How to sort each row in pandas dataframe and get indices instead?

Question

I have a pandas DataFrame containing probabilities. What I need is a new DataFrame where each row contains the indices of the sorted row. Example:

df = pd.DataFrame([[0.5,0.7,0.1],[0.1,0.7,0.5]])
df
     0    1    2
0  0.5  0.7  0.1
1  0.1  0.7  0.5

new_df = some sort operation on df
new_df
    0  1  2
0   1  0  2
1   2  0  1

I know how to do it the hard way, using multiple loops and such. I found references to sorting values but no "sorting" function that returned the indices. Anyone some suggestion?

use .index to get the indices for each column after sort (ensure NOT inplace), then build the required dataframe! — anurag
– anurag, Commented Jan 6, 2021 at 11:12
how do you have 2,1,0 in the second row? shouldnt it be 1,2,0? — anky
– anky, Commented Jan 6, 2021 at 11:28
@anky: He just has a typo in his definition of df (the output suggests df[1][1] was 0.5 instead of the 0.7 stated in the example). I'd probably just do df.transform(lambda s: np.argsort(-s.values), axis=1) — sim
– sim, Commented Jan 6, 2021 at 11:35

adir abargil · Accepted Answer · 2021-01-06 11:56:39Z

2

you can use numpy for that with argsort:

df = pd.DataFrame([[0.5,0.7,0.1],[0.1,0.7,0.5]])
array = df.values.argsort(axis=1)[:,::-1]
new_df  = pd.DataFrame(array)

output new_df:

    0   1   2
0   1   0   2
1   1   2   0

Note:

as commented by @anky there is something that doesnt make sense in the ouput you show,, also i assumed you want descending order and thats why the [:,::-1] slice in the result/

UPDATE

as @anky suggested in comments here it is still using the same idea of argsort, this is more strightforward solution then df.values.argsort(axis=1)[:,::-1]:

np.argsort(-df)

edited Jan 6, 2021 at 11:56

answered Jan 6, 2021 at 11:37

adir abargil

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