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Manage the type of a variable in the show implementation
by Christian Sperandio 23 Jul '14

23 Jul '14

Hi, I wrote this kind of code: data DataLog a = SingleData a | MultipleData [a] instance (Show a) => Show (DataLog a) where show (SingleData v) = show v show (MultipleData vs) = join "|" vs where join _ [] = [] join _ (x:[]) = show x join s (x:xs) = show x ++ s ++ join s xs It works but when the DataLog is String typed I got this result: "My Test"|"Another Value" I'd prefer have: My Test|Another Value I know the quote problem comes from the show function on String value. How could I do a show for no-string values and return directly the value for strings? Thanks. Chris

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Haskell in the Digital Humanities
by Niels-Oliver Walkowski 22 Jul '14

22 Jul '14

Dear Haskell Community, I am coming to you with a question which I try to solve for 2 month now, but still struggling, so I tried the step to refer to you. The question is simple and was raised by many people before, probably everyone who began with another language and at some point got to know Haskell: Should I learn Haskell? So I read a lot of things in the www that had the same question or that could support decision making. The reason why I couldn't manage to make a decision rest on – as so often – in the particular case or at least what seems particular for me. I would like to briefly outline why I would love to learn Haskell, what are the reasons why I am hesitating and how all this relates to my background. I would be tremendously grateful for feedback that will make a decision easier for me and hope that it is ok to ask such a question on a beginner's mailinglist. Background: I am a humanist, working in the so called Digital Humanities realm. I don't have any information technology education though I started coding in PHP at the beginning of University. In DH like in many other areas of 'data driven science' Python is 'Lengua Franca'. That might relate to the fact that coding is means not goal. In fact, often coding is more glueing things together than really code something and the ecosystem of scripts and libraries is as broad as scientific discourse itself. It is easy to reach a goal a goal (which is important when you want to test hypothesis) but things are often dirty of cause. Interest in Haskell: I got to know ideas of functional programing half a year ago when I digged a little bit into machine theory. Functional Programing expresses exactly how I think about computation and furthermore I enjoy very much to think within a functional logic about things I want to do. Additionally I think that the functional approach has some important advantages for coding in science at least on the publication level because it expresses ideas in a clear defined way that is not the same with an imperative approach. I started to read a little bit about functional programing in Python since Python is multi-paradigm but very quickly my impression was that it is something you do not really want to do, it makes writing code more verbose, you do not really have the benefits of functional programing and often feels like workaround, so I was brought back to Haskell again Concerns: As mentioned before coding in Digital Humanities is definitely always means for an end. On the other hand I like to to my things straight forward and on a certain abstraction level (maybe because I was trained in Philosophy ;-) ) General concerns rest on the fear that the Haskell ecosystem is not big enough for the area in which I code. I work in the area of text analysis (not identical with computational linguistics), topic modeling, I need sophisticated visualisations, I work with XML (state of the art in DH) and a lot with so called 'dirty data' which relates to the whole NoSQL world – especially Graph and JSON based. Managing Web technologies is important because of the importance of publishing. Approaches like the IPython Notebook are significant (I got to know that there is IHaskell which is IPython with a Haskell kernel but I don't know how valid and stable this project is). I absolutely do not need to have a library for any tiny task and I like to approach things from a higher level of abstraction but of cause there is a point where implementing something on your own is not reasonably time efficient anymore, especially when the goal is something completely different. And being not natively educated in informatics there is also a limit in capacity at a certain point. If you could provide me with feedback or with your judgment to help me evaluate the Haskell environment around my area and social field in a more substantial way and to get a more sophisticated feeling for or against the idea that learning Haskell will work out in my case, it would be wonderful. (A blog article about using Haskell in DH that summarizes the discussion and promotes the awareness of Haskell in the DH community would also something I might consider afterwards) If you feel I misused the list, please apologize. best, Niels-Oliver Walkowski

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Critique My First Project - CampBX API Client
by Pavan Rikhi 19 Jul '14

