Question

好吧，看来我对哈斯克尔的态度很糟糕，我发誓我一直困扰着所有这些问题。无论如何，我需要拿两个列表，压缩它们，然后用总和来映射。所以这就是我到目前为止所做的事情并没有奏效：

zipMapList :: [a] -> [a] -> [a]
zipMapList x y = do
    let zipList = zip x y
    let mapList = map + zipList
    mapList

好的，这是我在尝试加载文件时遇到的错误：

HW2.hs:18:25: error:
• Couldn't match expected type ‘(a1 -> b) -> [a1] -> [b]’
              with actual type ‘[(a, a)]’
• In the second argument of ‘(+)’, namely ‘zipList’
  In the expression: map + zipList
  In an equation for ‘mapList’: mapList = map + zipList
• Relevant bindings include
    mapList :: (a1 -> b) -> [a1] -> [b] (bound at HW2.hs:18:9)
    zipList :: [(a, a)] (bound at HW2.hs:17:9)
    y :: [a] (bound at HW2.hs:16:14)
    x :: [a] (bound at HW2.hs:16:12)
    zipMapList :: [a] -> [a] -> [a] (bound at HW2.hs:16:1)
    18 |     let mapList = map + zipList    |                             
HW2.hs:19:5: error:
• Couldn't match expected type ‘[a]’
              with actual type ‘(a0 -> b0) -> [a0] -> [b0]’
• Probable cause: ‘mapList’ is applied to too few arguments
  In a stmt of a 'do' block: mapList
  In the expression:
    do let zipList = zip x y
       let mapList = map + zipList
       mapList
  In an equation for ‘zipMapList’:
      zipMapList x y
        = do let zipList = ...
             let mapList = ...
             mapList
• Relevant bindings include
    zipList :: [(a, a)] (bound at HW2.hs:17:9)
    y :: [a] (bound at HW2.hs:16:14)
    x :: [a] (bound at HW2.hs:16:12)
    zipMapList :: [a] -> [a] -> [a] (bound at HW2.hs:16:1)
19 |     mapList    |     ^^^^^^^

我只是没有得到错误的含义，我的意思是我得到一个列表[a]和另一个列表[a]然后我将它们压缩并映射然后得到另一个列表，它没有＆＃t; t让我理解为什么我会得到元组错误，有人能伸出援助之手吗？

Answer 1

map + zipList尝试将map与zipList相加，如x+y中所示。在那里，您希望将+作为函数传递，因此它应该是map (+) zipList。

但是等等。 zipList是对的列表，(+)不会将一对作为输入：

-- simplified types for clarity
(+) :: Int -> Int -> Int
-- but we want
add :: (Int, Int) -> Int

可能的解决方案是定义我们自己的add

add (x,y) = x+y
-- ...
map add zipList

或使用lambda

map (\ (x,y) -> x+y) zipList

或取消+：

map (uncurry (+)) zipList

另请注意，您的类型有误，您需要a作为数字类型才能求和。

zipMapList :: Num a => [a] -> [a] -> [a]
zipMapList xs ys = map (uncurry (+)) (zip xs yz)
-- or, exploiting zipWith
zipMapList xs ys = zipWith (+) xs yz

Answer 2

map是map :: (a -> b) -> [a] -> [b]类型的函数，zip类型为zip :: [a] -> [b] -> [(a, b)]。当您编写map + zipList时，您尝试将(a -> b) -> [a] -> [b]类型的元素添加到[(a,b)]类型的元素中 - 您能看到它是如何运作的吗？

至于你如何实现你想要的，这将是算法：

编写一个匿名函数或命名函数来对压缩元组的元素求和。请注意，(+)函数的类型是什么 - 它不适用于元组，而是两个不同的输入。因此，在进行添加之前，需要将元组分成2个输入
使用上述功能将map应用于压缩列表。

Haskell：Map和Zip A List

2 个答案: