sort_values() with key in Python

Question

I have a dataframe where the column names are times (0:00, 0:10, 0:20, ..., 23:50). Right now, they're sorted in a string order (so 0:00 is first and 9:50 is last) but I want to sort them after time (so 0:00 is first and 23:50 is last).

If time is a column, you can use

df = df.sort(columns='Time',key=float)

But 1) that only works if time is a column itself, rather than the column names, and 2) sort() is deprecated so I try to abstain from using it.

I'm trying to use

df = df.sort_index(axis = 1)

but since the column names are in string format, they get sorted according to a string key. I've tried

df = df.sort_index(key=float, axis=1) 

but that gives an error message:

Traceback (most recent call last):
  File "<ipython-input-112-5663f277da66>", line 1, in <module>
      df.sort_index(key=float, axis=1)
TypeError: sort_index() got an unexpected keyword argument 'key'

Does anyone have ideas for how to fix this? So annoying that sort_index() - and sort_values() for that matter - don't have the key argument!!

Answer 1

Try sorting the columns with the sorted builtin function and passing the output to the dataframe for indexing. The following should serve as a working example:

import pandas as pd


records = [(2, 33, 23, 45), (3, 4, 2, 4), (4, 5, 7, 19), (4, 6, 71, 2)]
df = pd.DataFrame.from_records(records, columns = ('0:00', '23:40', '12:30', '11:23'))
df
#    0:00  23:40  12:30  11:23
# 0     2     33     23     45
# 1     3      4      2      4
# 2     4      5      7     19
# 3     4      6     71      2

df[sorted(df,key=pd.to_datetime)]

#    0:00  11:23  12:30  23:40
# 0     2     45     23     33
# 1     3      4      2      4
# 2     4     19      7      5
# 3     4      2     71      6

I hope this helps

Answer 2

Just prepend a leading zero to one-digit hours. This should be the simplest solution as you can simply sort lexically then.

E.g. 5:30 -> 05:30.