哈斯克爾組合和排列

Question

我有三個詞在列表[「a」，「b」，「c」]。我想找到所有可能的組合設置5,6等

例如用於一套5我將不得不上述

**[ [aaaaa],[aaaab],[aaaac], [aaabc] , ..... ]** etc 3^5 = 243 combinations 

AAAAAA將基本上是「一」，「一」，「一」 「一」，「一」 ....

Answer 1

replicateM你想要做什麼：

> import Control.Monad 
> replicateM 5 ["a", "b", "c"] 
[["a","a","a","a","a"],["a","a","a","a","b"],["a","a","a","a","c"],["a","a","a","b","a"],["a","a","a","b","b"],["a","a","a","b","c"],["a","a","a","c","a"],["a","a","a","c","b"],["a","a","a","c","c"],["a","a","b","a","a"],["a","a","b","a","b"],["a","a","b","a","c"],["a","a","b","b","a"],["a","a","b","b","b"],["a","a","b","b","c"]...] 

Answer 2

當然，nanothief的回答給出了最短的解決方案，但它可能是有益和好玩自己做。

有很多方法可以爲笛卡爾積寫一個函數。例如。您可以使用列表理解：

prod :: [[a]] -> [[a]] -> [[a]] 
prod as bs = [a ++ b | a <- as, b <- bs] 

凡(++) :: [a] -> [a] -> [a] - 見Data.List。另一種可能是使用列表的Applicative實例：

import Control.Applicative 

prod as bs = (++) <$> as <*> bs 

現在，你需要反覆應用此操作。的摺疊可以做到這一點，例如：

rep :: Int -> [[a]] -> [[a]] 
rep n as = foldr1 prod $ replicate n as 

rep 3 ['a','b','c'] 
--["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"] 

瞭解這個解決方案可能比服用replicateM快捷更有價值。也就是說，你可以使用Hoogle輕鬆找到後者。

-

更多關於函子和應用型見定義fmap（<$>）和ap（<*>）。 Functors, Applicatives, And Monads In Pictures也可以是一個很好的資源。