MultiIndex 中的新级别 DataFrame 基于现有列级别值

Question

Let's say I have a DataFrame like this:

df = pd.DataFrame(data = [[1,2,3,4,5,6], [3,4,5,6,7,8]], 
                  columns = pd.MultiIndex.from_product([('A1', 'B1', 'A2'), (10,20)], names=['level_0','level_1']))

Here's how it looks like: DataFrame image

I want to add a new level in the columns which contains 1 where level_0 value contains "1" and and 2 where level_0 value contains "2" . So, basically:

• Where level_0 == "A1" --> new_level = 1
• Where level_0 == "B1" --> new_level = 1
• Where level_0 == "A2" --> new_level = 2

Any suggestions on how to do it?

Answer 1

You could extract the values with a regex ( (\d+)$ = last digits of the value) and rework the MultiIndex with MultiIndex.from_arrays :

values = df.columns.get_level_values('level_0').str.extract('(\d+)$', expand=False)
# ['1', '1', '1', '1', '2', '2']

df.columns = pd.MultiIndex.from_arrays([*zip(*df.columns.to_list()), values],
                                       names=[*df.columns.names, 'level_2']
                                      )

NB. this generalizes to any XXX00 value

output:

level_0 A1    B1    A2   
level_1 10 20 10 20 10 20
level_2  1  1  1  1  2  2
0        1  2  3  4  5  6
1        3  4  5  6  7  8

Answer 2

Use lsit comprehension for extract number from first level values and create new MultiIndex by MultiIndex.from_tuples :

import re

df.columns = pd.MultiIndex.from_tuples([(re.findall(r'(\d+)$', x[0])[0], *x) 
                                         for x in df.columns.tolist()], 
                                       names=('new_level',*df.columns.names))
print (df)

new_level  1           2   
level_0   A1    B1    A2   
level_1   10 20 10 20 10 20
0          1  2  3  4  5  6
1          3  4  5  6  7  8