- Beginners - Haskell.org

Counting Fruits
by Adolfo Builes 20 Feb '09

20 Feb '09

Hello everyone , I wanted to do a small program which read a txt with fruit's name in each line and then print how many fruits I have of each type. something like these: apple apple and then [(apple,2)] I came up whit this import qualified Data.Map as Map import Data.List import System.IO main = do file <- readFile "fruits.txt" let answer = proccessFile $ lines file putStrLn (show answer) proccessFile :: [String] -> [(String,Int)] proccessFile file = Map.toAscList $ parseFile Map.empty file where parseFile fruits [] = fruits parseFile fruits_map (x:xs) = parseFile (Map.insertWith (+) x 1 fruits_map) xs It works, but I would like to know how would you do it ?, Share different points of view, different code. Was it a good idea to use a Map ?, Did I separate the code in a proper way, I mean pure - impure ? How can we improve the performance ? Best Regards for everyone. Adolfo

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Haskell Tutorial Code Error
by Pete Ryland 20 Feb '09

20 Feb '09

Hi, Sorry to post to a list, but there are no other contact addresses on the haskell.org site or in the tutorial itself. Any replies, please also CC me as I do not subscribe to any haskell.org lists. I noticed an error in the Haskell Tutorial in section 8.3. The code for the Tree parser will not work. Specifically, the following is broken: readsTree :: (Read a) => ReadS (Tree a) readsTree s = [(Branch l r, x) | ("<", t) <- lex s, (l, u) <- readsTree t, ("|", v) <- lex u, (r, w) <- readsTree v, (">", x) <- lex w ] ++ [(Leaf x, t) | (x, t) <- reads s ] It will not read a string like "<<1|2>|3>" because of the "<<" which the lexer will see as one token. Here is another implementation: showsTree :: (Show a) => Tree a -> ShowS showsTree (Leaf x) = ("Leaf " ++) . shows x showsTree (Branch l r) = ("Branch (" ++) . showsTree l . (") (" ++) . showsTree r . (')':) readsTree :: (Read a) => ReadS (Tree a) readsTree s = [(Branch l r, v) | ("Branch", t) <- lex s, (l, u) <- readsTree t, (r, v) <- readsTree u ] ++ [(Leaf x, u) | ("Leaf", t) <- lex s, (x, u) <- reads t ] ++ [(x, v) | ("(", t) <- lex s, (x, u) <- readsTree t, (")", v) <- lex u ] instance Show a => Show (Tree a) where showsPrec _ x = showsTree x instance Read a => Read (Tree a) where readsPrec _ s = readsTree s This is almost equivalent to the derived one, but will accept strings like "Branch Leaf 3 Branch Branch Leaf 1 Leaf 2 Branch Leaf 5 Leaf 4" (that is, without the parentheses). Regards, Pete

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Sequential IO processing
by Sergey V. Mikhanov 19 Feb '09

19 Feb '09

Hi community, I am making my first steps in Haskell, reading Haskell wikibook and now stuck with one of the excercises, namely this one: Implement a function sequenceIO :: [IO a] -> IO [a]. Given a list of actions, this function runs each of the actions in order and returns all their results as a list. This is what I came with: ioOne :: Num a => IO a ioOne = do print guid return guid where guid = 2 ioTwo :: Num a => IO a ioTwo = do print guid return guid where guid = 3 sequenceIO :: Num a => [IO a] -> IO [a] sequenceIO [] = return [] sequenceIO (x : xs) = do result <- x return result : sequenceIO xs First two functions are there because of the invocation that I've planned: sequenceIO [getGuid, getNextGuid]. However, this could not be compiled (GHC): Couldn't match expected type `[m a]' against inferred type `IO [a]' In the second argument of `(:)', namely `sequenceIO xs' In the expression: return result : sequenceIO xs In the expression: do result <- x return result : sequenceIO xs Fine, I thought, something wrong with the type of the 'sequenceIO xs' (becasue I am sure the type of 'result' is fine). So I wrote another program to check what happens to the result of IO action evaluation (namely, which type is assigned): bar :: Num a => IO a bar = do print guid return guid where guid = 2 foo = do result <- bar result This could not be compiled either: No instance for (Num (IO b)) arising from a use of `bar' at auxil.hs:8:19-21 Possible fix: add an instance declaration for (Num (IO b)) In a stmt of a 'do' expression: result <- bar In the expression: do result <- bar result In the definition of `foo': foo = do result <- bar result I am a newbie, so I am interpreting this like "Haskell could not construct Num from the result of invocation of bar, which is of type IO a". But why do I need this at all? When doing console I/O with 'result <- getLine', I do not need to reconstruct String from the result. What am I doing wrong? Where is the failure in reasoning? Thanks, Sergey

