“Groupby Fillna” Kod odpowiedzi

Groupby Fillna

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
Gifted Gecko

Groupby Fillna

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
Gifted Gecko

Groupby Fillna Ffill

df['three'] = df.groupby(['one','two'], sort=False)['three']
                .apply(lambda x: x.ffill().bfill())
print (df)
   one  two  three
0    1    1   10.0
1    1    1   10.0
2    1    1   10.0
3    1    2   20.0
4    1    2   20.0
5    1    2   20.0
6    1    3    NaN
7    1    3    NaN
Gifted Gecko

Groupby Fillna Ffill

print (df)
   one  two  three
0    1    1   10.0
1    1    1   40.0
2    1    1    NaN
3    1    2    NaN
4    1    2   20.0
5    1    2    NaN
6    1    3    NaN
7    1    3    NaN

df['three'] = df.groupby(['one','two'], sort=False)['three']
                .apply(lambda x: x.fillna(x.mean()))
print (df)
   one  two  three
0    1    1   10.0
1    1    1   40.0
2    1    1   25.0
3    1    2   20.0
4    1    2   20.0
5    1    2   20.0
6    1    3    NaN
7    1    3    NaN
Gifted Gecko

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