我如何找到所有可能的方法来组合列表中的项目而无需重复？

Question

I have six categories: A, B, C, D, E, and F.

I want to find out all the unique ways I can combine the categories, without repetition.

For example, if I combine the first three categories, I will get A, A, A, B, C, D. If I combine B, C, D, E, I will get A, B, B, B, B, C.

I tried itertools. itertools.product comes close, but there is a lot of repetition. For example, I get A, B, A, A, C, D, but I also get B, A, B, B, D, C, which is a duplicate in my case. The order matters, replacement matters, the count matter, but the character does not matter.

Answer 1

Since you only have 6 categories you can use itertools.product and then filter your result according to your criteria.

Your examples are somewhat confusing as I'm not sure how you get 'AAABCD' from the first three categories 'ABC' which doesn't contain 'D', or how you get 'ABBBBC' by combining 'BCDEI' which doesn't contain 'A'. However assuming you want to get all the unique combinations of some subset of 'ABCDEF', of length 6, up to symbol replacement you can do this.

from itertools import product

CATEGORIES = 'ABCDEF'

def combinations(cats):
    # use itertools to get all combinations 
    all_combs = product(cats,repeat=len(CATEGORIES))
    valid_combs = set()

    # For every possible combination find the order in which the characters appear
    for s in all_combs:
        s = ''.join(s)
        order = []
        for c in s:
            if c not in order: 
                order.append(c)

        # replace the character by ones following a set predetermined order
        for i,c in enumerate(order):
            replace_char = CATEGORIES[i].lower()
            s = s.replace(c, replace_char)

        # add to set to remove duplicates
        s = s.upper()
        valid_combs.add(s)
    return list(valid_combs)

usage

combinations('AB') 
['ABABBB', 'ABABBA', 'AABBBB', 'ABAABB', 'ABBAAA', 'AABAAA', 'AABABB', 'AAABAB', 'AABABA', 'AABAAB', 'ABAAAB', 'AABBAB', 'AAAAAB', 'ABBAAB', 'ABBABA', 'ABBABB', 'AAAABA', 'ABAAAA', 'AAABAA', 'ABAABA', 'ABBBAB', 'AAABBB', 'ABBBBA', 'AAABBA', 'AABBAA', 'ABABAA', 'AAAAAA', 'ABBBBB', 'ABABAB', 'ABBBAA', 'AABBBA', 'AAAABB']

The rationale of this is that if 'ABAACD' and 'BABBDC' belong to the same equivalence class, then the member where the characters appear in order is a unique representative of that equivalence class.

This isn't very efficient though, so for a much larger list of categories you may need to construct the list directly.