19 Jul '14

Hiya, so I've spent the last 2 months or so learning Haskell. I've finished the UPenn course, LYAH and a little bit of RWH. I wanted to get a project going so I tried making an API client for the CampBX Bitcoin Market[1]. I started out by following along with the FPComplete Mailchimp API Tutorial[2]. I'm asking you beautiful people for any feedback on it so I can be sure that I'm on the right track here, stuff like code style/idioms, project layout, public interface, implementation, docs, ideas for new features, anything you want to tell me, etc.: https://github.com/prikhi/campbx-haskell Typical usage would be something like this[3]: main :: IO () main = do cfg <- defaultCampBXConfig _ <- runCampBX cfg $ do totalAskVolume <- calculateAskVolume <$> getDepth liftIO . putStrLn $ "Total Ask Volume: " ++ show totalAskVolume return () calculateAskVolume :: Depth -> BTCAmount calculateAskVolume depthList = sum . map askPrice . asks $ depthList where askPrice (Ask (_, q)) = q There are still some things I want to work on: * Define Asks and Bids[4] using Record Syntax. The JSON[5] for a Bid/Ask comes back as a 2 item Array but the generically derived instance[6] expects an object. I haven't completely wrapped my head around Aeson's Array parsing... * Write tests. I'm imagining they would be more "given this JSON, make sure the data structure is created correctly" instead of property-based testing. * Use something other than Doubles to represent Amounts + Prices. Is there a standard library for accurate math with decimals(I need up to 8 decimal places)? I suppose I could always just use Integers to represent Satoshis(the smallest subunit of bitcoins). Also I tried adding "default-extensions: OverloadedStrings" to the .cabal file and removing the pragmas from the source files, but the package wouldn't build :( [1] https://campbx.com/api.php [2] https://www.fpcomplete.com/school/to-infinity-and-beyond/competition-winner… [3] https://github.com/prikhi/campbx-haskell/blob/master/bin/CampBXMain.hs [4] https://github.com/prikhi/campbx-haskell/blob/master/src/Web/CampBX/Types.h… [5] http://sleepanarchy.com/p/K1wuSm [6] https://github.com/prikhi/campbx-haskell/blob/master/src/Web/CampBX/Types.h…

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More simple continuation questions
by martin 17 Jul '14

17 Jul '14

Hello all, (1) when reading about continuations, there is this thing which needs an "other function" to be passed to produce a final result. And there is this "other function". Which of the two is "the continuation"? http://stackoverflow.com/questions/9050725/call-cc-implementation sais " ... instead, they are passed a function that represents the 'next step' in the computation - the 'continuation' " Which sound like the "other function" is called "continuation". But other articles hint in the opposite direction. (2) When comparing continuations with callbacks, it struck me that the type is newtype Cont r a = Cont { runCont :: (a -> r) -> r } So the type of the final result is r. But why does the function (a->r) need to return an r? With regular callbacks this does not seem to be the case. After returning from the callback the surrounding function is free to do anything it wants with the return value and return a value of a different type. Why is that so?

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cabal sandbox add-source + cabal repl (for tracking Idris git version)
by Andreas Reuleaux 14 Jul '14

