index and column for the max value in pandas dataframe

Question

I have a python dataframe df with five columns and five rows. I would like to get the row and column name of the max three values

Example:

df = 

  A   B  C  D  E  F
1 00 01 02 03 04 05
2 06 07 08 09 10 11
3 12 13 14 15 16 17
4 18 19 20 21 22 23
5 24 25 26 27 28 29

The output show say something like [5,F],[5,E],[5,D]

Answer 1

You could use unstack before sorting:

>>> df
    A   B   C   D   E   F
1   0   1   2   3   4   5
2   6   7   8   9  10  11
3  12  13  14  15  16  17
4  18  19  20  21  22  23
5  24  25  26  27  28  29
>>> df.unstack()
A  1     0
   2     6
   3    12
   4    18
   5    24
B  1     1
   2     7
   3    13
   4    19
   5    25
[...]
F  1     5
   2    11
   3    17
   4    23
   5    29

and so

>>> df2 = df.unstack().copy()
>>> df2.sort()
>>> df2[-3:]
D  5    27
E  5    28
F  5    29
>>> df2[-3:].index
MultiIndex
[(D, 5.0), (E, 5.0), (F, 5.0)]

or even

>>> df.unstack()[df.unstack().argsort()].index[-3:]
MultiIndex
[(D, 5.0), (E, 5.0), (F, 5.0)]

[I didn't bother reversing the order: sticking [::-1] at the end should do it.]

Answer 2

I am not going to pretend these are the most efficient way of dealing with this problem, but I though they are worth mentioning:

df

    A   B   C   D   E   F
1   0   1   2   3   4   5
2   6   7   8   9  10  11
3  12  13  14  15  16  17
4  18  19  20  21  22  23
5  24  25  26  27  28  29

Using df.max() to get the maximum value of each column and then sorting values and getting the biggest numbers. Then masking them against the original df and returning the values. A list comprehension can is finally used to get the indices:

df_2 = df[df.max().sort_values(ascending=True).tail(3).eq(df)]
[(i, df_2[i].first_valid_index()) for i in df_2.columns if df_2[i].first_valid_index() != None]

Output:

[('D', 5), ('E', 5), ('F', 5)]

or

s = df_2.apply(pd.Series.first_valid_index).dropna()
list(zip(s.index, s.astype(int)))

Output:

[('D', 5), ('E', 5), ('F', 5)]