Using itertools product but needing combination behavior

Question

I am trying to generate a list of possible strings of characters from input, accounting for wildcards "?" and "*". The ordering of list items should not matter, meaning if ABC already exists in the list, then I do not want to add ACB, or any other ordering (processing speed issue). The code I am using is below:

import itertools
from itertools import permutations

##################################################### 
def getWordsFromTiles(tiles, word):
#####################################################   
    return all(word.count(i) <= tiles.count(i) for i in word)

#####################################################
Main
#####################################################     
    chars = A?*
    
    wilds = [
        ('A','B','C','D','E','F','G','H','I','J','K',
         'L','M','N','O','P','Q','R','S','T','U','V',
         'W','X','Y','Z')
        if char == "?" or char == "*" else (char) for char in chars]
    
    for p in itertools.product(*wilds):
        x = ''.join(p)
        hits.extend([word for word in data if getWordsFromTiles(x.upper(), word) and word not in hits])

This generates a list like the following:

A,A,A
A,A,B
A,A,C
.....
A,B,A

I actually do not care about the order of these list, so I would like to not have A,B,A when I have already generated "A,A,B. Any ideas how to implement this?

Answer 1

actually do not care about the order of these list, so I would like to not have A,B,A when I have already generated "A,A,B".

Is this what you are looking for?

from itertools import combinations_with_replacement

s = ['A','B','C','D']

list(combinations_with_replacement(s,3))

[('A', 'A', 'A'),
 ('A', 'A', 'B'),
 ('A', 'A', 'C'),
 ('A', 'A', 'D'),
 ('A', 'B', 'B'),
 ('A', 'B', 'C'),
 ('A', 'B', 'D'),
 ('A', 'C', 'C'),
 ('A', 'C', 'D'),
 ('A', 'D', 'D'),
 ('B', 'B', 'B'),
 ('B', 'B', 'C'),
 ('B', 'B', 'D'),
 ('B', 'C', 'C'),
 ('B', 'C', 'D'),
 ('B', 'D', 'D'),
 ('C', 'C', 'C'),
 ('C', 'C', 'D'),
 ('C', 'D', 'D'),
 ('D', 'D', 'D')]