提升功能与高阶功能之间的区别 [英] Difference between lifting and higher order functions

问题描述

当人们谈论map，fold或bind时，我通常会听到提升"一词，但基本上每个高阶函数都不都是某种提升吗?

I usually hear the term lifting, when people are talking about map, fold, or bind, but isn't basically every higher order function some kind of lifting?

为什么filter不能成为从a -> Bool到[a] -> [a]的提升，甚至连bool函数(模拟if语句的函数)也可以被认为是从a -> a到Bool -> a的提升.如果不是，那么为什么Applicative类型类中的ap被认为是提升?

Why can't filter be a lift from a -> Bool to [a] -> [a], heck even the bool function (which models an if statement) can be considered a lift from a -> a to Bool -> a. And if they are not, then why is ap from the Applicative type class considered a lift?

如果重要的事情从... a ...变为... f a ...，则ap也不适合这种情况:f (a -> b) -> f a -> f b

If the important thing is going from ... a ... to ... f a ..., then ap wouldn't fit the case either: f (a -> b) -> f a -> f b

推荐答案

我很惊讶没有人回答这个问题.

I'm surprised no one has answered this already.

提升功能的作用是将功能提升到上下文中(通常是Functor或Monad).因此，将类型a -> b的函数放到List上下文中将导致类型为List[a] -> List[b]的函数.如果您考虑一下，这正是map(或Haskell中的fmap)所做的事情.实际上，它是Functor定义的一部分.

A lifting function's role is to lift a function into a context (typically a Functor or Monad). So lifting a function of type a -> b into a List context would result in a function of type List[a] -> List[b]. If you think about it this is exactly what map (or fmap in Haskell) does. In fact, it is part of the definition of a Functor.

但是，函子只能提升一个参数的功能.我们还希望能够处理其他Arities的功能.例如，如果我们具有类型a -> b -> c的函数，则不能使用map.这是图片中更一般的提升操作的地方.在Haskell中，对于这种情况，我们有一个lift2:

However, a Functor can only lift functions of one argument. We also want to be able to deal with functions of other arities as well. For example if we have a function of type a -> b -> c we cannot use map. This is where a more general lifting operation comes into the picture. In Haskell we have a lift2 for this case:

lift2:: (a -> b -> c) -> (M[a] -> M[b] -> M[c])

其中M[a]是使用给定类型a参数化的某些特定Monad(如List).

where M[a] is some particular Monad (like List) parameterized with a given type a.

还有lift的其他变体，也为其他工具定义了.

There are additional variants of lift defined as well for other arities.

这也是为什么filter不是提升函数的原因，因为它不符合所需的类型签名；您没有将a -> bool类型的功能提升为M[a] -> M[bool].但是，它是一个高阶函数.

This is also why filter is not a lifting function as it doesn't fit the type signature required; you are not lifting a function of type a -> bool to M[a] -> M[bool]. It is, however, a higher-ordered function.

如果您想了解有关解除Haskell Wiki的更多信息，请参见上面的好文章

If you want to read more about lifting the Haskell Wiki has a good article on it

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