python - 尽管有 "if statement"，但重复项最终出现在列表中

Question

好的，我正在研究一个代码，它可以给出您输入的乱序字母的所有可能组合。在这里:

import random, math

words = []
original = raw_input("What do you need scrambled? ")

def word_scramble(scrambled):

    original_length = len(scrambled)
    loops = math.factorial(original_length)

    while loops > 0:
        new_word = []

        used_numbers = []

        while len(new_word) < original_length:

            number = random.randint(0, original_length - 1)

            while number in used_numbers:
                number = random.randint(0, original_length - 1)

            while number not in used_numbers:
                used_numbers.append(number)
                new_word.append(scrambled[number])

        if new_word not in words:
            words.append(("".join(str(x) for x in new_word)))
            loops -= 1

word_scramble(original)

print ("\n".join(str(x) for x in words))

问题是，它仍然会给出重复项，即使它不应该这样做。例如，我可以输入“imlk”，有时会得到两次“milk”，但仍然只给我 24 个排列，这意味着一些排列被排除在外。的:

if new_word not in words:
    words.append(("".join(str(x) for x in new_word)))
    loops -= 1

应该防止重复出现在列表中。所以我不太确定问题出在哪里。很抱歉，问题的主要标题是如此模糊/奇怪。我真的不确定如何更好地表达它。

Answer 1

itertools.permutations 怎么样？

import itertools
original = raw_input("What do you need scrambled? ")
result = [''.join(s) for s in itertools.permutations(original)]