使用基于另一列的GROUPBY的最小最大归一化来归一化一列数据帧

Normalize a column of dataframe using min max normalization based on groupby of another column(使用基于另一列的GROUPBY的最小最大归一化来归一化一列数据帧)

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问题描述

数据帧如图所示

Name     Job      Salary
john   painter    40000
peter  engineer   50000
sam     plumber   30000
john    doctor    500000
john    driver    20000
sam    carpenter  10000
peter  scientist  100000

如何按列名分组并对每个组上的薪资列应用规范化?

预期结果:

Name     Job      Salary
john   painter    0.041666
peter  engineer   0
sam     plumber   1
john    doctor    1
john    driver    0
sam    carpenter  0
peter  scientist  1

我已尝试以下操作

data = df.groupby('Name').transform(lambda x: (x - x.min()) / x.max()- x.min())

但是,这会产生

         Salary
0 -19999.960000
1 -50000.000000
2  -9999.333333
3 -19999.040000
4 -20000.000000
5 -10000.000000
6 -49999.500000

推荐答案

您马上就到了。

>>> df                                                                                                                 
    Name        Job  Salary
0   john    painter   40000
1  peter   engineer   50000
2    sam    plumber   30000
3   john     doctor  500000
4   john     driver   20000
5    sam  carpenter   10000
6  peter  scientist  100000
>>>                                                                                                                    
>>> result = df.assign(Salary=df.groupby('Name').transform(lambda x: (x - x.min()) / (x.max()- x.min())))
>>> # alternatively, df['Salary'] = df.groupby(... if you don't need a new frame       
>>> result                                                                                                               
    Name        Job    Salary
0   john    painter  0.041667
1  peter   engineer  0.000000
2    sam    plumber  1.000000
3   john     doctor  1.000000
4   john     driver  0.000000
5    sam  carpenter  0.000000
6  peter  scientist  1.000000

所以基本上,您只是忘了用括号将x.max() - x.min()括起来。


请注意,使用一系列矢量化操作可以更快地完成此操作。

>>> grouper = df.groupby('Name')['Salary']                                                                             
>>> maxes = grouper.transform('max')                                                                                   
>>> mins = grouper.transform('min')                                                                                    
>>>                                                                                                                    
>>> result = df.assign(Salary=(df.Salary - mins)/(maxes - mins))                                                       
>>> result                                                                                                             
    Name        Job    Salary
0   john    painter  0.041667
1  peter   engineer  0.000000
2    sam    plumber  1.000000
3   john     doctor  1.000000
4   john     driver  0.000000
5    sam  carpenter  0.000000
6  peter  scientist  1.000000

计时:

>>> # Setup
>>> df = pd.concat([df]*1000, ignore_index=True)                                                                       
>>> df.Name = np.arange(len(df)//4).repeat(4) # 4 names per group                                                      
>>> df                                                                                                                 
      Name        Job  Salary
0        0    painter   40000
1        0   engineer   50000
2        0    plumber   30000
3        0     doctor  500000
4        1     driver   20000
...    ...        ...     ...
6995  1748    plumber   30000
6996  1749     doctor  500000
6997  1749     driver   20000
6998  1749  carpenter   10000
6999  1749  scientist  100000

[7000 rows x 3 columns]
>>>
>>> # Tests @ i5-6200U CPU @ 2.30GHz
>>> %timeit df.groupby('Name').transform(lambda x: (x - x.min()) / (x.max()- x.min()))                                 
1.19 s ± 20.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
>>> %%timeit 
...: grouper = df.groupby('Name')['Salary'] 
...: maxes = grouper.transform('max') 
...: mins = grouper.transform('min') 
...: (df.Salary - mins)/(maxes - mins) 
...:  
...:                                                                                                                   
3.04 ms ± 94.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

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