Question

如何组合两种不同类型的列表并在Haskell中遍历结果？

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制作示例，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'，则输出将为"'A'3"，而不是"A3"。

Answer 1

show :: Char -> String确实会在角色周围加上单引号：

*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

所以在这里我们构造一个字符列表，其中c作为头（第一个字符），show i（整数的表示）作为尾部。

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;

(:) <$> cs

此处<$>（其中fmap为中缀show <$> xs）将构建一个应用列表仿函数，而map show xs就像["1","2","3"]产生<*>并且["1","2","3"]我们只需将应用列表应用于["A1","A2","A3","B1","B2","B3","C1","C2","C3"]生成$.each( myObj, function( key, value ){ ... if( sthg... ){ myObj.somethingWentHorriblyWrong = true; return false; } }); if( myObj.somethingWentHorriblyWrong ){ // ... do something, not forgetting to go: delete myObj.somethingWentHorriblyWrong; }

如何在haskell中组合两种不同类型的列表

2 个答案: