从Python中的字母枚举所有长度为K的可能字符串 [英] enumerating all possible strings of length K from an alphabet in Python

问题描述

可能重复：

有没有最好的方法来生成所有可能的三个字母的关键字

如何从字母L枚举所有长度为K的字符串，其中L只是一个字符列表？例如。如果 L = ['A'，'B'，'C'] 和 K = 2 ，我会希望枚举所有可能由字母'A'，'B'组成的长度为2的字符串， C 。它们可以重复使用，因此'AA'是有效的。

how can I enumerate all strings of length K from an alphabet L, where L is simply a list of characters? E.g. if L = ['A', 'B', 'C'] and K = 2, I'd like to enumerate all possible strings of length 2 that can be made up with the letters 'A', 'B', 'C'. They can be reused, so 'AA' is valid.

从本质上讲，这是置换的置换我明白。如果对此有一个更正确的技术术语，请让我知道...。您可以通过从字母L中选择任何字母，并可能以对订单（因此， AB 与 BA 并不相同。）是否存在更清晰的陈述方式？

This is essentially permutations with replacement, as far as I understand. if theres a more correct technical term for this, please let me know.... its essentially all strings of length K that you can make by choosing ANY letter from the alphabet L, and possibly reusing letters, in a way that is sensitive to order (so AB is NOT identical to BA according to this.) is there a clearer way to state this?

无论如何我相信解决方案是：

in any case i believe the solution is:

[ ''.join(x) for x in product(L, repeat=K) ]

但是我很感兴趣在其他答案中，尤其是天真的方法与快速的Python方法，以及对速度注意事项的讨论。

but i am interested in other answers to this, esp. naive approaches versus fast Pythonic ones, and discussions of speed considerations.

推荐答案

这是 Python文档

EDIT2：当然正确的答案是产品，谢谢您的评论

of course the right answer is the product, thanks for the comment

print  [''.join(x) for x in product('ABC', repeat=3)]

打印27个元素

['AAA', 'AAB', 'AAC', 'ABA', 'ABB', 'ABC', 'ACA', 'ACB', 'ACC', 'BAA', 'BAB', 
'BAC', 'BBA', 'BBB', 'BBC', 'BCA', 'BCB', 'BCC', 'CAA', 'CAB', 'CAC', 'CBA', 
'CBB', 'CBC', 'CCA', 'CCB', 'CCC']

@ agf在
之前给出了正确的答案

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