Re: [Haskell-cafe] How to fold on types?
Try folding over data type constructor with $?
вторник, 25 декабря 2012 г. пользователь Magicloud Magiclouds писал:
Forgot to mention, solution without TemplateHaskell.
On Tue, Dec 25, 2012 at 4:59 PM, Magicloud Magiclouds
magicloud.magiclo...@gmail.com javascript:_e({}, 'cvml',
'magicloud.magiclo...@gmail.com'); wrote:
Say I have things like:
data LongDec = LongDef a b c ... x y z
values = [ 'a', 'b', 'c', ... 'x', 'y', 'z' ]
Now I want them to be LongDef 'a' 'b' 'c' ... 'x' 'y' 'z'.
In form, this is something like folding. But since the type changes, so
code like following won't work:
foldl (\def value - def value) LongDef values
Is it possible to do this in some way?
--
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山高哪阻野云飞
And for G+, please use magiclouds#gmail.com.
--
竹密岂妨流水过
山高哪阻野云飞
And for G+, please use magiclouds#gmail.com.
--
Best
Timur DeTeam Amirov
Moscow, Russia
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Re: [Haskell-cafe] How to fold on types?
Thinking from subway (: foldl ($) LongDef values ? вторник, 25 декабря 2012 г. пользователь Тимур Амиров писал: Try folding over data type constructor with $? вторник, 25 декабря 2012 г. пользователь Magicloud Magiclouds писал: Forgot to mention, solution without TemplateHaskell. On Tue, Dec 25, 2012 at 4:59 PM, Magicloud Magiclouds magicloud.magiclo...@gmail.com wrote: Say I have things like: data LongDec = LongDef a b c ... x y z values = [ 'a', 'b', 'c', ... 'x', 'y', 'z' ] Now I want them to be LongDef 'a' 'b' 'c' ... 'x' 'y' 'z'. In form, this is something like folding. But since the type changes, so code like following won't work: foldl (\def value - def value) LongDef values Is it possible to do this in some way? -- 竹密岂妨流水过 山高哪阻野云飞 And for G+, please use magiclouds#gmail.com. -- 竹密岂妨流水过 山高哪阻野云飞 And for G+, please use magiclouds#gmail.com. -- Best Timur DeTeam Amirov Moscow, Russia -- Best Timur DeTeam Amirov Moscow, Russia ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] How to fold on types?
Magiclouds asked how to build values of data types with many
components from a list of components. For example, suppose we have
data D3 = D3 Int Int Int deriving Show
v3 = [1::Int,2,3]
How can we build the value D3 1 2 3 using the list v3 as the source
for D3's fields? We can't use (foldl ($) D3 values) since the type
changes throughout the iteration: D3 and D3 1 have different type.
The enclosed code shows the solution. It defines the function fcurry
such that
t1 = fcurry D3 v3
-- D3 1 2 3
gives the expected result (D3 1 2 3).
The code is the instance of the general folding over heterogeneous
lists, search for HFoldr in
http://code.haskell.org/HList/Data/HList/HList.hs
{-# LANGUAGE MultiParamTypeClasses, FlexibleInstances, FlexibleContexts #-}
{-# LANGUAGE TypeFamilies, DataKinds, PolyKinds, ScopedTypeVariables #-}
{-# LANGUAGE UndecidableInstances #-}
-- `Folding' over the data type: creating values of data types
-- with many components from a list of components
-- UndecidableInstances is a bit surprising since everything is decidable,
-- but GHC can't see it.
-- Extensions DataKinds, PolyKinds aren't strictly needed, but
-- they make the code a bit nicer. If we already have them,
-- why suffer avoiding them.
module P where
-- The example from MagicCloud's message
data D3 = D3 Int Int Int deriving Show
v3 = [1::Int,2,3]
type family IsArrow a :: Bool
type instance IsArrow (a-b) = True
type instance IsArrow D3 = False
-- add more instances as needed for other non-arrow types
data Proxy a = Proxy
class FarCurry a r t where
fcurry :: (a-t) - [a] - r
instance ((IsArrow t) ~ f, FarCurry' f a r t) = FarCurry a r t where
fcurry = fcurry' (Proxy::Proxy f)
class FarCurry' f a r t where
fcurry' :: Proxy f - (a-t) - [a] - r
instance r ~ r' = FarCurry' False a r' r where
fcurry' _ cons (x:_) = cons x
instance FarCurry a r t = FarCurry' True a r (a-t) where
fcurry' _ cons (x:t) = fcurry (cons x) t
-- Example
t1 = fcurry D3 v3
-- D3 1 2 3
-- Let's add another data type
data D4 = D4 Int Int Int Int deriving Show
type instance IsArrow D4 = False
t2 = fcurry D4 [1::Int,2,3,4]
-- D4 1 2 3 4
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Re: [Haskell-cafe] How to fold on types?
