Haskell 单子函数映射
我有以下代码:Haskell 单子函数映射,haskell,monads,Haskell,Monads,我有以下代码: type Mapper a k v = a -> [(k,v)] type Reducer k v = k -> [v] -> [v] mapReduce :: Ord k => Mapper a k v -> Reducer k v -> [a] -> [(k,[v])] mapReduce m r = reduce r . shuffleKeys . concatMap (map listifyVal . m) where l
type Mapper a k v = a -> [(k,v)]
type Reducer k v = k -> [v] -> [v]
mapReduce :: Ord k => Mapper a k v -> Reducer k v -> [a] -> [(k,[v])]
mapReduce m r = reduce r . shuffleKeys . concatMap (map listifyVal . m)
where listifyVal (k,v) = (k,[v])
shuffleKeys = Map.fromListWith (++)
reduce r = Map.toList . Map.mapWithKey r
我得到了一个Monad
类型
type MapperM m a k v = a -> m [(k,v)]
type ReducerM m k v = k -> [v] -> m [v]
我需要将现有代码转换为使用Monad
mapReduceM :: (Ord k, Monad m) => MapperM m a k v -> ReducerM m k v -> [a] -> m [(k,[v])]
我一直在转换表达式concatMap(map listifyVal.m)
我编写了这个helper函数
listifyValM:: (Ord k, Monad m) => m(k,v) -> m(k,[v])
listifyValM mkv = do
(k,v) <- mkv
return (k,[v])
但是我甚至不能让这个(非常部分的)简单的事情作为一个开始。
我得到:
我可能在
monad
上缺少一些关于映射的基本信息 从头开始编写mapReduceM
可能比尝试机械地转换mapReduce
更容易。你想要:
mapReduceM :: (Monad m, Ord k) => MapperM m a k v -> ReducerM m k v -> [a] -> m [(k,[v])]
mapReduceM m r as = ...
因此,让我们假设以下类型的参数并尝试构建函数:
m :: a -> m [(k,v)]
r :: k -> [v] -> m [v]
as :: [a]
combineByKey :: [(K, V)] -> Map K [V]
combineByKey = Map.fromListWith (++) . map (\(k,v) -> (k,[v]))
创建一个使用具体类型的骨架文件会很有帮助,这样我们就可以在GHCi中键入检查表达式:
import Data.Map (Map)
import qualified Data.Map as Map
data A
data K = K deriving (Eq, Ord)
data V
data M a
instance Functor M
instance Applicative M
instance Monad M
m :: A -> M [(K,V)]
m = undefined
r :: K -> [V] -> M [V]
r = undefined
as :: [A]
as = undefined
显然,我们需要将的元素作为
传递给唯一可以接受它们的可用函数,即m
。将一元函数a->mb
应用于列表[a]
的标准方法是使用mapM
或遍历执行遍历。(在过去,前者用于monad,后者用于applicative,但现在所有monad都是applicative,traverse
是首选。)具体来说,加载此骨架后,我们可以在GHCi中键入检查此表达式:
> :t traverse m as
traverse m as :: M [[(K, V)]]
要折叠列表列表,我们需要在“monad”下应用concat
。幸运的是,单子是函子,因此fmap
(或其同义词()
)可以做到:
> :t concat <$> traverse m as
concat <$> traverse m as :: M [(K, V)]
在单子下:
> :t combineByKey . concat <$> traverse m as
combineByKey . concat <$> traverse m as :: M (Map K [V])
> Map.toList <$> (sequence $ Map.mapWithKey r mymap)
Map.toList <$> (sequence $ Map.mapWithKey r mymap) :: M [(K, [V])]
在这里,do
notation可以帮助我们完成这个过程:
do mymap <- combineByKey . concat <$> traverse m as
...
如果我们使用mapWithKey
应用减速器:
> Map.mapWithKey r mymap
Map.mapWithKey r mymap :: Map K (M [V])
我们有一个一元元素的结构。因为映射
是可遍历的,所以我们可以使用序列
将monad拉到外部:
> sequence $ Map.mapWithKey r mymap
sequence $ Map.mapWithKey r mymap :: M (Map K [V])
这几乎就是我们希望从mapReduceM
得到的返回值。我们只需将地图更改为单子下的列表:
> :t combineByKey . concat <$> traverse m as
combineByKey . concat <$> traverse m as :: M (Map K [V])
> Map.toList <$> (sequence $ Map.mapWithKey r mymap)
Map.toList <$> (sequence $ Map.mapWithKey r mymap) :: M [(K, [V])]
可以将其转换为无点形式,如下所示:
import Control.Monad
mapReduceM :: (Monad m, Ord k) => MapperM m a k v -> ReducerM m k v -> [a] -> m [(k,[v])]
mapReduceM m r =
traverse m >=> -- apply the mapper
pure . combineByKey . concat >=> -- combine keys
sequence . Map.mapWithKey r >=> -- reduce
pure . Map.toList -- convert to list
where combineByKey :: (Ord k) => [(k, v)] -> Map k [v]
combineByKey = Map.fromListWith (++) . map (\(k,v) -> (k,[v]))
从头开始编写mapReduceM
可能比尝试机械地转换mapReduce
更容易。你想要:
mapReduceM :: (Monad m, Ord k) => MapperM m a k v -> ReducerM m k v -> [a] -> m [(k,[v])]
mapReduceM m r as = ...
