Pandas: Get top n columns based on a row values

Question

Having a dataframe with a single row, I need to filter it into a smaller one with filtered columns based on a value in a row.
What's the most effective way?

df = pd.DataFrame({'a':[1], 'b':[10], 'c':[3], 'd':[5]})

a	b	c	d
1	10	3	5

For example top-3 features:

b	c	d
10	3	5

Answer 1

Use sorting per row and select first 3 values:

df1 = df.sort_values(0, axis=1, ascending=False).iloc[:, :3]
print (df1)
    b  d  c
0  10  5  3

Solution with Series.nlargest:

df1 = df.iloc[0].nlargest(3).to_frame().T
print (df1)
    b  d  c
0  10  5  3

Answer 2

You can transpose T, and use nlargest():

new = df.T.nlargest(columns = 0, n = 3).T

print(new)

   b  d  c
0  10  5  3

Answer 3

You can use np.argsort to get the solution. This Numpy method, in the below code, gives the indices of the column values in descending order. Then slicing selects the largest n values' indices.

import pandas as pd
import numpy as np

# Your dataframe
df = pd.DataFrame({'a':[1], 'b':[10], 'c':[3], 'd':[5]})

# Pick the number n to find n largest values
nlargest = 3

# Get the order of the largest value columns by their indices
order = np.argsort(-df.values, axis=1)[:, :nlargest]

# Find the columns with the largest values
top_features = df.columns[order].tolist()[0]

# Filter the dateframe by the columns
top_features_df = df[top_features]

top_features_df

output:

    b   d   c
0   10  5   3