- Beginners - Haskell.org

How to link two Types
by PICCA Frederic-Emmanuel 22 Feb '17

22 Feb '17

Hello, I wrote this code data DataFrameH5 a = DataFrameH5 (Nxs a) -- Nexus file (DataSource H5) -- gamma (DataSource H5) -- delta (DataSource H5) -- wavelength PoniGenerator -- ponie generator class Frame t a where len :: t -> IO (Maybe Int) row :: t -> Int -> MaybeT IO (DifTomoFrame a DIM1) instance Frame (DataFrameH5 DataFrameH5Path) DataFrameH5Path where len (DataFrameH5 _ _ (DataSourceH5 _ d) _ _) = lenH5Dataspace d row d@(DataFrameH5 nxs' g d' w ponigen) idx = do n <- lift $ len d let eof = fromJust n - 1 == idx let mu = 0.0 let komega = 0.0 let kappa = 0.0 let kphi = 0.0 gamma <- g `atIndex'` (ix1 0) delta <- d' `atIndex'` (ix1 idx) wavelength <- w `atIndex'` (ix1 0) let source = Source (head wavelength *~ nano meter) let positions = concat [mu, komega, kappa, kphi, gamma, delta] -- print positions let geometry = Geometry K6c source positions Nothing let detector = ZeroD m <- lift $ geometryDetectorRotationGet geometry detector poniext <- lift $ ponigen (MyMatrix HklB m) idx return $ DifTomoFrame { difTomoFrameNxs = nxs' , difTomoFrameIdx = idx , difTomoFrameEOF = eof , difTomoFrameGeometry = geometry , difTomoFramePoniExt = poniext } has you can see my t type contains also the a reference to the a one So when I create the instance, I need to write two times the DataFrameH5Path I would like to know how to write the same class with only class Frame t where len :: t -> IO (Maybe Int) row :: t -> Int -> MaybeT IO (DifTomoFrame <extract type a from type t> DIM1) thanks for your help Frederic

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Type error when using splitOn function.
by S. H. Aegis 22 Feb '17

22 Feb '17

Hello. I'm new to Haskell and this is the first time I use Data.Text module. And using stack on OSX 10.12.3 I'm try several times, but fail. and I don't understand what error message says. How can I fix this? Thank you a lot. Code is Main.hs : module Main where import Lib main :: IO () main = do sam <- readSam rxDxData <- readCSV print $ makeRxDxList rxDxData Lib.hs : module Lib -- ( someFunc -- ) where where import Data.Text as T import Text.Regex.TDFA import Prelude hiding (take, drop, map, lines) type RowSAM = Text type RowRxDx = Text --makeRxDxList :: Functor f => f Text -> f [Text] --makeRxDxList rowRxDx = fmap (\x -> splitOn (pack ",") x) rowRxDx -- This code pass a compile. makeRxDxList rowRxDx = map (\x -> splitOn (pack ",") pack x) $ lines rowRxDx (whole code is below...) Error Message is /Users/shaegis/Documents/Haskell/samChecker3/src/Lib.hs:25:35: error: • Couldn't match expected type ‘Char -> Char’ with actual type ‘[Text]’ • The function ‘splitOn’ is applied to three arguments, but its type ‘Text -> Text -> [Text]’ has only two In the expression: splitOn (pack ",") pack x In the first argument of ‘map’, namely ‘(\ x -> splitOn (pack ",") pack x)’ /Users/shaegis/Documents/Haskell/samChecker3/src/Lib.hs:25:54: error: • Couldn't match expected type ‘Text’ with actual type ‘String -> Text’ • Probable cause: ‘pack’ is applied to too few arguments In the second argument of ‘splitOn’, namely ‘pack’ In the expression: splitOn (pack ",") pack x In the first argument of ‘map’, namely ‘(\ x -> splitOn (pack ",") pack x)’ /Users/shaegis/Documents/Haskell/samChecker3/src/Lib.hs:25:64: error: • Couldn't match expected type ‘Text’ with actual type ‘[Text]’ • In the second argument of ‘($)’, namely ‘lines rowRxDx’ In the expression: map (\ x -> splitOn (pack ",") pack x) $ lines rowRxDx In an equation for ‘makeRxDxList’: makeRxDxList rowRxDx = map (\ x -> splitOn (pack ",") pack x) $ lines rowRxDx ------------------------------------------------------------------------------------------------------ Lib.hs module Lib -- ( someFunc -- ) where where import Data.Text as T import Text.Regex.TDFA import Prelude hiding (take, drop, map, lines) type RowSAM = Text type SAM = [Text] type Case = Text type RowRxDx = Text type RxDx = [Text] type RxDxList = [[Text]] type Rx = Text type Dx = Text type MediName = Text type Message = Text type Date = Text type PtName = Text --makeRxDxList :: Functor f => f Text -> f [Text] --makeRxDxList rowRxDx = fmap (\x -> splitOn (pack ",") x) rowRxDx makeRxDxList rowRxDx = map (\x -> splitOn (pack ",") pack x) $ lines rowRxDx pickupMediName :: RxDx -> MediName pickupMediName rxDx = rxDx !! 0 pickupDx :: RxDx -> Dx pickupDx rxDx = rxDx !! 2 pickupRx :: RxDx -> Rx pickupRx rxDx = rxDx !! 1 pickupPtName :: Case -> PtName pickupPtName ptCase = take 3 $ drop 45 ptCase pickupCaseDate :: Case -> Date pickupCaseDate ptCase = take 8 $ drop (348 + 2) ptCase isErrorRxDx :: Rx -> Dx -> Case -> Bool isErrorRxDx rxCode dxCode ptCase = case isExistRx rxCode ptCase of True -> if (isExistDx dxCode ptCase) then False else True False -> False isExistDx :: Dx -> Case -> Bool isExistDx dxCode ptCase = (unpack ptCase) =~ (unpack dxCode) isExistRx :: Rx -> Case -> Bool isExistRx rxCode ptCase = rxCode `isInfixOf` ptCase splitIntoCase :: RowSAM -> SAM splitIntoCase = splitOn $ pack "AH021" readCSV :: IO Text readCSV = pack <$> readFile "/Users/shaegis/Documents/Haskell/samChecker3/RxDxData.csv" readSam :: IO Text readSam = pack <$> readFile "/Users/shaegis/Documents/Haskell/samChecker3/BoHomUTF8.dat"

