haskell - 转换为 Pointfree 风格？

Question

Learn You a Haskell使用以下 Prob 类型讨论“制作 Monad”:

import Data.Ratio

newtype Prob a = Prob { getProb :: [(a,Rational)] } deriving Show  

Prob代表一个a类型，然后是一个Rational代表这个a被使用的概率。

让我们看一个 Prob 实例:

*Main> Prob [('a', 1%2), ('b', 1%2)]
Prob {getProb = [('a',1 % 2),('b',1 % 2)]}

LYAH 提出了一个练习来弄清楚如何将类型为 Prob(Prob Char) 的 thisSituation 转换为 Prob Char:

thisSituation :: Prob (Prob Char)
thisSituation = Prob
  [( Prob [('a', 1%2),('b',1%2)], 1%4)
  ,( Prob [('c', 1%2),('d',1%2)], 3%4)
  ]

这是我想出的:

flatten :: Prob (Prob a) -> Prob a
flatten pp = Prob $ convert $ getProb pp

convert :: [(Prob a, Rational)] -> [(a, Rational)]
convert xs = concat $ map f xs

f :: (Prob a, Rational) -> [(a, Rational)]
f (p, r) = map (mult r) (getProb p)

mult :: Rational -> (a, Rational) -> (a, Rational)
mult r (x, y) = (x, r*y)

我试过 point-free 是这样的:

flatten :: Prob (Prob a) -> Prob a
flatten = Prob $ convert $ getProb 

但是出现了这个错误:

*Main> :l MakingMonad.hs
[1 of 1] Compiling Main             ( MakingMonad.hs, interpreted )

MakingMonad.hs:37:11:
    Couldn't match expected type `Prob (Prob a) -> Prob a'
                with actual type `Prob a0'
    In the expression: Prob $ convert $ getProb
    In an equation for `flatten': flatten = Prob $ convert $ getProb

MakingMonad.hs:37:28:
    Couldn't match expected type `[(Prob a0, Rational)]'
                with actual type `Prob a1 -> [(a1, Rational)]'
    In the second argument of `($)', namely `getProb'
    In the second argument of `($)', namely `convert $ getProb'
    In the expression: Prob $ convert $ getProb
Failed, modules loaded: none.

我可以让 flatten 无点吗？如果是这样，请告诉我如何。如果不是，请解释原因。

Answer 1

当你在flatten中使用$时，你得到的代码看起来像

flatten = Prob $ convert $ getProb
==> Prob (convert (getProb))

这不是你想要的。

你想要 Prob 。转变 。获取概率