来自（a - > [b]）的Haskell函数 - > [a - > b] [英] Haskell function from (a -> [b]) -> [a -> b]

问题描述

我有一个函数 seperateFuncs ，使得

seperateFuncs :: [a -> b] -> (a -> [b])
seperateFuncs xs = \x -> map ($ x) xs

我想知道是否存在逆向，即是否有函数

I was wondering whether the converse existed, i.e. is there a function

joinFuncs :: (a -> [b]) -> [a -> b]

我认为不是（主要是因为列表不是固定长度），但也许我会证明是错的。
问题是有一些数据类型 f ，它有一个函数::（a - > fb） - > f（a - > b）？

I think not (mainly because lists are not fixed length), but perhaps I'll be proved wrong. The question then is there some datatype f which has a function :: (a -> f b) -> f (a -> b)?

推荐答案

您可以将 seperateFuncs 归纳为 Applicative $（code> Monad ）很干净：

You can generalize seperateFuncs to Applicative (or Monad) pretty cleanly:

seperateFuncs :: (Applicative f) => f (a -> b) -> (a -> f b)
seperateFuncs f x = f <*> pure x

使用无点式书写，您有 seperateFuncs =（ 。）（<*>）），所以你基本上需要 unap。 （。extract），给出下面的定义，如果你用有意义的风格写出它：

Written in point-free style, you have seperateFuncs = ((. pure) . (<*>)), so you basically want unap . (. extract), giving the following definition if you write it in pointful style:

joinFuncs :: (Unapplicative f) => (a -> f b) -> f (a -> b)
joinFuncs f = unap f (\ g -> f (extract g))

这里我将 Unapplictaive 定义为：

Here I define Unapplictaive as:

class Functor f => Unapplicactive f where
    extract  :: f a -> a
    unap     :: (f a -> f b) -> f (a -> b)

获取

To get the definitions given by leftaroundabout, you could give the following instances:

instance Unapplicative [] where
    extract = head
    unap f = [\a -> f [a] !! i | i <- [0..]]

instance Unapplicative ((->) c) where
    extract f = f undefined
    unap f = \x y -> f (const y) x

我认为很难想出一个有用的函数 f ::（fa - > fb） - >对于不同于（ - >） f ，f（a - > b） c $ c>。

I think it's hard to come up with a "useful" function f :: (f a -> f b) -> f (a -> b) for any f that isn't like (->).

这篇关于来自（a - > [b]）的Haskell函数 - > [a - > b]的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！