通过两个输入列表映射函数 [英] Mapping a function over two input lists

问题描述

我有一个功能，我想用几组输入来测试。假设这个函数是

  f :: a  - > b  - > c 
  

现在我有两个输入列表：

  inputA :: [a] 
 inputB :: [[b]] 
  

对于 inputA !!我，我想评估 f $输入！我为 inputB !!中的每个元素列出了。 I 。我知道我需要几个 map 这样的应用程序来执行此操作，但是我很难绕过解决方案。

我最近的尝试是

  map f inputA< $> inputB 
  

给出以下错误：

无法匹配预期的类型 a0 - > b0'与实际类型 [b1]'

在地图'

调用的返回类型中可能的原因： map'应用于太多参数

在（< $>）'的第一个参数中，即 map f inputA'

在表达式中：map f inputA inputB

我该如何解决这个问题？我不一定需要一个完整的解决方案。推动（甚至推动）有益的方向肯定会受到赞赏。

其他想法：

  map f inputA :: [b  - > c] 
  

我认为这是正确的方向。现在我需要将每个函数映射到 inputB 中的每个输入列表上。为了澄清，我想要通过 i map f inputA 中的 i th函数输入 inputB 中的输入列表以获得结果 outputC :: [[c]] 。

解决方案

如果我正确地理解了你的话，像这样： $ b

  mapNested ::（a  - > b  - > c） - > [a]  - > [[b]]  - > [[c]] 
 mapNested f [] _ = [] $ b $ mapNested f _ [] = [] $ b $ mapNested f（x：xs）ys = concatMap（map（fx））ys ：mapNested f xs ys 
 
 Main> mapNested（+）[1，2，3] [[1,2,3]，[4,5,6]，[7,8,9]] 
 [[2,3,4,5， 6,7,8,9,10]，[3,4,5,6,7,8,9,10,11]，[4,5,6,7,8,9,10,11,12] ] 
  

如果这不是您要查找的内容，您能否提供示例输入和输出？

编辑

或者这是你想要的吗？

mapNested ::（a - > b - > c） - > [a] - > [[b]] - > [[c]]
mapNested f xs = zipWith map（map f xs）

Main> mapNested（，）[1，2，3] [[1,2,3]，[4,5,6]，[7,8,9]]
[[（1,1），（1 ，2），（1,3）]，[（2,4），（2,5），（2,6）]，[（3,7），（3,8），（3,9）] ]


I have a function that I want to test with several sets of inputs. Let's say the function is

f :: a -> b -> c

Now I have two lists of inputs:

inputA :: [a]
inputB :: [[b]]

For inputA !! i, I want evaluate f $ input !! i for each element of the list at inputB !! i. I know I need several applications of map to do this, but I am having difficulty wrapping my head around a solution.

My most recent attempt is

map f inputA <$> inputB

which gives the following error:

Couldn't match expected type a0 -> b0' with actual type[b1]'
In the return type of a call of map'
Probable cause:map' is applied to too many arguments
In the first argument of (<$>)', namelymap f inputA'
In the expression: map f inputA inputB

How should I go about solving this problem? I don't necessarily want a complete solution. A push (or even a shove) in a helpful direction would definitely be appreciated.

Additional thoughts:

map f inputA :: [b -> c]

I think this is the right direction. Now I need to map each of the functions over each list of inputs in inputB.

To clarify, I want to map the ith function in map f inputA over the ith list of inputs in inputB to get a result outputC :: [[c]].

解决方案

If I'm understanding you correctly, something like:

mapNested :: (a -> b -> c) -> [a] -> [[b]] -> [[c]]
mapNested f [] _ = []
mapNested f _ [] = []
mapNested f (x:xs) ys = concatMap (map (f x)) ys : mapNested f xs ys

Main> mapNested (+) [1, 2, 3] [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
[[2,3,4,5,6,7,8,9,10],[3,4,5,6,7,8,9,10,11],[4,5,6,7,8,9,10,11,12]]

If this isn't what you're looking for, could you provide an example input and output?

EDIT

Or is this what you're wanting?

mapNested :: (a -> b -> c) -> [a] -> [[b]] -> [[c]]
mapNested f xs = zipWith map (map f xs)

Main> mapNested (,) [1, 2, 3] [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
[[(1,1),(1,2),(1,3)],[(2,4),(2,5),(2,6)],[(3,7),(3,8),(3,9)]]

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