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question on typeclasses and applicatives
by Alec Benzer 03 Sep '10

03 Sep '10

I came up with some code when trying to understand applicatives: import Control.Applicative import qualified Data.Map as M instance Applicative (M.Map String) where pure x = M.fromList [("",x)] fs <*> xs = M.fromList [(k1 ++ " " ++ k2,v1 v2) | k1 <- M.keys fs, k2 <- M.keys xs, v1 <- M.elems fs, v2 <- M.elems xs] 1. When I :load this in ghci it gives me some error about using (M.Map String) here, and tells me it'll work if I use the -XFlexibleInstances flag. Why is this type of behavior disabled by default? Is it potentially dangerous in some way? 2. When running the following: fromList [("double",(*2))] <*> fromList[("two",2),("seven",7)] I get: fromList [("double seven",4),("double two",4)] instead of what I'd expect: fromList [("double seven",14),("double two",4)] Although this: (*2) <$> fromList[("two",2),("seven",7)] gives what I'd expect: fromList [("seven",14),("two",4)] Why is this happening? I can't seem to figure it out.

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How does one sort with Data.Vector.Generic.Mutable?
by max smith 03 Sep '10

03 Sep '10

Hi, I have been trying to use Haskell to sort a long list of data (strings, floats etc) that is read from a large file (say a few million lines). I've been trying to use a Data.Vector.Generic.Mutable object and a sort algorithm from Data.Vector.Algorithms but I've been unable to get it to work since I haven't been able to find or create my own examples. The examples in the excellent Vector tutorial, cover filling a vector with random numbers but due to my limited understanding of monads, I can't figure out how to use the IO monad and fill a vector from a file. Does anyone have an example of how to do this. Also, is Data.Vector.Generic.Mutable the best way to read a large amount of data from a file and sort it? Thanks for any help, Max

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Ranges and List Comprehensions in SQL
by Tom Murphy 03 Sep '10

03 Sep '10

Hi Everyone, Is there a "From SQL"-type function that "restores" a range or list comprehension? Example: Let's say I want to keep track of which episodes of a TV show I've seen. I have an SQL table, in which is a record: id (INTEGER): 30 title (VARCHAR): "The Simpsons" episodes_watched (Some data format): [1..4], [14], [18..21], [23..25] Then, when I pull the record, in Haskell, the "Episodes Watched" is already one list: [1,2,3,4,14,18,19,21,23,24,25] , or a series of lists that I can append together: [1,2,3,4], [14], [18,19,20,21], [23,24,25] Note in the example that I would like to be able to store multiple ranges within a single record. Thanks so much for any help! Tom

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Missing or recursive dependencies
by Lyndon Maydwell 03 Sep '10

03 Sep '10

Hi beginners. I'm getting an odd problem quite a lot recently: ghc -fglasgow-exts --make -O2 -W X.hs -v -o X package Cabal-1.8.0.6-ec9be469687b5a514f4b7e8e2b8343c7 is shadowed by package Cabal-1.8.0.6-c995b90190c27e1dfa2ab3d5434fabef package QuickCheck-2.1.1.1-c7435cb0d5b5de72fe9540c48335606d is unusable due to missing or recursive dependencies: ghc-6.12.3-66a382195c8a71849653439b67021fd1 package bin-package-db-0.0.0.0-0dffb74a73bb78b5dc02ca941bbcbea0 is unusable due to missing or recursive dependencies: Cabal-1.8.0.6-ec9be469687b5a514f4b7e8e2b8343c7 package category-extras-0.53.5-ee54cd5c810a6c004a1b7a276d0e0076 is unusable due to missing or recursive dependencies: ghc-6.12.3-66a382195c8a71849653439b67021fd1 package comonad-random-0.1.2-897a5da683d269c3b2db3e03ac43279b is unusable due to missing or recursive dependencies: category-extras-0.53.5-ee54cd5c810a6c004a1b7a276d0e0076 ... I've asked on irc, and the consensus is that something is very wrong, however we weren't able to diagnose the cause. It would probably be worth looking at on-list for future reference. Does anyone have any ideas?

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problems with GADTs
by Michael Litchard 02 Sep '10

02 Sep '10

I'm learnined how to use GADTs, but for the moment I'm stuck on the following problem. some context first > newtype Cost = Cost Int > newtype ID = ID Int > newtype Student = Student String > newtype Activity = Activity String > data Table k v = PopTable > { tableName :: String > , tableList :: [(k, v)] > } deriving (Show, Read) > type Students = Table ID Student > type ActivityCosts = Table Activity Cost > type StudentActivities = Table Activity ID I want to do something like this > loaders = [("students", loadStudentIDs), ("costs", loadActivityCosts), ("activities", loadStudentActivities)] But I can't have list with multiple types. Okay so ski (from #haskell) suggested I use a GADT. This is new territory for me. Here's what it looks like right now. data ProgramName :: * -> * where PNStudents :: ProgramName Students PNCosts :: ProgramName ActivityCosts PNActivities :: ProgramName StudentActivities Now I should be able to use this to define a data structure called "loaders" that contains three tuples ("students", loadStudentIDs), ("costs", loadActivityCosts), and ("activities", loadStudentActivities). So, what would "loaders" look like? I've done some reading but it's not sunk in yet.

