Python pandas find the column names in descending order for each row

Question

I have the below pandas data frame and I am trying to find the column names in descending order for each row.

A B C D A 1 3 2 1 B 2 1 5 0 C 1 0 1 9 D 2 0 1 2

For each row I am trying to get the column indexes in sorted order ( descending ) I should get : B C D A C A B D and so on Is this possible using pandas.?

I am trying something like this. test[2].sort_index(axis = 0, ascending = True)

Update 12/23

>>> df1
user_handle  1   2   3   4   5   6   8   9   10  12
user_handle
1             6   0   0   1   0   0   0   0   0   0
2             0  95   0   0   1   0   0   0   1   0
3             0   0   2   0   0   0   0   0   0   0
4             1   0   0  12   0   1   0   0   2   0
5             0   1   0   0   9   0   0   0   0   0
6             0   0   0   1   0  14   0   0   1   0
8             0   0   0   0   0   0   4   0   0   0
9             0   0   0   0   0   0   0  12   0   0
10            0   1   0   2   0   1   0   0  49   0
12            0   0   0   0   0   0   0   0   0   2

>>> t1 = np.flip(df1.values.argsort(), 1)


>>> pd.DataFrame(df1.columns[t1])
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "python3.6/site-packages/pandas/core/frame.py", line 303, in __init__
    dtype=dtype)
  File "python3.6/site-packages/pandas/core/frame.py", line 411, in _init_dict
    return _arrays_to_mgr(arrays, data_names, index, columns, dtype=dtype)
  File "python3.6/site-packages/pandas/core/frame.py", line 5506, in _arrays_to_mgr
    return create_block_manager_from_arrays(arrays, arr_names, axes)
  File "python3.6/site-packages/pandas/core/internals.py", line 4310, in create_block_manager_from_arrays
    mgr = BlockManager(blocks, axes)
  File "python3.6/site-packages/pandas/core/internals.py", line 2792, in __init__
    (block.ndim, self.ndim))
AssertionError: Number of Block dimensions (3) must equal number of axes (2)
>>>
KeyboardInterrupt

Answer 1

You can use :

temp = np.flip(df.values.argsort(),1)
ndf = pd.DataFrame(df.columns[temp])

   0  1  2  3
0  B  C  D  A
1  C  A  B  D
2  D  C  A  B
3  D  A  C  B

Answer 2

By using rank and unstack

df.stack().groupby(level=0).rank(method ='first',ascending =False).reset_index(level=1).set_index(0,append=True).unstack()
Out[309]: 
  level_1            
0     1.0 2.0 3.0 4.0
A       B   C   A   D
B       C   A   B   D
C       D   A   C   B
D       A   D   C   B

And using numpy

pd.DataFrame(df.columns.values[np.lexsort(([df.columns.values]*len(df),df.values))]).iloc[:, ::-1]
Out[351]: 
   3  2  1  0
0  B  C  D  A
1  C  A  B  D
2  D  C  A  B
3  D  A  C  B