Pandas : groupby() different columns in a single step

Question

I have a dataframe as below. "feature1", "feature2" and "feature3" are independent features. The possible values in these features as 1, 2, 3, 4 only. I need to find the size, grouped by the values for each features. Shown below is how I'm doing it now, for each feature separately. Is there a way to do it in a single step or more optimal way ? Thank you.

import pandas as pd
df = pd.DataFrame({'State'    : ['AZ', 'FL', 'AZ', 'FL', 'FL', 'FL', 'AZ', 'FL', 'AZ', 'AZ', 'FL', 'AZ', 'FL', 'AZ', 'AZ'],
               'feature1' : [1, 3, 2, 2, 4, 3, 3, 1, 2, 4, 3, 1, 1, 1, 1],
               'feature2' : [1, 2, 1, 2, 3, 2, 4, 1, 2, 4, 3, 1, 1, 1, 1],
               'feature3' : [1, 4, 1, 2, 1, 4, 3, 3, 2, 4, 3, 1, 1, 1, 1],})
(
    df
    .groupby(['State', 'feature1'])
    ['feature1']
    .size()
    .to_frame('N')
 )

Output for feature1:

Answer 1

Option 1: melt + value_counts

df.melt('State', var_name='feature').value_counts().sort_index()

Result

State  feature   value
AZ     feature1  1        4
                 2        2
                 3        1
                 4        1
       feature2  1        5
                 2        1
                 4        2
       feature3  1        5
                 2        1
                 3        1
                 4        1
FL     feature1  1        2
                 2        1
                 3        3
                 4        1
       feature2  1        2
                 2        3
                 3        2
       feature3  1        2
                 2        1
                 3        2
                 4        2
dtype: int64

Option 2: melt + crosstab

s = df.melt('State', var_name='feature')
pd.crosstab(s['State'], [s['feature'], s['value']])

Result

feature        feature1          feature2          feature3         
value          1  2  3  4        1  2  3  4        1  2  3  4
State                                                        
AZ             4  2  1  1        5  1  0  2        5  1  1  1
FL             2  1  3  1        2  3  2  0        2  1  2  2