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Re: [Haskell-beginners] simplifying algebraic expression
by Alexander.Vladislav.Popov 01 Sep '10

01 Sep '10

It has given me an infinite loop and out-of-memory exception. I tried different variations but not found the valid one and don't know what to do next. 2010/9/1 jean verdier <verdier.jean(a)gmail.com> > I think that you need to reduce inner expressions before simplification. > > s rdc (Add a b) = rdc a `Add` rdc b > > will not reduce the Add result. What you need is something like > > s rdc (Add a b) = rdc (rdc a `Add` rdc b) > > that might not terminate but is your target. > > > On Wed, 2010-09-01 at 15:19 +0600, Alexander.Vladislav.Popov wrote: > > Hi. > > > > Please, tell me where I'm wrong and how to improve my approach. > > > > I'm trying to simplify algebraic expressions this way: > > > > import Data.Ratio > > > > data Func = Const (Ratio Int) > > | Pow (Ratio Int) > > | Add Func Func > > | Mul Func Func > > > > instance Show Func where > > show (Const n) = "(" ++ show n ++ ")" > > show (Pow n) | n == 0 = "1" > > | n == 1 = "x" > > | otherwise = "(x**(" ++ show n ++ "))" > > show (Add t1 t2) ="(" ++ (show t1) ++ "+" ++ (show t2) ++ ")" > > show (Mul t1 t2) ="(" ++ (show t1) ++ "*" ++ (show t2) ++ ")" > > > > > > deriv (Const _) = Const 0 > > deriv (Pow 1) = Const 1 > > deriv (Pow n) = Const n `Mul` Pow (n-1) > > deriv (Add a b) = deriv a `Add` deriv b > > deriv (Mul a b) = Add (deriv a `Mul` b) (a `Mul` deriv b) > > > > p0 = Const 1 > > p1 = p0 `Add` (Mul (Pow 1) (Const 2)) > > p2 = p1 `Add` (Mul (Pow 2) (Const 3)) > > > > > > s rdc (Const x) = Const x > > s rdc (Pow 0) = Const 1 > > s rdc (Pow x) = Pow x > > s rdc (Add (Const a) (Const b)) = Const (a+b) > > s rdc (Mul (Const 0) _) = Const 0 > > s rdc (Mul _ (Const 0)) = Const 0 > > s rdc (Mul (Const a) (Const b)) = Const (a*b) > > s rdc (Mul (Pow n) (Pow m)) = Pow (n+m) > > > > s rdc (Add x (Const 0)) = rdc x > > s rdc (Add (Const 0) x) = rdc x > > s rdc (Mul (Const m) (Mul (Const n) x)) = rdc $ Mul (Const (n*m)) (rdc > > x) > > s rdc (Mul x (Const 1)) = rdc x > > s rdc (Mul x (Const a)) = rdc $ Mul (Const a) (rdc x) > > s rdc (Mul (Const 1) x) = rdc x > > s rdc (Mul x (Add a b)) = Mul (rdc x) (rdc a) `Add` Mul (rdc x) (rdc > > b) > > s rdc (Mul (Add a b) x) = Mul (rdc a) (rdc x) `Add` Mul (rdc b) (rdc > > x) > > s rdc (Mul a b) = rdc a `Mul` rdc b > > s rdc (Add a b) = rdc a `Add` rdc b > > > > fix f = f (fix f) > > > > The result I got is : > > > > *Main> fix s $ deriv p2 > > (((2 % 1)+(0 % 1))+(((6 % 1)*x)+(0 % 1))) > > > > instead of the anticipated expression ((2 % 1)+((6 % 1)*x)). > > And worst of all, I must apply (fix s) repeatedly to achieve correct > > answer: > > > > *Main> fix s $ fix s $ deriv p2 > > ((2 % 1)+((6 % 1)*x)). > > > > I'll be very much appriciated for any help and useful links. > > _______________________________________________ > > Beginners mailing list > > Beginners(a)haskell.org > > http://www.haskell.org/mailman/listinfo/beginners > > >

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simplifying algebraic expression
by Alexander.Vladislav.Popov 01 Sep '10

