Haskell：映射和压缩列表

Question

好吧，看来我对 Haskell 很不满意，我发誓我一直被所有这些问题所困扰。无论如何，我需要取两个列表，压缩它们，然后用总和映射。所以这是我到目前为止所拥有的，但它不起作用：

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

好的，这是我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    |     ^^^^^^^

我只是不明白错误的含义，我的意思是我得到了一个列表 [a] 和另一个列表 [a]，然后我压缩它们并映射然后得到另一个列表，这对我来说没有意义为什么我收到元组错误，有人可以帮忙吗？

Answer 1

map + zipList 尝试将map 与zipList 相加，如x+y。在那里，您想将+ 作为函数传递，所以它应该是map (+) zipList。

但是等等。 zipList 是对的列表，(+) 不接受对作为输入：

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

一个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

还要注意你的类型是错误的，你需要a是一个数字类型来求和。

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 是一个类型为map :: (a -> b) -> [a] -> [b] 的函数，而zip 的类型为zip :: [a] -> [b] -> [(a, b)]。当您编写map + zipList 时，您正在尝试将(a -> b) -> [a] -> [b] 类型的元素添加到[(a,b)] 类型的元素中——您知道这是怎么回事吗？

至于你如何实现你想要的，这将是算法：

1. 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，您需要先将元组分成 2 个输入，然后才能进行加法
2. 使用上述函数将map 应用于压缩列表。