如何在Haskell中合并两种不同类型的列表

Question

如何合并两种不同类型的列表并遍历Haskell中的结果？

例如：

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

我尝试使用Int和Char进行示例，例如：

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

但是，这不起作用，因为如果我将此功能用于combine 3 'A' "'A'3" combine 3 'A' ，则输出将为"'A'3" ，而不是"A3" 。

Answer 1

show :: Char -> String实际上会将单引号引起来：

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

但是由于type String = [Char] ，我们可以使用：

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

因此，在这里我们构造了一个以c为首（第一个字符）的字符列表，并以show i （整数的表示形式）为尾。

现在，我们可以使用列表推导来组合两个列表：

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

然后生成输出：

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

Answer 2

您可以执行以下操作；

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"]

这里(:) <$> cs （其中<$>是infix fmap ）将构造一个应用列表函子，而show <$> xs就像map show xs产生["1","2","3"]并按<*>我们仅将应用列表应用于["1","2","3"]生成["A1","A2","A3","B1","B2","B3","C1","C2","C3"]