- Beginners - Haskell.org

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 02 Sep '10

02 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 30 Aug '10

30 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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Changing evalList to parList creates a space leak?
by Travis Erdman 29 Aug '10

29 Aug '10

Ok, since I last visited, as a result of the help from contributors on this list, I've reached a nice spot. The serial implementation of my processing algorithm works quite quickly (including the tree pruning that I mentioned last time) and the space leaks have been eliminated without resort to the dreadful foldl'rnf. Here are simplified versions of key components of my program: data Tree a = Node a [Tree a] calcTree :: Tree MyData -> InputData -> Tree MyData calcTree (Node cargo subtrees) input = Node (process cargo input) recurseChildren where process a b = .... recurseChildren = map (\t -> calcTree t input) subtrees `using` evalList rseq newTree = foldl' calcTree starttree (take n inputdatalist) The <`using` evalList rseq> term at the end of recurseChildren was key to solving a space leak without resorting to foldl'rnf (which is a performance killer). This is also the place in my code where it makes the most sense to introduce some parallelism (ie to traverse the subtrees in parallel). However, when I simply change the evalList to a parList, a space leak is created anew, which I didn't anticipate at all. (of course, I also made the change of compiling with -threaded and -feager-blackholing flags ... but if I compile with those flags and leave it as evalList, there is no space leak). I'm using parallel-3.1.0.0, so it's not the space leak from parallel-1.1.x. Any ideas why this simple change should introduce a space leak? How might I get to where I want to go? thanks again, Travis

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randomize the order of a list
by A Smith 28 Aug '10

28 Aug '10

I've abeen recommended on good authority(my son, a Cambridge Pure maths graduate, and Perl/Haskell expert) , and backed by a Google search that Fisher-Yates shuffle is the one to use, as it produces total unbiased results with every combination equally possible. As with most things with computers,don't reinvent the eheel, it's almost certainly been done before by someone brighter that you, Fisher,Yates. & Knuth! -- Andrew Smith B.Sc(Hons),MBA Edinburgh,Scotland On 27 August 2010 21:02, Gaius Hammond <gaius(a)gaius.org.uk> wrote: > Hi all, > > > > I am trying to randomly reorder a list (e.g. shuffle a deck of cards) . My > initial approach is to treat it as an array, generate a list of unique > random numbers between 0 and n - 1, then use those numbers as new indexes. I > am using a function to generate random numbers in the State monad as > follows: > > > > randInt∷ Int → State StdGen Int > randInt x = do g ← get > (v,g') ← return $ randomR (0, x) g > put g' > return v > > > > This is pretty much straight from the documentation. My function for the > new indexes is: > > > > -- return a list of numbers 0 to x-1 in random order > randIndex∷ Int → StdGen → ([Int], StdGen) > randIndex x = runState $ do > let randIndex' acc r > | (length acc ≡ x) = acc > | (r `elem` acc) ∨ (r ≡ (−1)) = do > r' ← randInt (x − 1) > randIndex' acc r' > | otherwise = do > r' ← randInt (x − 1) > randIndex' r:acc r' > in > randIndex' [] (−1) > > > > This fails to compile on > > > > > Couldn't match expected type `[a]' > against inferred type `State StdGen b' > In a stmt of a 'do' expression: r' <- randInt (x - 1) > In the expression: > do { r' <- randInt (x - 1); > randIndex' acc r' } > > > > > I can see what's happening here - it's treating randIndex' as the second > argument to randInt instead of invisibly putting the State in there. Or am I > going about this completely the wrong way? > > > Thanks, > > > > G > > > > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://www.haskell.org/mailman/listinfo/beginners >

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