14 Jul '14

Being somehow new to cabal sandboxes, I would expect that sources that I add with cabal sandbox add-source /some/path are available to me in my local (sandbox aware) cabal repl, this is however not the case, or at least I haven't figured out how to make this this work. I am trying to use Idris as a library - roughly as described in http://brianmckenna.org/blog/idris_library, (and in particular the "cabal sandbox add-source..." step, described there) but my question is really more Haskell and cabal sandbox than Idris related, therefore I am asking here. I have successfully installed Idris from git in a sandbox (For completeness's sake here are the individual steps, including some test steps with highlight-versions, that are not strictly necessary). I install only the idris deps via cabal (i.e. from hackage), idris itself via make from its git sources: -------------------- (within my dir ~/idr, i.e. creating ~/idr/dev here) git clone git://github.com/idris-lang/Idris-dev dev cd dev cabal sandbox init cabal update cabal install highlight-versions (activate my sandbox with some bash magic, i.e. include /home/reuleaux/idr/dev/.cabal-sandbox/bin in my PATH, so that the command highlight-versions is found:) cabal install --dry-run --only-dependencies idris | highlight-versions (now the wet run:) cabal install --only-dependencies idris make (which installs idris from its git sources) I have two more wrappers in my sandbox for that purpose: (really some bash magic...) * ghc: exec "/usr/bin/ghc" -no-user-package-db -package-db=/home/reuleaux/idr/dev/.cabal-sandbox/x86_64-linux-ghc-7.6.3-packages.conf.d "$@" * ghc-pkg: exec "/usr/bin/ghc-pkg" --no-user-package-db --package-db=/home/reuleaux/idr/dev/.cabal-sandbox/x86_64-linux-ghc-7.6.3-packages.conf.d "$@" I know that "cabal sandbox hc-pkg" is the regular wrapper for ghc-pkg, but idris make expects a ghc-pkg command (not cabal sandbox hc-pkg) Now within ~/dev I can e.g. call * "idris" or * cabal repl and then > :m +Idris.Parser i.e. load some idris modules on the repl -------------------- So far, so good. Now I would like to track that dev version of idris in a separate directory / sandbox, I am therefore using "cabal sandbox add-source...": mkdir ~/foo cd foo cabal sandbox init cabal sandbox add-source ~/idr/dev cabal update (I don't mind reinstalling the deps, not sure if this is really necessary:) cabal install --only-dependencies idris I would think that within foo, I should be able to just do e.g. cabal repl > :m +Idris.Parser (as cabal repl is sandbox aware, as I learned) but nope !? Prelude > :m +Idris.Parser <no location info>: Could not find module `Idris.Parser' It is not a module in the current program, or in any known package. Prelude > Well I can see the sources with [sb] reuleaux@mirabelle ~/foo $ cabal sandbox list-sources Source dependencies registered in the current sandbox ('/home/reuleaux/foo/.cabal-sandbox'): /home/reuleaux/idr/dev To unregister source dependencies, use the 'sandbox delete-source' command. [sb] reuleaux@mirabelle ~/foo $ But that's about it. Similarly: How would I use these added sources with ghc? Do I have to use both: * /home/reuleaux/idr/dev/.cabal-sandbox/x86_64-linux-ghc-7.6.3-packages.conf.d * and /home/reuleaux/idr/foo/.cabal-sandbox/x86_64-linux-ghc-7.6.3-packages.conf.d in my -package-db ? I am happy to give more details of my installation, if necessary (I am on debian jessie, my cabal is v 1.20) Thanks in advance. -Andreas

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Simple Continuation question
by martin 13 Jul '14

13 Jul '14

Hello all, I just started trying to understand Continuations, but my very first exercise already left me mystified. import Control.Monad.Cont resultIs :: Int -> Cont String String resultIs i = ContT $ f where f :: (String -> a) -> a f k = k ("result=" ++ show i) If resultIs returns a Cont String String, then f should be (String->String)->String, but that doesn't compile. Why is that so?

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Help with improving a program
by Marcelo Lacerda 11 Jul '14

11 Jul '14

Hi I'm just starting with haskell and want some help with it. I tried to solve the Store Credit[1] problem from google code jam just to practice, the result was a slow code[2] that I find hard to read. Can you guys give me some directions on how to improve it? [1] - https://code.google.com/codejam/contest/351101/dashboard#s=p0 [2] - http://pastebin.com/jNGxGP5H

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foldM efficiency
by Matthew Moppett 11 Jul '14

11 Jul '14

I recently wrote the following code to solve a Project Euler problem: coin :: Int -> Int -> [Int] coin coinvalue stash = [stash, stash + coinvalue .. 200] main = print . length . foldl (>>=) [0] $ map coin [200, 100, 50, 20, 10, 5, 2] The program (correctly) works out how many ways there are to make up £2 from £2, £1, 50p, 20p, 10p, 5p, and 1p coins. Looking at the code, I realised that (foldl (>>=) [0] . map) does a very similar job to foldM from Control.Monad. So I tried rewriting the code as follows: import Control.Monad coin :: Int -> Int -> [Int] coin stash coinvalue = [stash, stash + coinvalue .. 200] main = print . length $ foldM coin 0 [200, 100, 50, 20, 10, 5, 2] This works, but it's about four times slower: it takes about 0.065 seconds on my computer, while the first version takes about 0.017 seconds. To put it another way, if I define foldM as follows: foldM :: Monad m => (a -> b -> m a) -> a -> [b] -> m a foldM f x = foldl (>>=) (return x) . map (flip f) ... and use that definition instead of the standard Control.Monad version, the code runs about four times faster. Here's the Control.Monad version (copied from Hackage): foldM :: (Monad m) => (a -> b -> m a) -> a -> [b] -> m a foldM _ a [] = return a foldM f a (x:xs) = f a x >>= \fax -> foldM f fax xs Why is my version so much faster here? And under what circumstances (if any) could I expect the Control.Monad foldM to give better performance than my version?

