top k columns with values in pandas dataframe for every row

Question

I have a pandas dataframe like the following:

   A  B  C  D
0  7  2  5  2
1  3  3  1  1
2  0  2  6  1
3  3  6  2  9

There can be 100s of columns, in the above example I have only shown 4.

I would like to extract top-k columns for each row and their values.

I can get the top-k columns using:

pd.DataFrame({n: df.T[column].nlargest(k).index.tolist() for n, column in enumerate(df.T)}).T

which, for k=3 gives:

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

But what I would like to have is:

   0  1  2  3  4  5
0  A  7  C  5  B  2
1  A  3  B  3  C  1
2  C  6  B  2  D  1
3  D  9  B  6  A  3

Is there a pand(a)oic way to achieve this?

Answer 1

You can use numpy solution:

• numpy.argsort for columns names
• array already sort (thanks Jeff), need values by indices
• interweave for new array
• DataFrame constructor

k = 3
vals = df.values
arr1 = np.argsort(-vals, axis=1)

a = df.columns[arr1[:,:k]]
b = vals[np.arange(len(df.index))[:,None], arr1][:,:k]

c = np.empty((vals.shape[0], 2 * k), dtype=a.dtype)
c[:,0::2] = a
c[:,1::2] = b
print (c)
[['A' 7 'C' 5 'B' 2]
 ['A' 3 'B' 3 'C' 1]
 ['C' 6 'B' 2 'D' 1]
 ['D' 9 'B' 6 'A' 3]]

df = pd.DataFrame(c)
print (df)
   0  1  2  3  4  5
0  A  7  C  5  B  2
1  A  3  B  3  C  1
2  C  6  B  2  D  1
3  D  9  B  6  A  3

Answer 2

>>> def foo(x):
...     r = []
...     for p in zip(list(x.index), list(x)):
...             r.extend(p)
...     return r
... 
>>> pd.DataFrame({n: foo(df.T[row].nlargest(k)) for n, row in enumerate(df.T)}).T
   0  1  2  3  4  5
0  A  7  C  5  B  2
1  A  3  B  3  C  1
2  C  6  B  2  D  1
3  D  9  B  6  A  3

Or, using list comprehension:

>>> def foo(x):
...     return [j for i in zip(list(x.index), list(x)) for j in i]
... 
>>> pd.DataFrame({n: foo(df.T[row].nlargest(k)) for n, row in enumerate(df.T)}).T
   0  1  2  3  4  5
0  A  7  C  5  B  2
1  A  3  B  3  C  1
2  C  6  B  2  D  1
3  D  9  B  6  A  3