Haskell Monad-列表中的Monad如何工作? [英] Haskell Monad - How does Monad on list work?

问题描述

为了理解Monad，我提出了以下定义:

In order to understand Monad, I came up with the following definitions:

class Applicative' f where
 purea :: a -> f a
 app :: f (a->b) -> f a -> f b

class Applicative' m =>  Monadd m where
 (>>|) :: m a -> (a -> m b) -> m b

instance Applicative' [] where
 purea x = [x]
 app gs xs = [g x | g <- gs, x <- xs]

instance Monadd [] where
 (>>|) xs f = [ y | x <-xs, y <- f x]

它按预期工作:

(>>|) [1,2,3,4] (\x->[(x+1)])
[2,3,4,5]

我不确定它是如何工作的. 例如:

I am not sure how it is working though. For example:

[ y | y <- [[1],[2]]]
[[1],[2]]

对[1,2,3]的每个列表元素的应用程序(\x->([x+1])如何导致[2,3,4]而不是[[2],[3],[4]]

How does application (\x->([x+1]) to each list element of [1,2,3] result in [2,3,4] and not [[2],[3],[4]]

或者很简单地，我的困惑似乎是由于不了解该语句[ y | x <-xs, y <- f x]的实际工作方式

Or quite simply my confusion seems to stem from not understanding how this statement [ y | x <-xs, y <- f x] actually works

推荐答案

LYAH ， Quora 以及更多描述单子的列表.

Wadler, School of Haskell, LYAH, HaskellWiki, Quora and many more describe the list monad.

比较:

• (=<<) :: Monad m => (a -> m b) -> m a -> m b用于具有
的列表
• concatMap :: (a -> [b]) -> [a] -> [b] 用于m = [].

• (=<<) :: Monad m => (a -> m b) -> m a -> m b for lists with
• concatMap :: (a -> [b]) -> [a] -> [b] for m = [].

常规的(>>=)绑定运算符的参数已翻转，但否则只是一个后缀concatMap.

The regular (>>=) bind operator has the arguments flipped, but is otherwise just an infix concatMap.

或者很简单地，我的困惑似乎是由于不了解该语句的实际工作原理而产生的:

Or quite simply my confusion seems to stem from not understanding how this statement actually works:

(>>|) xs f = [ y | x <- xs, y <- f x ]

由于列表推导与列表的Monad实例等效，因此此定义是一种作弊行为.基本上，您说的是某种东西，就像它是Monad一样，是Monadd，所以剩下两个问题:了解列表推导和仍然了解Monad.

Since list comprehensions are equivalent to the Monad instance for lists, this definition is kind of cheating. You're basically saying that something is a Monadd in the way that it's a Monad, so you're left with two problems: Understanding list comprehensions, and still understanding Monad.

可以理解列表的理解力，以便更好地理解:

List comprehensions can be de-sugared for a better understanding:

• 删除语法糖:Haskell中的列表理解

对于您而言，该语句可以用多种其他方式编写:

In your case, the statement could be written in a number of other ways:

• 使用符号:

• Using do-notation:

(>>|) xs f = do x <- xs
                y <- f x
                return y

建议使用(>>=)运算符:

De-sugared into using the (>>=) operator:

(>>|) xs f = xs >>= \x ->
             f x >>= \y ->
             return y

这可以缩短(每行重写一次):

This can be shortened (one rewrite per line):

  (>>|) xs f = xs >>= \x -> f x >>= \y -> return y -- eta-reduction
≡ (>>|) xs f = xs >>= \x -> f x >>= return         -- monad identity
≡ (>>|) xs f = xs >>= \x -> f x                    -- eta-reduction
≡ (>>|) xs f = xs >>= f                            -- prefix operator
≡ (>>|) xs f = (>>=) xs f                          -- point-free
≡ (>>|) = (>>=)

因此，通过使用列表推导，您尚未真正声明一个新的定义，而是仅依赖现有的定义.如果需要，您可以定义instance Monadd []而不依赖现有的Monad实例或列表理解:

So from using list comprehensions, you haven't really declared a new definition, you're just relying on the existing one. If you wanted, you could instead define your instance Monadd [] without relying on existing Monad instances or list comprehensions:

• 使用concatMap:

instance Monadd [] where
  (>>|) xs f = concatMap f xs

进一步说明:

Spelling that out a little more:

instance Monadd [] where
  (>>|) xs f = concat (map f xs)

进一步说明:

Spelling that out even more:

instance Monadd [] where
  (>>|) [] f = []
  (>>|) (x:xs) f = let ys = f x in ys ++ ((>>|) xs f)

Monadd类型类应具有与return类似的内容.我不确定为什么会丢失它.

The Monadd type class should have something similar to return. I'm not sure why it's missing.

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