[FIXED] How can I find pairs and triplets of values in a pandas dataframe

Issue

I have a pandas dataframe in Python that contains the following pair of columns.

I need to count how many times pairs and triplets of combination of data appear with and without considering the order. As an example, let's say that I have a dataframe with two columns, Classification and Individual and the following token data

data = {

    'Classification': [1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5],
    'Individual': ['A', 'A', 'B', 'B', 'A', 'A', 'B', 'C', 'C', 'C', 'A', 'A', 'A', 'B', 'B', 'A', 'A', 'B', 'B', 'B', 'C', 'C', 'C', 'C', 'C', 'A', 'A', 'B', 'B', 'B']
}

Now, I want to arrive to the following results

Clasification   ValueSeries TimesClassification PercentageClassification    

1   AB  5   1
2   AB  5   1
3   AC  2   0.4
3   AB  5   1   
3   ABC 3   0.6
4   AB  5   1
4   BC  2   0.4
4   ABC 3   0.6
5   AC  2   0.4
5   AB  5   1
5   ABC 3   0.6

this is, for each value of clasification the unnordered pairs and triplets contained within.

Solution

The exact logic is not fully clear, but you can use itertools to produce the combinations of Classification, then apply a value_counts and groupby.transform to compute the counts:

from itertools import chain, combinations

def powerset(s):
    s = set(s)
    return list(chain.from_iterable(combinations(s, r)
                                    for r in range(2, len(s)+1))
               )

out = df.groupby('Classification')['Individual'].agg(powerset).explode()

out = (out
    .reset_index(name='ValueSeries')
    .merge(out.value_counts().rename('TimesClassification'),
           how='left',
           left_on='ValueSeries', right_index=True)
    .assign(PercentageClassification=lambda d: d['TimesClassification']
            / d.groupby('Classification')['TimesClassification'].transform('max')
           )
)

Output:

    Classification ValueSeries  TimesClassification  PercentageClassification
0                1      (A, B)                    5                       1.0
1                2      (A, B)                    5                       1.0
2                3      (C, A)                    3                       0.6
3                3      (C, B)                    3                       0.6
4                3      (A, B)                    5                       1.0
5                3   (C, A, B)                    3                       0.6
6                4      (C, A)                    3                       0.6
7                4      (C, B)                    3                       0.6
8                4      (A, B)                    5                       1.0
9                4   (C, A, B)                    3                       0.6
10               5      (C, A)                    3                       0.6
11               5      (C, B)                    3                       0.6
12               5      (A, B)                    5                       1.0
13               5   (C, A, B)                    3                       0.6

Answered By - mozway