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Equality error
by PATRICK BROWNE 10 Jul '14

10 Jul '14

{- Hi, I am trying to get some Haskell code to run from a research paper[1] The relevant fragment is shown below. The issue is that I cannot get the code to run using the original Eq instance. I have replaced the Eq instance with a set of deriving statements on the data types and this works fine. But I am anxious to know why I get the error when using the original code. Also, it seems reasonable to be able to define equality based on names. With the original Eq Instance I get the following error: Context reduction stack overflow;... In the expression: elem car cars In an equation for `element': element car cars = elem car cars In the instance declaration for `Collection Cars Car' I am not too sure how to interpret this message. Any clarification of the error and possible fix would be appreciated. Thanks, Pat [1] ifgi.uni-muenster.de/~sumitsen/sen_Font05.pdf -} {-# LANGUAGE MultiParamTypeClasses, FlexibleInstances,FunctionalDependencies, TypeSynonymInstances,UndecidableInstances,TypeSynonymInstances ,OverlappingInstances #-} import Data.List class Collection collection single where empty :: collection addOne :: single -> collection -> collection remove :: single -> collection -> collection element :: single -> collection -> Bool doToAll :: (single -> single) -> collection -> collection class Named object name | object -> name where name :: object -> name -- **** in original code **** -- All named objects can be compared for equality if names can instance (Eq name, Named object name) => Eq object where object1 == object2 = (name object1) == (name object2) -- could add (Eq, Show) here data Edge = Edge Node Node deriving Show -- (Eq, Show) type Name = [Char] data Node = Node Name deriving Show -- (Eq, Show) data Car = Car Node deriving Show -- (Eq, Show) type Cars = [Car] instance Named Car Name where name c = "a" instance Collection Cars Car where empty = [] -- ::Cars addOne car cars = car:cars remove car cars = delete car cars element car cars = elem car cars doToAll f cars = map f cars -- addOne (Car (Node "a")) [Car (Node "a")] -- addOne (Car (Node "a")) []::Cars -- addOne (Car (Node "b")) [(Car (Node "a"))] -- remove (Car (Node "a")) (addOne (Car (Node "b")) [(Car (Node "a"))]) -- This email originated from DIT. If you received this email in error, please delete it from your system. Please note that if you are not the named addressee, disclosing, copying, distributing or taking any action based on the contents of this email or attachments is prohibited. www.dit.ie Is ó ITBÁC a tháinig an ríomhphost seo. Má fuair tú an ríomhphost seo trí earráid, scrios de do chóras é le do thoil. Tabhair ar aird, mura tú an seolaí ainmnithe, go bhfuil dianchosc ar aon nochtadh, aon chóipeáil, aon dáileadh nó ar aon ghníomh a dhéanfar bunaithe ar an ábhar atá sa ríomhphost nó sna hiatáin seo. www.dit.ie Tá ITBÁC ag aistriú go Gráinseach Ghormáin – DIT is on the move to Grangegorman <http://www.dit.ie/grangegorman>

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Functors in Haskell
by Dimitri DeFigueiredo 06 Jul '14

06 Jul '14

Hi All, I'm trying to merge in my head the concepts of: - functors from category theory; with, - the Functor typeclass in haskell. Here's what I have come up with: A Functor F in haskell is a type constructor that provides a functor map (fmap) and that maps Hask (the category of haskell types) into a category generated by the Functor. For example, Maybe maps the category of all haskell types "a" (i.e. Hask) into the category of all types of the form "Maybe a" by providing an implementation of fmap. Neglecting the fact that apparently Hask isn't a category, does this sound reasonable? I think the insight for me here is that functors in haskell are type constructors mapping Hask into another category. Is that a good way to think about it? Thanks! Dimitri

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