如何在Haskell中合並兩種不同類型的列表

Question

如何合並兩種不同類型的列表並遍歷Haskell中的結果？

例如：

input: [1,2,3]  ['A','B','C'],
output: ["A1","A2","A3","B1","B2","B3","C1","C2","C3"].

我嘗試使用Int和Char進行示例，例如：

combine :: Int -> Char -> String
combine a b = show b ++ show a

但是，這不起作用，因為如果我將此功能用於combine 3 'A' "'A'3" combine 3 'A' ，則輸出將為"'A'3" ，而不是"A3" 。

Answer 1

show :: Char -> String實際上會將單引號引起來：

*Main> show 'A'
"'A'"

但是由於type String = [Char] ，我們可以使用：

combine :: Int -> Char -> String
combine i c = c : show i

因此，在這里我們構造了一個以c為首（第一個字符）的字符列表，並以show i （整數的表示形式）為尾。

現在，我們可以使用列表推導來組合兩個列表：

combine_list :: [Int] -> [Char] -> [String]
combine_list is cs = [combine i c | c <- cs, i <- is]

然后生成輸出：

*Main> combine_list [1,2,3]  ['A','B','C']
["A1","A2","A3","B1","B2","B3","C1","C2","C3"]

Answer 2

您可以執行以下操作；

combiner :: [Char] -> [Int] -> [String]
combiner cs xs = (:) <$> cs <*> (show <$> xs)

*Main> combiner ['A','B','C'] [1,2,3]
["A1","A2","A3","B1","B2","B3","C1","C2","C3"]

這里(:) <$> cs （其中<$>是infix fmap ）將構造一個應用列表函子，而show <$> xs就像map show xs產生["1","2","3"]並按<*>我們僅將應用列表應用於["1","2","3"]生成["A1","A2","A3","B1","B2","B3","C1","C2","C3"]