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Re: [Haskell-beginners] Beginners Digest, Vol 104, Issue 9
by Praveen Velliengiri 15 Feb '17

15 Feb '17

Thanks a lot.🙂 On 15 February 2017 at 17:30, <beginners-request(a)haskell.org> wrote: > Send Beginners mailing list submissions to > beginners(a)haskell.org > > To subscribe or unsubscribe via the World Wide Web, visit > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > or, via email, send a message with subject or body 'help' to > beginners-request(a)haskell.org > > You can reach the person managing the list at > beginners-owner(a)haskell.org > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of Beginners digest..." > > > Today's Topics: > > 1. OPERATOR OVERLOADING AS IN C++ (Praveen Velliengiri) > 2. Re: OPERATOR OVERLOADING AS IN C++ (Francesco Ariis) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Tue, 14 Feb 2017 18:21:37 +0530 > From: Praveen Velliengiri <praveenvelliengiri(a)gmail.com> > To: beginners(a)haskell.org > Subject: [Haskell-beginners] OPERATOR OVERLOADING AS IN C++ > Message-ID: > <CAL56b7SMDdF+doiqyLxTM85Tgv+R=ooMrP2TaLmAHSuaBtOo+A@mail. > gmail.com> > Content-Type: text/plain; charset="utf-8" > > I'm new to functional programming, Whether it is possible to do operator > overloading in Haskell ? I go through the Idea of TYPECLASSES in Learn you > haskell tutorial. But I can't understand the idea Can anyone suggest me > some idea regarding Operator overloading and Type classes in Haskell. > Thank you guys > Praveen v >

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OPERATOR OVERLOADING AS IN C++
by Praveen Velliengiri 14 Feb '17

14 Feb '17

I'm new to functional programming, Whether it is possible to do operator overloading in Haskell ? I go through the Idea of TYPECLASSES in Learn you haskell tutorial. But I can't understand the idea Can anyone suggest me some idea regarding Operator overloading and Type classes in Haskell. Thank you guys Praveen v

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how do typeclasses work again?
by Nicholls, Mark 12 Feb '17