{-# LANGUAGE DeriveDataTypeable, ScopedTypeVariables #-}
Hi MagicCloud,
A worse, but perhaps simpler alternative to Oleg's solution uses Data.Dynamic:
import Data.Dynamic
data LongDec a = LongDec a a a a a a a a
deriving (Show, Typeable)
values = abcdefgh
mkLongDec :: forall a. Typeable a = [a] - Maybe (LongDec a)
mkLongDec = (fromDynamic =) .
foldl
(\f x - do
f' - f
dynApply f' (toDyn x))
(Just (toDyn (\x - LongDec (x :: a
main = do
print (mkLongDec values)
print (mkLongDec [1 .. 8 :: Integer])
*Main main
Just (LongDec 'a' 'b' 'c' 'd' 'e' 'f' 'g' 'h')
Just (LongDec 1 2 3 4 5 6 7 8)
There is no check that all arguments of LongDec are the same
type (in this case a specific instance of Typeable): you'd only
be able to get Nothing out of mkLongDec was defined as:
data LongDec a = LongDec a Int a a a Char
Regards,
Adam Vogt
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Re: [Haskell-cafe] How to fold on types?
On 12/25/2012 09:59 AM, Magicloud Magiclouds wrote:
Say I have things like:
data LongDec = LongDef a b c ... x y z
values = [ 'a', 'b', 'c', ... 'x', 'y', 'z' ]
Now I want them to be LongDef 'a' 'b' 'c' ... 'x' 'y' 'z'.
In form, this is something like folding. But since the type changes, so
code like following won't work:
foldl (\def value - def value) LongDef values
Is it possible to do this in some way?
--
竹密岂妨流水过
山高哪阻野云飞
And for G+, please use magiclouds#gmail.com http://gmail.com/.
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This hack works, in case that helps:
{-# LANGUAGE FlexibleInstances, MultiParamTypeClasses #-}
data LongDec = LongDef Char Char Char Char Char Char
deriving Show
values = [ 'a', 'b', 'c', 'x', 'y', 'z' ]
class Apply a b c where
apply :: b - [a] - c
instance Apply a b b where
apply = const
instance (Apply a b c) = Apply a (a - b) c where
apply f (x:xs) = apply (f x) xs
main = print (apply LongDef values :: LongDec)
It requires an explicit type annotation to fix type parameter 'c'. It
cannot be a function type. (I am not sure why though.)
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Re: [Haskell-cafe] How to fold on types?
You guys are great! Thanks.
On Wed, Dec 26, 2012 at 9:04 AM, Timon Gehr timon.g...@gmx.ch wrote:
On 12/25/2012 09:59 AM, Magicloud Magiclouds wrote:
Say I have things like:
data LongDec = LongDef a b c ... x y z
values = [ 'a', 'b', 'c', ... 'x', 'y', 'z' ]
Now I want them to be LongDef 'a' 'b' 'c' ... 'x' 'y' 'z'.
In form, this is something like folding. But since the type changes, so
code like following won't work:
foldl (\def value - def value) LongDef values
Is it possible to do this in some way?
--
竹密岂妨流水过
山高哪阻野云飞
And for G+, please use magiclouds#gmail.com http://gmail.com/.
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This hack works, in case that helps:
{-# LANGUAGE FlexibleInstances, MultiParamTypeClasses #-}
data LongDec = LongDef Char Char Char Char Char Char
deriving Show
values = [ 'a', 'b', 'c', 'x', 'y', 'z' ]
class Apply a b c where
apply :: b - [a] - c
instance Apply a b b where
apply = const
instance (Apply a b c) = Apply a (a - b) c where
apply f (x:xs) = apply (f x) xs
main = print (apply LongDef values :: LongDec)
It requires an explicit type annotation to fix type parameter 'c'. It
cannot be a function type. (I am not sure why though.)
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--
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山高哪阻野云飞
And for G+, please use magiclouds#gmail.com.
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