如何通过pandas中的groupby输出来填充？

Question

我有一个有4列(A,B,C,D)的数据帧.D有一些NaN条目.我想用具有相同值A,B,C的D的平均值填充NaN值.

例如,如果A,B,C,D的值分别是x,y,z和Nan,那么我希望将NaN值替换为A的平均值,其中A,B,C的值为分别是x,y,z.

Answer 1

df['D'].fillna(df.groupby(['A','B','C'])['D'].transform('mean')) 会比 apply

In [2400]: df
Out[2400]:
   A  B  C    D
0  1  1  1  1.0
1  1  1  1  NaN
2  1  1  1  3.0
3  3  3  3  5.0

In [2401]: df['D'].fillna(df.groupby(['A','B','C'])['D'].transform('mean'))
Out[2401]:
0    1.0
1    2.0
2    3.0
3    5.0
Name: D, dtype: float64

In [2402]: df['D'] = df['D'].fillna(df.groupby(['A','B','C'])['D'].transform('mean'))

In [2403]: df
Out[2403]:
   A  B  C    D
0  1  1  1  1.0
1  1  1  1  2.0
2  1  1  1  3.0
3  3  3  3  5.0

细节

In [2396]: df.shape
Out[2396]: (10000, 4)

In [2398]: %timeit df['D'].fillna(df.groupby(['A','B','C'])['D'].transform('mean'))
100 loops, best of 3: 3.44 ms per loop


In [2397]: %timeit df.groupby(['A','B','C'])['D'].apply(lambda x: x.fillna(x.mean()))
100 loops, best of 3: 5.34 ms per loop

Answer 2

我想你需要:

df.D = df.groupby(['A','B','C'])['D'].apply(lambda x: x.fillna(x.mean()))

样品:

df = pd.DataFrame({'A':[1,1,1,3],
                   'B':[1,1,1,3],
                   'C':[1,1,1,3],
                   'D':[1,np.nan,3,5]})

print (df)
   A  B  C    D
0  1  1  1  1.0
1  1  1  1  NaN
2  1  1  1  3.0
3  3  3  3  5.0

df.D = df.groupby(['A','B','C'])['D'].apply(lambda x: x.fillna(x.mean()))
print (df)
   A  B  C    D
0  1  1  1  1.0
1  1  1  1  2.0
2  1  1  1  3.0
3  3  3  3  5.0