01 Sep '10

Hi. Please, tell me where I'm wrong and how to improve my approach. I'm trying to simplify algebraic expressions this way: import Data.Ratio data Func = Const (Ratio Int) | Pow (Ratio Int) | Add Func Func | Mul Func Func instance Show Func where show (Const n) = "(" ++ show n ++ ")" show (Pow n) | n == 0 = "1" | n == 1 = "x" | otherwise = "(x**(" ++ show n ++ "))" show (Add t1 t2) ="(" ++ (show t1) ++ "+" ++ (show t2) ++ ")" show (Mul t1 t2) ="(" ++ (show t1) ++ "*" ++ (show t2) ++ ")" deriv (Const _) = Const 0 deriv (Pow 1) = Const 1 deriv (Pow n) = Const n `Mul` Pow (n-1) deriv (Add a b) = deriv a `Add` deriv b deriv (Mul a b) = Add (deriv a `Mul` b) (a `Mul` deriv b) p0 = Const 1 p1 = p0 `Add` (Mul (Pow 1) (Const 2)) p2 = p1 `Add` (Mul (Pow 2) (Const 3)) s rdc (Const x) = Const x s rdc (Pow 0) = Const 1 s rdc (Pow x) = Pow x s rdc (Add (Const a) (Const b)) = Const (a+b) s rdc (Mul (Const 0) _) = Const 0 s rdc (Mul _ (Const 0)) = Const 0 s rdc (Mul (Const a) (Const b)) = Const (a*b) s rdc (Mul (Pow n) (Pow m)) = Pow (n+m) s rdc (Add x (Const 0)) = rdc x s rdc (Add (Const 0) x) = rdc x s rdc (Mul (Const m) (Mul (Const n) x)) = rdc $ Mul (Const (n*m)) (rdc x) s rdc (Mul x (Const 1)) = rdc x s rdc (Mul x (Const a)) = rdc $ Mul (Const a) (rdc x) s rdc (Mul (Const 1) x) = rdc x s rdc (Mul x (Add a b)) = Mul (rdc x) (rdc a) `Add` Mul (rdc x) (rdc b) s rdc (Mul (Add a b) x) = Mul (rdc a) (rdc x) `Add` Mul (rdc b) (rdc x) s rdc (Mul a b) = rdc a `Mul` rdc b s rdc (Add a b) = rdc a `Add` rdc b fix f = f (fix f) The result I got is : *Main> fix s $ deriv p2 (((2 % 1)+(0 % 1))+(((6 % 1)*x)+(0 % 1))) instead of the anticipated expression ((2 % 1)+((6 % 1)*x)). And worst of all, I must apply (fix s) repeatedly to achieve correct answer: *Main> fix s $ fix s $ deriv p2 ((2 % 1)+((6 % 1)*x)). I'll be very much appriciated for any help and useful links.

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confusion about parameterized types.
by Michael Litchard 31 Aug '10

31 Aug '10

Here is a record I defined > module PrepData where > data Value = Cost Int | ID Int > type Tname = String > data StudentsOrActivites = Students String | Activities String > data Table soa v = PopTable > { tableName :: Tname > , tableColumns :: [(soa, v)] > } deriving (Show, Read) I'm trying to define another record in terms of Table > data ProcessData = > ProcessData { flatfile :: String > , processfunc :: String -> what I would like processfunc to be is a function that takes a String and returns a Table. The type variables are tripping me up. Could someone help me work out the syntax, and the clarifying of my intent?

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Results of Bug Report
by David Webster 29 Aug '10

29 Aug '10

I submitted a bug report to Haskell-Platform a few days ago. The ticket was assigned an owner who accepted it. How will I know if the problem has been corrected? David Webster

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instances of different kinds
by Greg 29 Aug '10

29 Aug '10

Hi-- More silly typeclass questions. I'm not sure the right way to ask it, so I'll start with a failed code snippet: data Foo a = Foo a class TwoPi a where div2pi :: (Floating b) => a -> b instance (Floating a) => TwoPi (Foo a) where div2pi (Foo a) = a / (2*pi) instance TwoPi Float where div2pi a = a / (2*pi) This code is obviously meaningless, but I'm trying to figure out how you can create instances of a typeclass for data types of different kinds. I have a similar piece of code that works: data Foo a = Foo a class Testable a where isPos :: a -> Bool instance (Ord b, Num b) => Testable (Foo b) where isPos (Foo b) = b > 0 instance Testable Float where isPos a = a > 0 One obvious difference is that the type of isPos is a -> Bool, with a defined type as the return. I'd rather not commit to a specific Floating type up front (I'd prefer sometimes Float sometimes Double, depending on the 'a' in Foo a, but trying to declare it as Float doesn't help me. This fails: data Foo a = Foo a class TwoPi a where div2pi :: a -> Float instance (Floating b) => TwoPi (Foo b) where div2pi (Foo b) = b / (2*pi) instance TwoPi Float where div2pi a = a / (2*pi) The errors I'm getting are various permutations of: Couldn't match expected type `Float' against inferred type `b' `b' is a rigid type variable bound by the instance declaration at gcbTestBad.hs:8:19 In the expression: b / (2 * pi) In the definition of `div2pi': div2pi (Foo b) = b / (2 * pi) In the instance declaration for `TwoPi (Foo b)' What is the difference between these last two cases ("a -> Bool" and "a -> Float"), and is there anyway to make "a -> b" work? Thanks-- Greg

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