python - 将列表拆分为列

Question

我有一个 df 像

uid                                  services
000c80b7d2b3643689b1e516918ec193    ['A']
001b292c588ec6cc11f57324d40e422d    ['B','A',C']
006696f65899fdd87ba4894c784716f9    ['C','B']

(服务列中未排序的列表)

我想重新映射列中的列表

uid                                  services      A   B  C 
000c80b7d2b3643689b1e516918ec193    ['A']          1   0  0
001b292c588ec6cc11f57324d40e422d    ['B','A',C']   1   1  1
006696f65899fdd87ba4894c784716f9    ['C','B']      0   1  1

谢谢

Answer 1

您可以使用 MultiLabelBinarizer先然后join :

from sklearn.preprocessing import MultiLabelBinarizer
mlb = MultiLabelBinarizer()

print (pd.DataFrame(mlb.fit_transform(df['services']),columns=mlb.classes_, index=df.index))
   A  B  C
0  1  0  0
1  1  1  1
2  0  1  1

df1 = pd.DataFrame(mlb.fit_transform(df['services']),columns=mlb.classes_, index=df.index)
df = df.join(df1)
print (df)
                                uid   services  A  B  C
0  000c80b7d2b3643689b1e516918ec193        [A]  1  0  0
1  001b292c588ec6cc11f57324d40e422d  [B, A, C]  1  1  1
2  006696f65899fdd87ba4894c784716f9     [C, B]  0  1  1

纯 Pandas 替代品 get_dummies和 groupby 按具有聚合 max 的列:

df1 = pd.get_dummies(pd.DataFrame(df['services'].values.tolist()), prefix='', prefix_sep='')
        .groupby(axis=1, level=0).max()
print (df1)
   A  B  C
0  1  0  0
1  1  1  1
2  0  1  1

df = df.join(df1)
print (df)
                                uid   services  A  B  C
0  000c80b7d2b3643689b1e516918ec193        [A]  1  0  0
1  001b292c588ec6cc11f57324d40e422d  [B, A, C]  1  1  1
2  006696f65899fdd87ba4894c784716f9     [C, B]  0  1  1

时间:

#3k rows 
df = pd.concat([df]*1000).reset_index(drop=True)

#John Galt solution
In [255]: %timeit (df.join(df.services.apply(lambda x: pd.Series({y:1 for y in x})).fillna(0).astype(int)))
1 loop, best of 3: 658 ms per loop

#user1717828 solution
In [256]: %timeit (df.join(df['services'].apply(lambda x: "|".join(x)).str.get_dummies()))
100 loops, best of 3: 16.8 ms per loop

#Jez solution1
In [257]: %timeit (df.join(pd.DataFrame(mlb.fit_transform(df['services']),columns=mlb.classes_, index=df.index)))
100 loops, best of 3: 4.66 ms per loop

#Jez solution2
In [258]: %timeit (df.join(pd.get_dummies(pd.DataFrame(df['services'].values.tolist()), prefix='', prefix_sep='').groupby(axis=1, level=0).max()))
100 loops, best of 3: 7.04 ms per loop

---

#30k rows
df = pd.concat([df]*10000).reset_index(drop=True)


#John Galt solution
In [260]: %timeit (df.join(df.services.apply(lambda x: pd.Series({y:1 for y in x})).fillna(0).astype(int)))
1 loop, best of 3: 6.68 s per loop

#user1717828 solution
In [261]: %timeit (df.join(df['services'].apply(lambda x: "|".join(x)).str.get_dummies()))
10 loops, best of 3: 138 ms per loop

#Jez solution1
In [262]: %timeit (df.join(pd.DataFrame(mlb.fit_transform(df['services']),columns=mlb.classes_, index=df.index)))
10 loops, best of 3: 39.8 ms per loop

#Jez solution2
In [263]: %timeit (df.join(pd.get_dummies(pd.DataFrame(df['services'].values.tolist()), prefix='', prefix_sep='').groupby(axis=1, level=0).max()))
10 loops, best of 3: 20.6 ms per loop