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Re: [Haskell-cafe] Re: permuting a list
by Alberto Ruiz 17 Feb '09

17 Feb '09

Heinrich Apfelmus wrote: > Jon Fairbairn wrote: >> Heinrich Apfelmus writes: >> >>> The answer is a resounding "yes" and the main idea is that shuffling a >>> list is *essentially the same* as sorting a list; the minor difference >>> being that the former chooses a permutation at random while the latter >>> chooses a very particular permutation, namely the one that sorts the input. >>> >>> For the full exposition, see >>> >>> http://apfelmus.nfshost.com/random-permutations.html >> I haven't been following the thread, but my initial reaction >> would have been something like use System.Random.randoms to >> get a list rs and then do (roughly) >> >> randomPerm = map snd . sortBy (compare `on` fst) . zip rs >> >> How bad is that? I mean, how unfair does it get? > > It's fair, but may duplicate elements, i.e. it doesn't necessarily > create a permutation. For example, rs could be something like > > rs = [5,3,3,3,2,4] > How about using random doubles? randomPerm xs = fmap (map snd . sort . flip zip xs) rs where rs = fmap (randoms . mkStdGen) randomIO :: IO [Double]

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Palindromic solution??
by Alan Cameron 16 Feb '09

16 Feb '09

>Date: Mon, 16 Feb 2009 16:32:23 +0100 >From: Miguel Pignatelli <miguel.pignatelli(a)uv.es> >Subject: [Haskell-beginners] Indentation of local functions >To: beginners(a)haskell.org >Message-ID: <90C28A9E-A67D-47CA-8416-5F1C5C27A1F9(a)uv.es> >Content-Type: text/plain; charset="us-ascii" > >Hi all, > >This is my first post in this forum, I'm pretty new to Haskell (although I >have some previous experience in functional programming with OCaml) > >I'm trying to write the typical function that determines if a list is a >palindrome. >the typical answer would be something like: > >isPalindrome xs = xs == (reverse xs) > >But I find this pretty inefficient (duplication of the list and double of >needed comparisons). >So I tried my own version using just indexes: > >isPalindrome xs = > isPalindrome' 0 (length xs) > where isPalindrome' i j = > if i == j -- line 43 > then True > else > if (xs !! i) == (xs !! (j-1)) > then isPalindrome' (i+1) (j-1) > else False > >But, when trying to load this in ghci it throws the following error: > >xxx.hs:43:12: parse error (possibly incorrect indentation) Failed, modules >loaded: none. >(Line 43 is marked in the code) > >I seems that the definition of isPalindrome' must be in one line. So, this >works as expected: > >isPalindrome xs = > isPalindrome' 0 (length xs) > where isPalindrome' i j = if i == j then True else if (xs !! i) == (xs !! (j-1)) then isPalindrome' (i+1) (j-1) else False > >Is there any way to make the local definition of isPalindrome' more >readable? > >Any help in understanding this would be appreciated > >Thanks in advance, > I have found one solution to your problem isPalindrome xs = isPalindrome' 0 (length xs) where isPalindrome' i j = if i == j -- line 43 then True else if (xs !! i) == (xs !! (j-1)) then isPalindrome' (i+1) (j-1) else False This loads without error but poses a second problem it generates an index too large exception. Alan Cameron

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multi parameter classes and instance declarations
by Markus Barenhoff 16 Feb '09

16 Feb '09

Good Afternoon, i'am trying to express something like this here: > -- CI.hs http://pastebin.com/f5b17cf57 > {-# OPTIONS -XMultiParamTypeClasses #-} > module CI(D,f) where > class B b > class (B b) => A a b where > f :: b > data D = D > instance B D > instance A D D where > f = D > -- CI.hs > -- MultiClassTest.hs http://pastebin.com/f1c481e8d > import CI > a = f > -- MultiClassTest.hs ghc output: MultiClassTest.hs:4:4: No instance for (CI.A a D) arising from a use of `f' at MultiClassTest.hs:4:4 Possible fix: add an instance declaration for (CI.A a D) In the expression: f In the definition of `a': a = f Failed, modules loaded: CI. when opening MultiClassTest.hs it looks like the instance declaration gets not important, but I think it might have a reson, that the ghc rejects that code... Regards Markus -- Markus Barenhoff - MÃŒnster Germany,Europe,Earth sip:alios@alios.org - iax2:alios@alios.org - xmpp:alios@jabber.ccc.de gpg:0xAE7C7759 fp:79 64 AA D9 B7 16 F5 06 6A 88 5F A9 4D 49 45 BB

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Re: [Haskell-cafe] Re: permuting a list
by Felipe Lessa 15 Feb '09

15 Feb '09

On Sun, Feb 15, 2009 at 9:47 AM, Heinrich Apfelmus <apfelmus(a)quantentunnel.de> wrote: > It's fair, but may duplicate elements, i.e. it doesn't necessarily > create a permutation. For example, rs could be something like > > rs = [5,3,3,3,2,4] > But our sort doesn't discard values when the keys are the same. For example, [1,2,3,4] == map snd . sortBy (compare `on` fst) . zip (repeat 1) $ [1,2,3,4] Nothing gets duplicated. Or did I miss something? -- Felipe.

