这是一个广泛支持你美丽想法的论点。
第一部分:mapMaybe
我在这里的计划是重述这个问题mapMaybe
,希望这样做能让我们更熟悉。为此,我将使用一些Either
-juggling 实用程序函数:
maybeToRight :: a -> Maybe b -> Either a b
rightToMaybe :: Either a b -> Maybe b
leftToMaybe :: Either a b -> Maybe a
flipEither :: Either a b -> Either b a
(我从relude中取了前三个名字,从errors中取了第四个名字。顺便说一下,errors分别提供maybeToRight
和rightToMaybe
asnote
和hush
in Control.Error.Util
。)
正如您所指出的,mapMaybe
可以定义为partition
:
mapMaybe :: Filterable f => (a -> Maybe b) -> f a -> f b
mapMaybe f = snd . partition . fmap (maybeToRight () . f)
至关重要的是,我们也可以反过来:
partition :: Filterable f => f (Either a b) -> (f a, f b)
partition = mapMaybe leftToMaybe &&& mapMaybe rightToMaybe
这表明根据mapMaybe
. 根据恒等律,这样做给了我们一个很好的借口来完全忘记trivial
:
-- Left and right unit
mapMaybe rightToMaybe . fmap (bwd elunit) = id -- [I]
mapMaybe leftToMaybe . fmap (bwd erunit) = id -- [II]
至于关联性,我们可以使用rightToMaybe
和leftToMaybe
将定律拆分为三个方程,一个对应于我们从连续分区中获得的每个分量:
-- Associativity
mapMaybe rightToMaybe . fmap (bwd eassoc)
= mapMaybe rightToMaybe . mapMaybe rightToMaybe -- [III]
mapMaybe rightToMaybe . mapMaybe leftToMaybe . fmap (bwd eassoc)
= mapMaybe leftToMaybe . mapMaybe rightToMaybe -- [IV]
mapMaybe leftToMaybe . fmap (bwd eassoc)
= mapMaybe leftToMaybe . mapMaybe leftToMaybe -- [V]
参数均值与我们在这里处理mapMaybe
的值无关。Either
既然如此,我们可以使用我们的Either
同构小库来打乱东西,并证明 [I] 等价于 [II],而 [III] 等价于 [V]。我们现在归结为三个方程:
mapMaybe rightToMaybe . fmap (bwd elunit) = id -- [I]
mapMaybe rightToMaybe . fmap (bwd eassoc)
= mapMaybe rightToMaybe . mapMaybe rightToMaybe -- [III]
mapMaybe rightToMaybe . mapMaybe leftToMaybe . fmap (bwd eassoc)
= mapMaybe leftToMaybe . mapMaybe rightToMaybe -- [IV]
参数化允许我们吞下fmap
[I]:
mapMaybe (rightToMaybe . bwd elunit) = id
然而,这简直就是……
mapMaybe Just = id
...这相当于来自witherable的Filterable
守恒定律/恒等律:
mapMaybe (Just . f) = fmap f
这Filterable
也有一个组成定律:
-- The (<=<) is from the Maybe monad.
mapMaybe g . mapMaybe f = mapMaybe (g <=< f)
我们也可以从我们的法律中推导出这一点吗?让我们从 [III] 开始,再一次让参数化发挥作用。这个比较棘手,所以我会完整地写下来:
mapMaybe rightToMaybe . fmap (bwd eassoc)
= mapMaybe rightToMaybe . mapMaybe rightToMaybe -- [III]
-- f :: a -> Maybe b; g :: b -> Maybe c
-- Precomposing fmap (right (maybeToRight () . g) . maybeToRight () . f)
-- on both sides:
mapMaybe rightToMaybe . fmap (bwd eassoc)
. fmap (right (maybeToRight () . g) . maybeToRight () . f)
= mapMaybe rightToMaybe . mapMaybe rightToMaybe
. fmap (right (maybeToRight () . g) . maybeToRight () . f)
mapMaybe rightToMaybe . mapMaybe rightToMaybe
. fmap (right (maybeToRight () . g) . maybeToRight () . f) -- RHS
mapMaybe rightToMaybe . fmap (maybeToRight () . g)
. mapMaybe rightToMaybe . fmap (maybeToRight () . f)
mapMaybe (rightToMaybe . maybeToRight () . g)
. mapMaybe (rightToMaybe . maybeToRight () . f)
mapMaybe g . mapMaybe f
mapMaybe rightToMaybe . fmap (bwd eassoc)
. fmap (right (maybeToRight () . g) . maybeToRight () . f) -- LHS
mapMaybe (rightToMaybe . bwd eassoc
. right (maybeToRight () . g) . maybeToRight () . f)
mapMaybe (rightToMaybe . bwd eassoc
. right (maybeToRight ()) . maybeToRight () . fmap @Maybe g . f)
-- join @Maybe
-- = rightToMaybe . bwd eassoc . right (maybeToRight ()) . maybeToRight ()
mapMaybe (join @Maybe . fmap @Maybe g . f)
mapMaybe (g <=< f) -- mapMaybe (g <=< f) = mapMaybe g . mapMaybe f
在另一个方向:
mapMaybe (g <=< f) = mapMaybe g . mapMaybe f
-- f = rightToMaybe; g = rightToMaybe
mapMaybe (rightToMaybe <=< rightToMaybe)
= mapMaybe rightToMaybe . mapMaybe rightToMaybe
mapMaybe (rightToMaybe <=< rightToMaybe) -- LHS
mapMaybe (join @Maybe . fmap @Maybe rightToMaybe . rightToMaybe)
-- join @Maybe
-- = rightToMaybe . bwd eassoc . right (maybeToRight ()) . maybeToRight ()
mapMaybe (rightToMaybe . bwd eassoc
. right (maybeToRight ()) . maybeToRight ()
. fmap @Maybe rightToMaybe . rightToMaybe)
mapMaybe (rightToMaybe . bwd eassoc
. right (maybeToRight () . rightToMaybe)
. maybeToRight () . rightToMaybe)
mapMaybe (rightToMaybe . bwd eassoc) -- See note below.
