September 2010 - Beginners

Re: [Haskell-beginners] Ranges and List Comprehensions in SQL
by Brent Yorgey 03 Sep '10

03 Sep '10

On Fri, Sep 03, 2010 at 11:21:27AM -0400, Tom Murphy wrote: > > > > I am not sure I understand what you are asking. Do you want something > > like this? > > > > data Range = Range Integer Integer > > > > toRanges :: [Integer] -> [Range] > > toRanges = ... > > > > toRanges should not be too hard to write but nothing like that exists > > as far as I know. > > > > > > I don't quite understand the functionality of toRanges. When I say "range," > I mean a list containing an enumeration, in the form: > > [2..300] [2..300] is just syntactic sugar for the list [2,3,4,5,6,7,8,9,......,300]. It isn't actually a real "thing" in Haskell. That's why I made a new data type to represent it: data Range = Range Integer Integer toList (Range x y) = [x..y] So for example (Range 2 300) represents the list [2..300]. > The record would actually be coming from an SQL database. > I would like to be able to store a large list in SQL records, without having > to store every element. I see. So the question seems more about database design than about Haskell. Toby has suggested a more relational way to model the data, but his solution does end up storing every element. But you could do something more like CREATE TABLE tv_show ( id INTEGER NOT NULL, -- etc title VARCHAR(80), PRIMARY KEY (id) ); CREATE TABLE episode_ranges ( tv_show_id INTEGER NOT NULL, start_episode INTEGER, end_episode INTEGER PRIMARY KEY (tv_show_id, start_episode) ); Or something like that (it's been a while since I've used SQL). The data would then be more like tv_show (30, "The Simpsons") episode_ranges (30,1,4), (30,14,14), (30,18,21), (30,23,25) -Brent

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question on typeclasses and applicatives
by Alec Benzer 02 Sep '10

02 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 02 Sep '10

02 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 02 Sep '10

02 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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