May 2013 - Beginners - Haskell.org

Unmaybe
by Adrian May 17 May '13

17 May '13

Hi there, I have a really annoying scrap of code: unmaybe Nothing = mempty unmaybe (Just dia) = dia It happened because I'm using Diagrams but building my diagram requires looking something up in a list using findIndex, which returns Maybe Int. How do I rid myself of this blotch? TIA, Adrian.

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Diagrams brain twister
by Adrian May 16 May '13

16 May '13

Hi all, I'm trying to draw a picture with diagrams (this isn't the gantt chart I was talking about before.) I have a load of objects strewn around a diagram according to their own sweet logic, and for *some* of them, I want to draw a horizontal line going from the right hand edge of the object to some globally fixed x coordinate, call it the "margin". So those lines are all different lengths because the objects are all over the place, but their right-hand ends should all be aligned vertically. This seems quite hard, because that sweet logic is already quite complicated and local to a set of objects in the immediate neighbourhood of the object in question. Somehow I have to tease out a selection of them and process each of them into this line whose properties depend on where the object is from the global perspective. But how? Adrian.

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How to improve lazyness of a foldl (and memory footprint)
by Giacomo Tesio 15 May '13

15 May '13

Hi, I'm trying to improve a small haskell program of mine. A more extended description with full source code is here: http://codereview.stackexchange.com/questions/26107/how-to-improve-readabil… The script transforms CSV files into other CSV files but looks like it's reading the whole input files before writing output files. I guess that the script can be improved in many ways, in readability and efficiency, thus any suggestion is wellcome as an occasion to learn. But what I can't understand is why this design doesn't work: transformFile :: FilePath -> ([String] -> a) -> (a -> IO r) -> IO r transformFile file operation continuation = withFile file ReadMode (\h -> hGetContents h >>= (continuation.operation.lines)) This function recieves a path, a function to left fold lines to a new list of objects and a function to persist the fold output to files. Here the relevant parts: importTrades :: FilePath -> FilePath -> IO () importTrades outDir csvFile = transformFile csvFile (foldTradingSample.getTickWriteTrades) (saveTradingSamples outDir) where getTickWriteTrades = filter (isBetween (9, 0) (18, 0)).(catMaybes.(map fromCSVLine)) foldTradingSample = foldl toTradingSample [] This is the folding function: toTradingSample :: [TradingSample] -> Tw.Trade -> [TradingSample] toTradingSample (current:others) twTrade | newEqt == equity current && newDay == day current = (current { trades = newTrades }):others | otherwise = current : toTradingSample others twTrade where newEqt = Tw.tSimbol twTrade newDay = Tw.tDate twTrade newTrade = fromTickWrite twTrade newTrades = trades current ++ [newTrade] toTradingSample [] twTrade = [TradingSample { equity = Tw.tSimbol twTrade , day = Tw.tDate twTrade , trades = [fromTickWrite twTrade] }] And this is the function that safe the fold results to files saveTradingSamples :: String -> [TradingSample] -> IO () saveTradingSamples folder samples = mapM_ (saveTradingSample folder) samples saveTradingSample :: String -> TradingSample -> IO () saveTradingSample folder sample = writeFile fileName contents where fileName = folder ++ "\\" ++ (equity sample) ++ "_" ++ (formatTime defaultTimeLocale "%F" $ day sample) ++ ".CSV" contents = tradingSampleToCSV sample What's wrong here? My insight is that the problem is in the signature of transform files, that requires to completely compute the list of TradingSample before calling saveTradingSamples. Is this the problem? How can I fix this? Giacomo

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HLint fails to give suggestions
by irfan hudda 15 May '13

15 May '13

I install HLint via cabal (output http://hpaste.org/88021) To check if it was working I used following code fun1 :: Int -> String fun1 1 = "hell" fun1 2 = 3 and ran HLint on it $ ~/.cabal/bin/hlint hlint_test.hs No suggestions and it ouputs no suggestions any ideas? Additionally $ ~/.cabal/bin/hlint -v HLint v1.8.45, (C) Neil Mitchell 2006-2012

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Invitation to connect on LinkedIn
by Daniel Díaz Casanueva 15 May '13

15 May '13

LinkedIn ------------ Daniel Díaz Casanueva requested to add you as a connection on LinkedIn: ------------------------------------------ Mihai, I'd like to add you to my professional network on LinkedIn. - Daniel Accept invitation from Daniel Díaz Casanueva http://www.linkedin.com/e/-3dpk6t-hgr1eda3-6o/uyFflI7IwvrmSA8-0MFfTI7lXX8s3… View profile of Daniel Díaz Casanueva http://www.linkedin.com/e/-3dpk6t-hgr1eda3-6o/rso/255311896/LIky/name/25134… ------------------------------------------ You are receiving Invitation emails. This email was intended for Mihai Maruseac. Learn why this is included: http://www.linkedin.com/e/-3dpk6t-hgr1eda3-6o/plh/http%3A%2F%2Fhelp%2Elinke… (c) 2012, LinkedIn Corporation. 2029 Stierlin Ct, Mountain View, CA 94043, USA.

