Sort pandas dataframe both on values of a column and index?

Question

Is it feasible to sort pandas dataframe by values of a column, but also by index?

If you sort a pandas dataframe by values of a column, you can get the resultant dat

Answer 1

Pandas 0.23 finally gets you there :-D

You can now pass index names (and not only column names) as parameters to sort_values. So, this one-liner works:

df = df.sort_values(by = ['MyCol', 'MyIdx'], ascending = [False, True])

And if your index is currently unnamed:

df = df.rename_axis('MyIdx').sort_values(by = ['MyCol', 'MyIdx'], ascending = [False, True])

Answer 2

As of pandas version 0.22.

You can temporarily set the column as an index, sort the index on that column and then reset. By default it will maintain the order of the existing index:

df = df.set_index('column_name', append=True).sort_index(level=1).reset_index(level=1)

I think the above could be done with 'inplace' options but I think it's easier to read as above.

Answer 3

You can use the ascending parameter in sort_index, but you must pass it as a list for it to work correctly as of pandas 0.22.0.

import pandas as pd
import numpy as np
df = pd.DataFrame({'idx_0':[2]*6+[1]*5,
                   'idx_1':[6,4,2,10,18,5,11,1,7,9,3],
                   'value_1':np.arange(11,0,-1),
                   'MyName':list('SORTEDFRAME')})

df = df.set_index(['idx_0','idx_1'])
df

Output:

            MyName  value_1
idx_0 idx_1                
2     6          S       11
      4          O       10
      2          R        9
      10         T        8
      18         E        7
      5          D        6
1     11         F        5
      1          R        4
      7          A        3
      9          M        2
      3          E        1

Sorting by values and index should get "FRAMESORTED" instead of "SORTEDFRAME"

df.sort_values('value_1', ascending=False)\
  .sort_index(level=0, ascending=[True])

Output:

            MyName  value_1
idx_0 idx_1                
1     11         F        5
      1          R        4
      7          A        3
      9          M        2
      3          E        1
2     6          S       11
      4          O       10
      2          R        9
      10         T        8
      18         E        7
      5          D        6

Note you must pass ascending parameter in sort_index as a list and not as a scalar. It will not work.

Answer 4

I solve this problem next way:

df.to_csv('df.csv', index = False)
df = df.read_csv('df.csv')

Answer 5

In pandas 0.23+ you can do it directly - see OmerB's answer. If you don't yet have 0.23+, read on.

---

I'd venture that the simplest way is to just copy your index over to a column, and then sort by both.

df['colFromIndex'] = df.index
df = df.sort(['count', 'colFromIndex'])

I'd also prefer to be able to just do something like df.sort(['count', 'index']), but of course that doesn't work.

Answer 6

To sort a column descending, while maintaining the index ascending:

import pandas as pd
df = pd.DataFrame(index=range(5), data={'c': [4,2,2,4,2]})
df.index = df.index[::-1]
print df.sort(column='c', ascending=False)

Output:

   c
1  4
4  4
0  2
2  2
3  2