How to use two key functions when sorting a MultiIndex dataframe?

Question

In this call to df.sort_index() on a MultiIndex dataframe, how to use func_2 for level two?

func_1 = lambda s: s.str.lower()
func_2 = lambda x: np.abs(x)
m_sorted = df_multi.sort_index(level=['one', 'two'], key=func_1)

The documentation says "For MultiIndex inputs, the key is applied per level", which is ambiguous.

---

import pandas as pd
import numpy as np
np.random.seed(3)

# Create multiIndex
choice = lambda a, n: np.random.choice(a, n, replace=True)
df_multi = pd.DataFrame({
    'one': pd.Series(choice(['a', 'B', 'c'], 8)),
    'two': pd.Series(choice([1, -2, 3], 8)),
    'A': pd.Series(choice([2,6,9,7] ,8))
    })
df_multi = df_multi.set_index(['one', 'two'])

# Sort MultiIndex
func_1 = lambda s: s.str.lower()
func_2 = lambda x: np.abs(x)
m_sorted = df_multi.sort_index(level=['one'], key=func_1)

Answer 1

sort_index takes a unique function as key that would be used for all levels.

That said, you could use a wrapper function to map the desired sorting function per level name:

def sorter(level, default=lambda x: x):
    return {
      'one': lambda s: s.str.lower(),
      'two': np.abs,
    }.get(level.name, default)(level)

df_multi.sort_index(level=['one', 'two'], key=sorter)

NB. in case of no match a default function is used that returns the level unchanged.

Another option with numpy.lexsort instead of sort_index:

# levels, functions in desired sorting order
sorters = [('one', lambda s: s.str.lower()), ('two', np.abs)]

out = df_multi.iloc[np.lexsort([f(df_multi.index.get_level_values(lvl))
                                for lvl, f in sorters[::-1]])]

lexsort uses the major keys last, thus the [::-1]

Output:

         A
one two   
a    1   6
    -2   2
     3   7
B    1   6
    -2   7
    -2   7
     3   2
     3   6