mapMaybe rightToMaybe . fmap (bwd eassoc)
-- mapMaybe rightToMaybe . fmap (bwd eassoc)
-- = mapMaybe rightToMaybe . mapMaybe rightToMaybe
(注意:虽然maybeToRight () . rightToMaybe :: Either a b -> Either () b
is not id
,但在推导中左侧的值无论如何都会被丢弃,因此将其剔除为 是公平的id
。)
因此 [III] 等价于witherable 's的组成定律Filterable
。
此时,我们可以使用合成法则来处理[IV]:
mapMaybe rightToMaybe . mapMaybe leftToMaybe . fmap (bwd eassoc)
= mapMaybe leftToMaybe . mapMaybe rightToMaybe -- [IV]
mapMaybe (rightToMaybe <=< leftToMaybe) . fmap (bwd eassoc)
= mapMaybe (letfToMaybe <=< rightToMaybe)
mapMaybe (rightToMaybe <=< leftToMaybe . bwd eassoc)
= mapMaybe (letfToMaybe <=< rightToMaybe)
-- Sufficient condition:
rightToMaybe <=< leftToMaybe . bwd eassoc = letfToMaybe <=< rightToMaybe
-- The condition holds, as can be directly verified by substiuting the definitions.
这足以表明您的课程相当于一个完善的公式Filterable
,这是一个非常好的结果。以下是对法律的回顾:
mapMaybe Just = id -- Identity
mapMaybe g . mapMaybe f = mapMaybe (g <=< f) -- Composition
正如枯萎的文档所指出的,这些是从Kleisli Maybe到Hask的函子的函子定律。
第二部分:Alternative 和 Monad
现在我们可以解决您的实际问题,这是关于替代单子的。您建议的实施partition
是:
partitionAM :: (Alternative f, Monad f) => f (Either a b) -> (f a, f b)
partitionAM
= (either return (const empty) =<<) &&& (either (const empty) return =<<)
按照我更广泛的计划,我将切换到mapMaybe
演示文稿:
mapMaybe f
snd . partition . fmap (maybeToRight () . f)
snd . (either return (const empty) =<<) &&& (either (const empty) return =<<)
. fmap (maybeToRight () . f)
(either (const empty) return =<<) . fmap (maybeToRight () . f)
(either (const empty) return . maybeToRight . f =<<)
(maybe empty return . f =<<)
所以我们可以定义:
mapMaybeAM :: (Alternative f, Monad f) => (a -> Maybe b) -> f a -> f b
mapMaybeAM f u = maybe empty return . f =<< u
或者,在无点拼写中:
mapMaybeAM = (=<<) . (maybe empty return .)
在上面的几段中,我注意到Filterable
定律说这mapMaybe
是从Kleisli Maybe到Hask的函子的态射映射。由于函子的组合是函子,并且(=<<)
是从Kleisli f到Hask的(maybe empty return .)
函子的态射映射,因此作为从Kleisli Maybe到Kleisli f的函子的态射映射就足够mapMaybeAM
合法了。相关的函子定律是:
maybe empty return . Just = return -- Identity
maybe empty return . g <=< maybe empty return . f
= maybe empty return . (g <=< f) -- Composition
这个恒等律成立,所以让我们关注组合一:
maybe empty return . g <=< maybe empty return . f
= maybe empty return . (g <=< f)
maybe empty return . g =<< maybe empty return (f a)
= maybe empty return (g =<< f a)
-- Case 1: f a = Nothing
maybe empty return . g =<< maybe empty return Nothing
= maybe empty return (g =<< Nothing)
maybe empty return . g =<< empty = maybe empty return Nothing
maybe empty return . g =<< empty = empty -- To be continued.
