如何将函数应用于 Pandas 中的多个多索引列？

原文标题 ：How to Apply a function to multiple multiindex columns in Pandas?

给定一个多索引列

         a                      ...                              
         E1                      ...        E3                    
         g1        g2        g3  ...        g1        g2        g3
0  0.548814  0.715189  0.602763  ...  0.437587  0.891773  0.963663
1  0.383442  0.791725  0.528895  ...  0.087129  0.020218  0.832620
2  0.778157  0.870012  0.978618  ...  0.118274  0.639921  0.143353
3  0.944669  0.521848  0.414662  ...  0.568434  0.018790  0.617635
4  0.612096  0.616934  0.943748  ...  0.697631  0.060225  0.666767
5  0.670638  0.210383  0.128926  ...  0.438602  0.988374  0.102045
6  0.208877  0.161310  0.653108  ...  0.158970  0.110375  0.656330
7  0.138183  0.196582  0.368725  ...  0.096098  0.976459  0.468651
8  0.976761  0.604846  0.739264  ...  0.296140  0.118728  0.317983
9  0.414263  0.064147  0.692472  ...  0.093941  0.575946  0.929296
[10 rows x 9 columns]

我想将由第二级过滤的多列（即E1，E2，E3）应用于函数（例如，ration_type1，ration_type2，或者在实际实现中可以更多）。

例如。假设我们要计算函数ration_type1和ration_type2下E1的第二级。那么我们只处理以下df

    a                    
         E1                    
         g1        g2        g3
0  0.548814  0.715189  0.602763
1  0.383442  0.791725  0.528895
  .................
8  0.976761  0.604846  0.739264
9  0.414263  0.064147  0.692472

为了概括所有第二级，我依赖于下面的列表理解

对于ration_type1和ration_type2中的每一个。

all_df1 = [ration_type1(df.loc[:, (slice(None), dgroup, slice(None))]) for dgroup in [`E1`, `E2`, `E3`]]


all_df2 = [ration_type2(df.loc[:, (slice(None), dgroup, slice(None))]) for dgroup in [`E1`, `E2`, `E3`]]

在将其连接回原始df之前。

但是，我想知道是否有比list comprehension方法更优雅和紧凑的方式。这是因为，在现实生活中，可以有更多的配给功能。

完整代码如下

 import numpy as np

import pandas as pd

np.random.seed(0)

arr = np.random.rand(10,9)

tuples = [('a', 'E1', 'g1'), ('a', 'E1', 'g2'), ('a', 'E1', 'g3'), ('a', 'E2', 'g1'), ('a', 'E2', 'g2'),
          ('a', 'E2', 'g3'), ('a', 'E3', 'g1'), ('a', 'E3', 'g2'), ('a', 'E3', 'g3')]
df = pd.DataFrame(data=arr, columns=pd.MultiIndex.from_tuples(tuples))

print(df)
def ration_type1(df):
    """
    (g3+g2)/g1
    # Ugly way since have to convert to numpy 1st
    """

    print(df)
    dration = 'ration_type1'
    l1, l2, _ = df.columns.tolist()[0]
    total = df.loc[:, (slice(None), slice(None), 'g2')].to_numpy() + \
            df.loc[:, (slice(None), slice(None), 'g3')].to_numpy()
    arr = total / df.loc[:, (slice(None), slice(None), 'g1')].to_numpy()

    return pd.DataFrame(data=arr, columns=pd.MultiIndex.from_tuples([(l1, l2, dration)]))


def ration_type2(df):
    """
    (g3+g2+g1)/g1
    # Ugly way since have to convert to numpy 1st
    """
    dration = 'ration_type2'
    l1, l2, _ = df.columns.tolist()[0]
    total = df.loc[:, (slice(None), slice(None), 'g1')].to_numpy() + \
            df.loc[:, (slice(None), slice(None), 'g2')].to_numpy() + \
            df.loc[:, (slice(None), slice(None), 'g3')].to_numpy()
    arr = total / df.loc[:, (slice(None), slice(None), 'g1')].to_numpy()

    return pd.DataFrame(data=arr, columns=pd.MultiIndex.from_tuples([(l1, l2, dration)]))


level1_name = list(set(df.columns.get_level_values(1)))

all_df1 = [ration_type1(df.loc[:, (slice(None), dgroup, slice(None))]) for dgroup in level1_name]
all_df2 = [ration_type2(df.loc[:, (slice(None), dgroup, slice(None))]) for dgroup in level1_name]

df1 = pd.concat(all_df1, axis=1)
df2 = pd.concat(all_df2, axis=1)

df=pd.concat([df,df1,df2],axis=1)

预期输出。

          a                      ...                                       
         E1                      ...                        E2           E3
         g1        g2        g3  ... ration_type2 ration_type2 ration_type2
0  0.548814  0.715189  0.602763  ...     3.401458     2.962896     5.240151
1  0.383442  0.791725  0.528895  ...     4.444124     2.754497    10.788191
2  0.778157  0.870012  0.978618  ...     3.375653     2.554145     7.622516
3  0.944669  0.521848  0.414662  ...     1.991363     5.650758     2.119612
4  0.612096  0.616934  0.943748  ...     3.549735     2.168255     2.042087
5  0.670638  0.210383  0.128926  ...     1.505949     3.960760     3.486126
6  0.208877  0.161310  0.653108  ...     4.899035     3.806001     5.822965
7  0.138183  0.196582  0.368725  ...     5.091008     2.138921    16.037821
8  0.976761  0.604846  0.739264  ...     2.376088    11.283905     2.474676
9  0.414263  0.064147  0.692472  ...     2.826423     2.391873    17.023361

[10 rows x 15 columns]

我正在考虑做一些像使用apply

# function for prepending 'Geek'
def multiply_by_2(number):
    return 2 * number
 
# executing the function
df[["Integers", "Float"]] = df[["Integers", "Float"]].apply(multiply_by_2)

但是，自从我的示例涉及多索引列以来，我遇到了困难（由于我的知识有限）