因此,让我们假设以下类型的参数并尝试构建函数:
m :: a -> m [(k,v)]
r :: k -> [v] -> m [v]
as :: [a]
combineByKey :: [(K, V)] -> Map K [V]
combineByKey = Map.fromListWith (++) . map (\(k,v) -> (k,[v]))
创建一个使用具体类型的骨架文件会很有帮助,这样我们就可以在GHCi中键入检查表达式:
import Data.Map (Map)
import qualified Data.Map as Map
data A
data K = K deriving (Eq, Ord)
data V
data M a
instance Functor M
instance Applicative M
instance Monad M
m :: A -> M [(K,V)]
m = undefined
r :: K -> [V] -> M [V]
r = undefined
as :: [A]
as = undefined
显然,我们需要将的元素作为
传递给唯一可以接受它们的可用函数,即m
。将一元函数a->mb
应用于列表[a]
的标准方法是使用mapM
或遍历执行遍历。(在过去,前者用于monad,后者用于applicative,但现在所有monad都是applicative,traverse
是首选。)具体来说,加载此骨架后,我们可以在GHCi中键入检查此表达式:
> :t traverse m as
traverse m as :: M [[(K, V)]]
要折叠列表列表,我们需要在“monad”下应用concat
。幸运的是,单子是函子,因此fmap
(或其同义词()
)可以做到:
> :t concat <$> traverse m as
concat <$> traverse m as :: M [(K, V)]
在单子下:
> :t combineByKey . concat <$> traverse m as
combineByKey . concat <$> traverse m as :: M (Map K [V])
> Map.toList <$> (sequence $ Map.mapWithKey r mymap)
Map.toList <$> (sequence $ Map.mapWithKey r mymap) :: M [(K, [V])]
在这里,do
notation可以帮助我们完成这个过程:
do mymap <- combineByKey . concat <$> traverse m as
...
如果我们使用mapWithKey
应用减速器:
> Map.mapWithKey r mymap
Map.mapWithKey r mymap :: Map K (M [V])
我们有一个一元元素的结构。因为映射
是可遍历的,所以我们可以使用序列
将monad拉到外部:
> sequence $ Map.mapWithKey r mymap
sequence $ Map.mapWithKey r mymap :: M (Map K [V])
这几乎就是我们希望从mapReduceM
得到的返回值。我们只需将地图更改为单子下的列表:
> :t combineByKey . concat <$> traverse m as
combineByKey . concat <$> traverse m as :: M (Map K [V])
> Map.toList <$> (sequence $ Map.mapWithKey r mymap)
Map.toList <$> (sequence $ Map.mapWithKey r mymap) :: M [(K, [V])]
可以将其转换为无点形式,如下所示:
import Control.Monad
mapReduceM :: (Monad m, Ord k) => MapperM m a k v -> ReducerM m k v -> [a] -> m [(k,[v])]
mapReduceM m r =
traverse m >=> -- apply the mapper
pure . combineByKey . concat >=> -- combine keys
sequence . Map.mapWithKey r >=> -- reduce
pure . Map.toList -- convert to list
where combineByKey :: (Ord k) => [(k, v)] -> Map k [v]
combineByKey = Map.fromListWith (++) . map (\(k,v) -> (k,[v]))
您编写了map listifyValM
,但由于您没有传递要映射的列表,因此它不知道需要“创建”问题函数的m
、k
和v
。它与Monad本身无关。谁能给我一个如何处理这个问题的提示吗?对于您遇到的错误,只需给您的helper函数一个带有类约束的类型签名:step0::(Monad m,Ord k)=>[m(k,v)]->[m(k,[v])
您编写的map listifyValM
,但是,由于您没有将列表传递给映射,因此它不知道什么是m
、k
和v
,它需要“创建”出问题的函数。它与Monad本身无关。有谁能给我一个如何处理这个问题的提示吗?对于您遇到的错误,只需给您的helper函数一个带有类约束的类型签名:step0::(Monad m,Ord k)=>[m(k,v)]->[m(k,[v])]