12 Feb '17

Sorry..I do haskell about once every 6 months for 2 hours...and then get on with my life. I always forget some nuance of typeclasses. Consider some simple typeclass > class Is isx x where > apply :: (x -> y) -> isx -> y We can make any type a member of it...mapping to itself > instance Is x x where > apply f = f But we can also make a tuple a member of it...and pull the 1st member.. > instance Is (x,y) x where > apply f (x,y) = f x Weird and largey useless...but I'm playing. Then construct a function to operate on it > foo2 :: (Is isx Integer) => isx -> String > foo2 = apply (\i -> "") And... • Could not deduce (Is isx x0) arising from a use of ‘apply’ from the context: Is isx Integer bound by the type signature for: foo2 :: Is isx Integer => isx -> String at prop.lhs:51:3-43 The type variable ‘x0’ is ambiguous Relevant bindings include foo2 :: isx -> String (bound at prop.lhs:52:3) These potential instances exist: instance Is x x -- Defined at prop.lhs:41:12 instance Is (x, y) x -- Defined at prop.lhs:45:12 • In the expression: apply (\ i -> "") In an equation for ‘foo2’: foo2 = apply (\ i -> "") What's it going on about? (my brain is locked in F# OO type mode) I've told it to expect a function "Integer -> String"...surely? Whats the problem. CONFIDENTIALITY NOTICE This e-mail (and any attached files) is confidential and protected by copyright (and other intellectual property rights). If you are not the intended recipient please e-mail the sender and then delete the email and any attached files immediately. Any further use or dissemination is prohibited. While MTV Networks Europe has taken steps to ensure that this email and any attachments are virus free, it is your responsibility to ensure that this message and any attachments are virus free and do not affect your systems / data. Communicating by email is not 100% secure and carries risks such as delay, data corruption, non-delivery, wrongful interception and unauthorised amendment. If you communicate with us by e-mail, you acknowledge and assume these risks, and you agree to take appropriate measures to minimise these risks when e-mailing us. MTV Networks International, MTV Networks UK & Ireland, Greenhouse, Nickelodeon Viacom Consumer Products, VBSi, Viacom Brand Solutions International, Be Viacom, Viacom International Media Networks and VIMN and Comedy Central are all trading names of MTV Networks Europe. MTV Networks Europe is a partnership between MTV Networks Europe Inc. and Viacom Networks Europe Inc. Address for service in Great Britain is 17-29 Hawley Crescent, London, NW1 8TT.

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Re: [Haskell-beginners] how do typeclasses work again?
by Dimitri DeFigueiredo 10 Feb '17

10 Feb '17

I would like to make a general comment that is likely not useful to the specific question asked (which I think has already been nicely answered by others). But I would like to point out this video of an excellent lecture on typeclasses given by Simon Peyton-Jones a few years ago. https://channel9.msdn.com/posts/MDCC-TechTalk-Classes-Jim-but-not-as-we-kno… Cheers, Dimitri -- 2E45 D376 A744 C671 5100 A261 210B 8461 0FB0 CA1F

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Semigroup Instances
by Atrudyjane 06 Feb '17

06 Feb '17

I'm currently studying semigroups and trying to figure out how to determine which type variables need a semigroup instance. Here are a couple of examples from Evan Cameron's github (https://github.com/leshow/haskell-programming-book/blob/master/src/Ch15ex.hs) (1) data Validation a b = Failure a | Success b deriving (Eq, Show) instance Semigroup a => Semigroup (Validation a b) where Success a <> Success b = Success a Failure a <> Success b = Success b Success a <> Failure b = Success a Failure a <> Failure b = Failure (a <> b) * Why doesn't 'b' need an instance of semigroup? (2) newtype AccumulateRight a b = AccumulateRight (Validation a b) deriving (Eq, Show) instance Semigroup b => Semigroup (AccumulateRight a b) where AccumulateRight (Success a) <>AccumulateRight (Failure b) =AccumulateRight (Success a) AccumulateRight (Failure a) <>AccumulateRight (Success b) =AccumulateRight (Success b) AccumulateRight (Failure a) <>AccumulateRight (Failure b) =AccumulateRight (Failure a) AccumulateRight (Success a) <> AccumulateRight (Success b) = AccumulateRight (Success (a <> b)) * Why doesn't 'a' need an instance of semigroup? Thank you, Andrea Sent with [ProtonMail](https://protonmail.com) Secure Email.

