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.

ABCDA 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 : BCDACABD 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

Answer 3

You can use argsort() and then index into columns :

(df.apply(lambda x: (-x).argsort(), axis=1) # use -x for sort descending
   .apply(lambda x: df.columns[x], axis=1))

   A  B  C  D
A  B  C  A  D
B  C  A  B  D
C  D  A  C  B
D  A  D  C  B

Note: If you don't want the index and columns in your final output, just add .values .

Answer 4

Using numpy argsort,

pd.DataFrame(df.columns[np.argsort(-df.values, axis=1)])


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