我怎样才能编写一个函数，使用一个带有字母字符串的列表来重新编写一个带有空格的列表，并使用请求的尽可能多的“for循环”？

Question

你好，stackoverflow社区！ 几周来，我一直在努力寻找一种方法来完成这项工作，而不寻求帮助（有点像个人的挑战），但我做不到，而且大学项目占用了我大部分的时间，不能做到这一点让我很沮丧，因为这看起来不太难，但我想不出怎么做。成为开发人员的第一课是学会与他人合作，对吗？所以我是来寻求帮助的。我有这个密码：

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)

输出如下：

['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，它可以计算出nonelistI中有多少个空格想要。就像传递一个参数，选择要替换多少个无。我想象这个函数是如何工作的，在我的脑海中它有3个参数（无列表的大小，要替换的字母列表，要替换的无的数目）。我对递归很熟悉，但想不出用它实现这个的方法。提前感谢：）

Answer 1

递归是一种方法：

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)

也可以使用标准工具：

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'的工作原理是一样的。你知道吗