-- Case 2: f a = Just b
maybe empty return . g =<< maybe empty return (Just b)
= maybe empty return (g =<< Just b)
maybe empty return . g =<< return b = maybe empty return (g b)
maybe empty return (g b) = maybe empty return (g b) -- OK.
因此,mapMaybeAM
对于maybe empty return . g =<< empty = empty
任何g
. 现在,如果empty
定义为absurd <$> nil ()
,正如您在此处所做的那样,我们可以证明f =<< empty = empty
对于任何f
:
f =<< empty = empty
f =<< empty -- LHS
f =<< absurd <$> nil ()
f . absurd =<< nil ()
-- By parametricity, f . absurd = absurd, for any f.
absurd =<< nil ()
return . absurd =<< nil ()
absurd <$> nil ()
empty -- LHS = RHS
直观地说,如果empty
真的是空的(因为它必须是,给定我们在这里使用的定义),将没有f
要应用的值,因此除了 .f =<< empty
之外不会产生任何结果empty
。
这里的另一种方法是研究Alternative
和Monad
类的交互。碰巧的是,有一个替代 monad 的类:MonadPlus
. 因此,重新设计的mapMaybe
样式可能如下所示:
-- Lawful iff, for any f, mzero >>= maybe empty mzero . f = mzero
mmapMaybe :: MonadPlus m => (a -> Maybe b) -> m a -> m b
mmapMaybe f m = m >>= maybe mzero return . f
虽然对于哪一套法律最适合有不同的意见MonadPlus
,但似乎没有人反对的法律之一是……
mzero >>= f = mzero -- Left zero
...这正是empty
我们在上面讨论的几段的属性。的合法性mmapMaybe
紧随左零定律。
(顺便提一下,Control.Monad
提供mfilter :: MonadPlus m => (a -> Bool) -> m a -> m a
,它与filter
我们可以使用定义的匹配mmapMaybe
。)
总之:
但是这种实施总是合法的吗?有时是否合法(对于“有时”的一些正式定义)?
是的,执行是合法的。这个结论取决于empty
确实是空的,因为它应该是空的,或者取决于左零MonadPlus
定律的相关替代单子,这归结为几乎相同的事情。
值得强调的Filterable
是,它不包含在MonadPlus
中,我们可以用以下反例来说明:
ZipList
: 可过滤,但不是单子。该Filterable
实例与列表的实例相同,即使该实例Alternative
不同。
Map
:可过滤,但既不是单子也不是应用程序。事实上,Map
甚至不能应用,因为没有合理的实现pure
. 但是,它确实有自己的empty
.
MaybeT f
:虽然它的Monad
和Alternative
实例需要f
是一个单子,并且一个独立的empty
定义至少需要Applicative
,但Filterable
实例只需要Functor f
(如果你将Maybe
一层滑入其中,任何东西都变得可过滤)。
第三部分:空
在这一点上,人们可能仍然想知道一个角色有多大empty
,或者nil
说,真的扮演了多大的角色Filterable
。它不是类方法,但大多数实例似乎都有它的合理版本。
我们可以确定的一件事是,如果可过滤类型有任何居民,其中至少有一个是空结构,因为我们总是可以取出任何居民并将所有内容过滤掉:
chop :: Filterable f => f a -> f Void
chop = mapMaybe (const Nothing)
, 的存在chop
并不意味着会有一个空 nil
值,或者chop
总是会给出相同的结果。例如,考虑 ,MaybeT IO
其Filterable
实例可能被认为是审查计算结果的一种方式IO
。该实例是完全合法的,即使chop
可以产生带有任意效果的不同MaybeT IO Void
值。IO
最后一点,您已经提到了使用强幺半群函子的可能性,因此通过制作/和/同构将Alternative
和联系起来。具有和作为互逆是可以想象的,但相当有限,因为它丢弃了关于大部分实例的元素排列的一些信息。至于另一个同构,是微不足道的,但有趣的是它意味着只有一个值,适用于相当大份额的实例。碰巧有这种情况的一个版本。如果我们要求,对于任何...Filterable
union
partition
nil
trivial
union
partition
union . partition
trivial . nil
nil . trivial
f Void
Filterable
MonadPlus
u
absurd <$> chop u = mzero
...然后替换第二mmapMaybe
部分,我们得到:
absurd <$> chop u = mzero
absurd <$> mmapMaybe (const Nothing) u = mzero
mmapMaybe (fmap absurd . const Nothing) u = mzero
mmapMaybe (const Nothing) u = mzero
u >>= maybe mzero return . const Nothing = mzero
u >>= const mzero = mzero
u >> mzero = mzero
此属性被称为 的右零定律MonadPlus
,尽管有充分的理由质疑其作为该特定类别的定律的地位。