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Applicative for State
by mike h 06 Feb '17

06 Feb '17

Hi, I have a State by another name, Stat, just to experiment and learn. newtype Stat s a = Stat { runStat :: s -> (a, s) } instance Functor (Stat s) where fmap f (Stat g) = Stat $ \s -> (f a, s) where (a, s) = g s instance Applicative (Stat s) where pure a = Stat $ \s -> (a, s) (Stat f) <*> (Stat g) = Stat $ \s -> (a, s) where (a, s) = undefined I really can’t get what the <*> in the Applicative should be! I just do see how I ‘get the f out of the Stat’ and then apply it. I’d be really grateful if someone would explain what it should be and the steps/reasoning needed to get there. Many thanks Mike

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Ambiguous type variable prevents the constraint `(Ord t0)' from being solved.
by Ivan Kush 06 Feb '17

06 Feb '17

I get this error (full message at the end of the mail). How could I correct my code? =================== Code: =================== module Intro where import Data.Bits -- for xor, .&. data Func a b = Empty | Leaf Int [(a, b)] | Branch Int Int (Func a b) (Func a b) applyd = let apply_listd l d x = case l of [] -> d x (a, b) : t -> let c = compare x a in if c == EQ then b else if c == GT then apply_listd t d x else d x in \f d x -> let k = 5 -- hash x - todo in let look t = case t of Leaf h l | h == k -> apply_listd l d x Branch p b l r | (k `xor` p) .&. (b - 1) == 0 -> -- (Branch p b l r) | ((k xor p) .&. (b - 1)) == 0 -> look (if k .&. b == 0 then l else r) _ -> d x in look f =================== Error: =================== Intro.hs:37:25: error: * Ambiguous type variable `t0' arising from a use of `apply_listd' prevents the constraint `(Ord t0)' from being solved. Relevant bindings include l :: [(t0, t)] (bound at Intro.hs:36:28) t :: Func t0 t (bound at Intro.hs:34:21) look :: Func t0 t -> t (bound at Intro.hs:34:16) x :: t0 (bound at Intro.hs:32:15) d :: t0 -> t (bound at Intro.hs:32:13) f :: Func t0 t (bound at Intro.hs:32:11) (Some bindings suppressed; use -fmax-relevant-binds=N or -fno-max-relevant-binds) Probable fix: use a type annotation to specify what `t0' should be. These potential instances exist: instance Ord Ordering -- Defined in `GHC.Classes' instance Ord Integer -- Defined in `integer-gmp-1.0.0.1:GHC.Integer.Type' instance Ord a => Ord (Maybe a) -- Defined in `GHC.Base' ...plus 22 others ...plus five instances involving out-of-scope types (use -fprint-potential-instances to see them all) * In the expression: apply_listd l d x In a case alternative: Leaf h l | h == k -> apply_listd l d x In the expression: case t of { Leaf h l | h == k -> apply_listd l d x Branch p b l r | (k `xor` p) .&. (b - 1) == 0 -> look (if k .&. b == 0 then l else r) _ -> d x } -- Best wishes, Ivan Kush

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Need help understanding the tell function in the Monad Writer example in LYAH
by Olumide 06 Feb '17

06 Feb '17

Hello List, Would someone kindly explain how the tell function interacts with the Writer Monad as show below: (Example taken from chapter 14 of LYAH http://learnyouahaskell.com/for-a-few-monads-more#reader ) multWithLog :: Writer [String] Int multWithLog = do a <- logNumber 3 b <- logNumber 5 tell ["Gonna multiply these two"] return (a*b) Result: ghci> runWriter multWithLog (15,["Got number: 3","Got number: 5","Gonna multiply these two"]) I know that tell function binds to an argument that is discarded but I don't know how its argument "Gonna multiply these two" is concatenated with the other the Writer created by logNumber 3 and logNumber 5. Also, I don't understand the paragraph following the example: "It's important that return (a*b) is the last line, because the result of the last line in a do expression is the result of the whole do expression. Had we put tell as the last line, () would have been the result of this do expression. We'd lose the result of the multiplication. However, the log would be the same." Regards, - Olumide

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