Sort DataFrame based on part of its index

What I would like to achieve

I have a DataFrame whose indices are “ID (int) + underscore (_) + name (str)”. I would like to sort the data based on the ID.

import pandas as pd

data = [[1, 2], [3, 4], [5, 6], [7, 8], [9, 10], [11, 12], [13, 14]]
index = ['11_ItemName', '0_ItemName', '1_ItemName', '2_ItemName', '10_ItemName', '20_ItemName', '101_ItemName']
columns = ['column1', 'column2']

df = pd.DataFrame(data, index=index, columns=columns)
print(df)
## Output
#              column1  column2
# 11_ItemName        1        2
# 0_ItemName         3        4
# 1_ItemName         5        6
# 2_ItemName         7        8
# 10_ItemName        9       10
# 20_ItemName       11       12
# 101_ItemName      13       14

# print(DO SOMETHING!)
## Expected output
#              column1  column2
# 0_ItemName         3        4
# 1_ItemName         5        6
# 2_ItemName         7        8
# 10_ItemName        9       10
# 11_ItemName        1        2
# 20_ItemName       11       12
# 101_ItemName      13       14

What I tested

I tried to use sort_index and failed.
https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.sort_index.html

To be honest, I’m stuck at the debug of lambda.

df(sort_index())
## Output
#              column1  column2
# 0_ItemName         3        4
# 101_ItemName      13       14
# 10_ItemName        9       10
# 11_ItemName        1        2
# 1_ItemName         5        6
# 20_ItemName       11       12
# 2_ItemName         7        8

print(df.sort_index(key=(lambda x: int(x.str.split('_')[0]))))
# TypeError: int() argument must be a string, a bytes-like object or a real number, not 'list'

Environment

Python 3.10.5
Pandas 1.4.3

Answer

Try df.sort_index with custom key=:

df = df.sort_index(
    key=lambda idx: idx.str.split("_")
    .str[0]
    .astype(int)
)
print(df)

Prints:

              column1  column2
0_ItemName          3        4
1_ItemName          5        6
2_ItemName          7        8
10_ItemName         9       10
11_ItemName         1        2
20_ItemName        11       12
101_ItemName       13       14