
                   'affinity': {0: 0.1, 1: 0.4, 2: 0.9, 3: 1.0},
                   'user': {0: 1, 1: 2, 2: 3, 3: 4}},
                    columns=['user','item','affinity'])
print df
   user  item  affinity
0     1    13       0.1
1     2    11       0.4
2     3    14       0.9
3     4    12       1.0

df1 = df.user.astype(str).str.get_dummies()
df1.columns = ['user' + str(x) for x in df1.columns]
print df1
   user1  user2  user3  user4
0      1      0      0      0
1      0      1      0      0
2      0      0      1      0
3      0      0      0      1

df2 = df.item.astype(str).str.get_dummies()
df2.columns = ['item' + str(x) for x in df2.columns]
print df2
   item11  item12  item13  item14
0       0       0       1       0
1       1       0       0       0
2       0       0       0       1
3       0       1       0       0

print pd.concat([df1,df2, df.affinity], axis=1)
   user1  user2  user3  user4  item11  item12  item13  item14  affinity
0      1      0      0      0       0       0       1       0       0.1
1      0      1      0      0       1       0       0       0       0.4
2      0      0      1      0       0       0       0       1       0.9
3      0      0      0      1       0       1       0       0       1.0

时间安排：

len(df) = 4:

In [49]: %timeit pd.concat([df1,df2, df.affinity], axis=1)
The slowest run took 4.91 times longer than the fastest. This could mean that an intermediate result is being cached 
1000 loops, best of 3: 690 µs per loop

len(df) = 40:

df = pd.concat([df]*10).reset_index(drop=True)

In [51]: %timeit pd.concat([df1,df2, df.affinity], axis=1)
The slowest run took 5.56 times longer than the fastest. This could mean that an intermediate result is being cached 
1000 loops, best of 3: 719 µs per loop

len(df) = 400:

df = pd.concat([df]*100).reset_index(drop=True)

In [43]: %timeit pd.concat([df1,df2, df.affinity], axis=1)
The slowest run took 4.55 times longer than the fastest. This could mean that an intermediate result is being cached 
1000 loops, best of 3: 748 µs per loop

len(df) = 4k:

df = pd.concat([df]*1000).reset_index(drop=True)

In [41]: %timeit pd.concat([df1,df2, df.affinity], axis=1)
The slowest run took 4.67 times longer than the fastest. This could mean that an intermediate result is being cached 
1000 loops, best of 3: 761 µs per loop

len(df) = 40k:

df = pd.concat([df]*10000).reset_index(drop=True)

%timeit pd.concat([df1,df2, df.affinity], axis=1)
1000 loops, best of 3: 1.83 ms per loop

len(df) = 400k:

df = pd.concat([df]*100000).reset_index(drop=True)

%timeit pd.concat([df1,df2, df.affinity], axis=1)
100 loops, best of 3: 15.6 ms per loop