允许重复的列表的所有可能组合的排列数量增加

Question

I am trying without success to understand how to use itertools to generate a list with all possible combinations of the elements of a list, with an increasing size of elements to pick and including repetitions. I would like to add also a separator:

lis = ['a','b','c']
separator = '/'
total_number_of_combinations = 3
permutation_list = ['a','b','c', 'a/a', 'a/b', 'a/c', 'b/a', 'b/b', 'b/c', 'c/a', 'c/b', 'c/c',
                    'a/a/a', 'a/a/b', 'a/a/c', 'a/b/a', 'a/b/b', 'a/b/c', 'a/c/a', 'a/c/b', 'a/c/c'
                    'b/a/a', 'b/a/b', 'b/a/c', 'b/b/a', 'b/b/b', 'b/b/c', 'b/c/a', 'b/c/b', 'b/c/c'
                    'c/a/a', 'c/a/b', 'c/a/c', 'c/b/a', 'c/b/b', 'c/b/c', 'c/c/a', 'c/c/b', 'c/c/c']

The list will have then len(lis)+len(lis)**2+len(lis)**3+...++len(lis)**n elements, with n=total_number_of_combinations . I need to keep the separator and the total_numbers_of_combinations changeables.

I need this in a list that can be check as a condition for filtering a pandas DataFrame (i will check dt[dt.my_col.isin(permutation_list)] )

I appreciate any help or pointing to a duplicated topic or even an explanation of how to correctly state this problem, because I did not found any topic that answer this question (maybe I am using the wrong keywords...). Maybe also there is a function from another module that does that, but I don't know.

UPDATE: Following the request of @Scott, here is my real case:

lis = ['BRUTELE','COCKPIT EST', 'CIRCET']
separator = ' / '
total_number_of_combinations = 10

so my final list need to have 88572 elements.

Answer 1

Are you looking for this?

[ '/'.join([*a]) for i in range(1,3) for a in itertools.combinations_with_replacement('ABCDEF', i) ]

Result is:

['A', 'B', 'C', 'D', 'E', 'F', 
'A/A', 'A/B', 'A/C', 'A/D', 'A/E', 'A/F', 'B/B', 'B/C', 'B/D', 'B/E', 'B/F', 'C/C', 'C/D', 'C/E', 'C/F', 'D/D', 'D/E', 'D/F', 'E/E', 'E/F', 'F/F']

The following will give you 39 entries:

['/'.join([*a]) for i in range(1,4) for a in itertools.product(['a','b','c'],repeat=i)]

Please notice, that your given reference only has 35 entries, since {'a/c/c', 'b/a/a', 'b/c/c', 'c/a/a'} are missing. I see no logic why these are missing, so I assume an error with your list.

Answer 2

What you are trying to get is a product, not a combination.

lis = ["BRUTELE", "COCKPIT EST", "CIRCET"]
separator = " / "
total_number_of_combinations = 3

result = list(
    itertools.chain.from_iterable(
        (separator.join(a) for a in itertools.product(lis, repeat=i))
        for i in range(1, total_number_of_combinations + 1)
    )
)

assert result == ['BRUTELE', 'COCKPIT EST', 'CIRCET', 'BRUTELE / BRUTELE', 'BRUTELE / COCKPIT EST', 'BRUTELE / CIRCET', 'COCKPIT EST / BRUTELE', 'COCKPIT EST / COCKPIT EST', 'COCKPIT EST / CIRCET', 'CIRCET / BRUTELE', 'CIRCET / COCKPIT EST', 'CIRCET / CIRCET', 'BRUTELE / BRUTELE / BRUTELE', 'BRUTELE / BRUTELE / COCKPIT EST', 'BRUTELE / BRUTELE / CIRCET', 'BRUTELE / COCKPIT EST / BRUTELE', 'BRUTELE / COCKPIT EST / COCKPIT EST', 'BRUTELE / COCKPIT EST / CIRCET', 'BRUTELE / CIRCET / BRUTELE', 'BRUTELE / CIRCET / COCKPIT EST', 'BRUTELE / CIRCET / CIRCET', 'COCKPIT EST / BRUTELE / BRUTELE', 'COCKPIT EST / BRUTELE / COCKPIT EST', 'COCKPIT EST / BRUTELE / CIRCET', 'COCKPIT EST / COCKPIT EST / BRUTELE', 'COCKPIT EST / COCKPIT EST / COCKPIT EST', 'COCKPIT EST / COCKPIT EST / CIRCET', 'COCKPIT EST / CIRCET / BRUTELE', 'COCKPIT EST / CIRCET / COCKPIT EST', 'COCKPIT EST / CIRCET / CIRCET', 'CIRCET / BRUTELE / BRUTELE', 'CIRCET / BRUTELE / COCKPIT EST', 'CIRCET / BRUTELE / CIRCET', 'CIRCET / COCKPIT EST / BRUTELE', 'CIRCET / COCKPIT EST / COCKPIT EST', 'CIRCET / COCKPIT EST / CIRCET', 'CIRCET / CIRCET / BRUTELE', 'CIRCET / CIRCET / COCKPIT EST', 'CIRCET / CIRCET / CIRCET']                                      

However, the number of resulting items is len(lis) ** total_number_of_combinations , which may be computationally expensive. A better method would be parsing a string by splitting at separators and testing the membership of each split string.