python - 我如何编写一个函数，它使用尽可能多的 'for loops' 来使用带有字母字符串的列表来对带有空格的列表进行字谜排列？

Question

您好 stackoverflow 社区! 几个星期以来，我一直在努力寻找一种方法来做到这一点而不寻求帮助(有点个人挑战)，但我做不到，而且大学项目占用了我大部分时间，无法做到这一点令人沮丧我，因为它似乎不太难，但想不出怎么做。成为开发者的第一课就是学会与他人合作，对吧？所以我来这里寻求帮助。我有这段代码:

l = [None]*8
k = ['m','i','k','e']
for a in k: 
   for b in k: 
       for c in k: 
          l[0] = a 
          l[1] = b 
          l[2] = c 
          print(l)

输出这个:

['m', 'm', 'm', None, None, None, None, None]
['m', 'm', 'i', None, None, None, None, None]
['m', 'm', 'k', None, None, None, None, None]
['m', 'm', 'e', None, None, None, None, None]
['m', 'i', 'm', None, None, None, None, None]
['m', 'i', 'i', None, None, None, None, None]
['m', 'i', 'k', None, None, None, None, None]
['m', 'i', 'e', None, None, None, None, None]
['m', 'k', 'm', None, None, None, None, None]
['m', 'k', 'i', None, None, None, None, None]
['m', 'k', 'k', None, None, None, None, None]
['m', 'k', 'e', None, None, None, None, None]
['m', 'e', 'm', None, None, None, None, None]
['m', 'e', 'i', None, None, None, None, None]
['m', 'e', 'k', None, None, None, None, None]
['m', 'e', 'e', None, None, None, None, None]
['i', 'm', 'm', None, None, None, None, None]
['i', 'm', 'i', None, None, None, None, None]
['i', 'm', 'k', None, None, None, None, None]
['i', 'm', 'e', None, None, None, None, None]
['i', 'i', 'm', None, None, None, None, None]
['i', 'i', 'i', None, None, None, None, None]
['i', 'i', 'k', None, None, None, None, None]
['i', 'i', 'e', None, None, None, None, None]
['i', 'k', 'm', None, None, None, None, None]
['i', 'k', 'i', None, None, None, None, None]
['i', 'k', 'k', None, None, None, None, None]
['i', 'k', 'e', None, None, None, None, None]
['i', 'e', 'm', None, None, None, None, None]
['i', 'e', 'i', None, None, None, None, None]
['i', 'e', 'k', None, None, None, None, None]
['i', 'e', 'e', None, None, None, None, None]
['k', 'm', 'm', None, None, None, None, None]
['k', 'm', 'i', None, None, None, None, None]
['k', 'm', 'k', None, None, None, None, None]
['k', 'm', 'e', None, None, None, None, None]
['k', 'i', 'm', None, None, None, None, None]
['k', 'i', 'i', None, None, None, None, None]
['k', 'i', 'k', None, None, None, None, None]
['k', 'k', 'm', None, None, None, None, None]
['k', 'k', 'i', None, None, None, None, None]
['k', 'k', 'k', None, None, None, None, None]
['k', 'k', 'e', None, None, None, None, None]
['k', 'e', 'm', None, None, None, None, None]
['k', 'e', 'i', None, None, None, None, None]
['k', 'e', 'k', None, None, None, None, None]
['k', 'e', 'e', None, None, None, None, None]
['e', 'm', 'm', None, None, None, None, None]
['e', 'm', 'i', None, None, None, None, None]
['e', 'm', 'k', None, None, None, None, None]
['e', 'm', 'e', None, None, None, None, None]
['e', 'i', 'm', None, None, None, None, None]
['e', 'i', 'i', None, None, None, None, None]
['e', 'i', 'k', None, None, None, None, None]
['e', 'i', 'e', None, None, None, None, None]
['e', 'k', 'm', None, None, None, None, None]
['e', 'k', 'i', None, None, None, None, None]
['e', 'k', 'k', None, None, None, None, None]
['e', 'k', 'e', None, None, None, None, None]
['e', 'e', 'm', None, None, None, None, None]
['e', 'e', 'i', None, None, None, None, None]
['e', 'e', 'k', None, None, None, None, None]
['e', 'e', 'e', None, None, None, None, None]

我想做的是有一个函数，可以对我想要的 nonelist 中的空格数执行此操作。就像传递一个参数来选择我想要的 None 数来代替。我想象着这个函数是如何工作的，在我看来它有 3 个参数(None 列表的大小，要替换的字母列表，要替换的 None 的数量)。我熟悉递归，但想不出使用它来实现它的方法。提前致谢:)

Answer 1

递归是要走的路:

def print_permutations(k, l, level = 0):
    if level == len(l):
        print(l)
    else:
        for a in k:
            l[level] = a
            print_permutations(k, l, level+1 )



l = [None]*8
k = ['m','i','k','e']

print_permutations(k, l)

或者您可以使用标准工具:

from itertools import combinations_with_replacement


k = ['m','i','k','e']

for p in combinations_with_replacement(k, 8):
    print(p) 
    # print(list(p)) if you want to print it as a list

顺便说一句。 k='mike' 的效果相同。