使用 itertools 排列和乘积返回所有可能的字符排序

Question

I am having a bit of difficulty merging itertools permutations and product to get the output (list) I desire. I am trying to generate all orderings of characters with consideration for wildcard chars (?, *).

For instance, if input is A?, I am trying to get the following output: AA AB BA AC CA AD DA... etc

The code below does a great job of generating all permutations, with the wildcards left in.

chars = "HELLO?"
for i in range(len(chars)+1):
    perms = map(''.join, permutations(chars,i))
    for perm in perms:
        print(perm)

And this code allows me to substitute wildcard chars with all 26 possible alphabetic characters.

chars = "HELLO?"
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 == "?" else (char) for char in chars]
 
 for p in itertools.product(*wilds):
     print(p)

What is the best way to combine these two segments (most efficient) to get the output I am looking for? Is there a better, more efficient way to do this?

Answer 1

You can just nest your two for loops, Append the values to a list if you want; but it will be big: if memory is a concern and you only need the values once I'd use a generator:

import itertools


def perm_with_wild_generator(chars):
    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 == "?" else (char) for char in chars]

    for p in itertools.product(*wilds):
        for i in range(len(p) + 1):
            perms = map(''.join, itertools.permutations(p, i))
            for perm in perms:
                yield perm


for c in perm_with_wild_generator("HELLO?"):
    print(c)

The duplicates you mention in comments are because the outputs of itertools.product which are separately permuted share 5/6 of their letters. This code replaces product (recursively to handle multiple wilds) and eliminates duplicates:

import itertools

def replace_wild(chars, wilds):
    if '?' not in chars:
        yield chars
    else:
        for wild in wilds:
            for w in replace_wild(chars.replace("?", wild, 1), wilds):
                yield w


def perm_with_wild_generator(chars):
    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'} - set(chars)

    for i in range(len(chars) + 1):
        perms = map(''.join, itertools.permutations(chars, i))
        for perm in perms:
            for w in replace_wild(perm, wilds | set(perm.replace('?', ""))):
                yield w


for c in perm_with_wild_generator("HELLO?"):
    print(c)

In python 3, using yield from would simplify a bit.