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permuting a list
by Jan Snajder 14 Feb '09

14 Feb '09

Hi! I'm trying to write a list permutation function, and there is in fact a nice explanation of how to do it here: http://sneakymustard.com/2008/12/23/shuffling-in-haskell But for the start I wanted to keep things simple and avoid monad transformers (since I'm not into this yet). Instead, I'd like to write a function of type: > permute :: [a] -> IO [a] and so this is what I did: > permute xs = do > let n = length xs - 1 > arr0 <- newListArray (0, n) xs > arr <- foldM swap arr0 [n..1] > getElems arr > where swap arr n = do > x <- readArray arr n > r <- randomRIO (0, n) > y <- readArray arr r > writeArray arr n y > writeArray arr r x > return arr Unfortunately, what I get is: > permute :: (MArray a1 a IO) => [a] -> IO [a] and so when I try to apply this function: > permute [1,2,3] this is what I get: <interactive>:1:0: No instance for (MArray a1 t IO) arising from a use of `permute' at <interactive>:1:0-14 Possible fix: add an instance declaration for (MArray a1 t IO) In the expression: permute [1, 2, 3] In the definition of `it': it = permute [1, 2, 3] How can I fix this? Thanx, jan

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Re: permuting a list
by Jan Snajder 13 Feb '09

13 Feb '09

Brent Yorgey wrote: > It seems everyone has just been reading the first few words of Jan's > email and not the actual content. Jan is clearly trying to write a > *random list shuffling* function, not a function to generate > permutations. Let's try to be helpful, people... > </rant> Thanks Brant, I forgot to mention explicitly that I need a random permutation. > Jan, this is tricky. The type of permute is indeed (MArray a1 a IO) > => [a] -> IO [a], but this is fine, it just means that there has to be > some sort of mutable array which can store the things you are trying > to shuffle. > This is not the problem. The problem seems to be that > Haskell has no way to know what sort of array you want to use. I was > able to get the code to work, but it's sort of sneaky: > > > {-# LANGUAGE FlexibleContexts, ScopedTypeVariables #-} I guess by 'sneaky' you mean this solution is GHC-specific? > > import Data.Array.MArray > > import Data.Array.IO > > import Control.Monad > > import System.Random > > > permute :: forall a. (MArray IOArray a IO) => [a] -> IO [a] > > permute xs = do > > let n = length xs - 1 > > arr0 <- (newListArray (0, n) xs :: IO (IOArray Int a)) > > arr <- foldM swap arr0 [n..1] > > getElems arr > > where swap arr n = do > > x <- readArray arr n > > r <- randomRIO (0, n) > > y <- readArray arr r > > writeArray arr n y > > writeArray arr r x > > return arr Ok, this seems to work! (after replacing '[n..1]' with [n,n-1,..1] as Daniel noted). Great! Why do I need 'forall a' ? Aren't type variables implicitly universaly quantified? j.

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cabal-install can't connect
by Kirk Martinez 12 Feb '09

12 Feb '09

I'm trying to use cabal-install through the firewall at my work and I am consistently denied: C:\>cabal update -v3 Downloading package list from server 'http://hackage.haskell.org/packages/archive' Sending: GET http://hackage.haskell.org/packages/archive/00-index.tar.gz HTTP/1.1 User-Agent: cabal-install/0.6.0 Creating new connection to hackage.haskell.org cabal: connect: failed (Connection refused (WSAECONNREFUSED)) I built cabal-install 0.6.0 and compiled with ghc 6.10.1 without a hitch. I set my http_proxy variable the same as other apps I use which are able to connect to other sites via http. I even patched cabal-install according to these instructions<http://www.nabble.com/cabal-install-sends-invalid-proxy-password-td20976277…>, but I still get this same "Connection refused". Seems it's not even getting to the proxy. This is on Windows XP SP2. Any help or suggestions would be greatly appreciated! Installing package dependencies by hand is something I thought I left behind in the 90's... Thanks, Kirk Martinez ________________________________ Confidentiality Notice. This message may contain information that is confidential or otherwise protected from disclosure. If you are not the intended recipient, you are hereby notified that any use, disclosure, dissemination, distribution, or copying of this message, or any attachments, is strictly prohibited. If you have received this message in error, please advise the sender by reply e-mail, and delete the message and any attachments. Thank you.

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