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find by
by Julian Arni 15 May '13

15 May '13

(Truncated again - trying plain text. Sorry for the spam.) I have a function that, simplifying a little, looks like this: fn :: [a] -> a fn = (head .) . takeWhile $ (\_ -> True) >From this, I want a function with the signature fn' :: [(x,a)] -> (x,a) such that: snd $ fn' (zip [x] [a]) = fn [a] I can see ways of doing this by altering fn, or by repeating some of fn in the definition of fn', or (because in this case I know that if fn xs = x, fn is returning the first x in xs and not any others), by doing something nasty like: fn' xs = xs !! fromMaybe 0 (findIndex (\(_,a) -> a == fn (snd $ unzip xs)) xs ) But it seems to me like there should be prettier solutions to this (that do not involve changing the definition of fn). After all, this doesn't seem like a rare pattern. Anyone know if in fact there's a better way to go about it?

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find by
by Julian Arni 15 May '13

15 May '13

(For some reason my previous post seems to have been truncated.) I have a function that, simplifying a little, looks like this: fn :: [a] -> a fn = (head .) . takeWhile $ (\_ -> True) >From this, I want a function with the signature fn' :: [(x,a)] -> (x,a) such that: snd $ fn' (zip [x] [a]) = fn [a] I can see ways of doing this by altering fn, or by repeating some of fn in the definition of fn', or (because in this case I know that if fn xs = x, fn is returning the first x in xs and not any others), by doing something nasty like: fn' xs = xs !! fromMaybe 0 (findIndex (\(_,a) -> a == fn (snd $ unzip xs)) xs ) But it seems to me like there should be prettier solutions to this (that *do not* involve changing the definition of fn). After all, this doesn't seem like a rare pattern. Anyone know if in fact there's a better way to go about it?

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find by
by Julian Arni 15 May '13

15 May '13

I have a function that, simplifying a little, looks like this: fn :: [a] -> a fn = (head .) . takeWhile $ (\_ -> True) >From this, I want a function with the signature fn' :: [(x,a)] -> (x,a) such that: snd $ fn' (zip [x] [a]) = fn [a] I can see ways of doing this by altering fn, or by repeating some of fn in the definition of fn', or (because in this case I know that if fn xs = x, fn is returning the first x in xs and not any others), by doing something nasty like: fn' xs = xs !! fromMaybe 0 (findIndex (\(_,a) -> a == fn (snd $ unzip xs)) xs ) But it seems to me like there should be prettier solutions to this (that *do not* involve changing the definition of fn). After all, this doesn't seem like a rare pattern. Anyone know if in fact there's a better way to go about it?

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Please help with this basic lexical analyzer program, implemented using the "Alex" tool
by Costello, Roger L. 15 May '13

15 May '13

Hi Folks, I am learning how to use the Haskell lexical analyzer tool called "Alex" [1]. I am trying to implement a lexical analyzer for this string (an email "From:" header): From: "John Doe" <john(a)doe.org> I want to break it up into this list of tokens: [ From, DisplayName "John Doe", Email, LocalName "john", Domain "doe.org" ] Below is my implementation. It works fine if the string doesn't contain a display name. That is, this works fine: let s = "From: <john(a)doe.org>" alexScanTokens s However, when I include a display name, I get this error message: [From*** Exception: lexical error That is, this results in an error: let s = "From: \"John Doe\" <john(a)doe.org>" alexScanTokens s I am guessing that this part of my "Alex" program is causing the error: \"[a-zA-Z ]+\" { \s -> DisplayName (init (tail s)) } In Alex the left side is a regular expression: \"[a-zA-Z ]+\" and the right side is the action to be taken when a string is found that matches the regular expression: { \s -> DisplayName (init (tail s)) } Any thoughts on what the problem might be? Here is my lexical analyzer program: ------------------------------------------------- { module Main (main) where } %wrapper "basic" $digit = 0-9 -- digits $alpha = [a-zA-Z] -- alphabetic characters tokens :- $white+ ; From: { \s -> From } \"[a-zA-Z ]+\" { \s -> DisplayName (init (tail s)) } \< { \s -> Email } [$alpha]+@ { \s -> LocalPart (init s) } [$alpha\.]+> { \s -> Domain (init s) } { -- Each action has type :: String -> Token -- The token type: data Token = From | DisplayName String | Email | LocalPart String | Domain String deriving (Eq,Show) main = do s <- getContents print (alexScanTokens s) } ------------------------------------------------- [1] The "Alex" lexical analyzer tool may be found at this URL: http://www.haskell.org/alex/doc/html/introduction.html

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recvFrom return value different from length of resulting string
by Nathan Hüsken 11 May '13

11 May '13

Hey, I am trying to write a small haskell application which forwards udp messages. The complete program is here: http://hpaste.org/87681 Notice the lines: 43 (msg,bytes,addr) <- recvFrom sock 1024 44 putStrLn $ "Received: " ++ show bytes ++ " = " ++ (show . length $ msg) If I understand correctly, bytes is the number of bytes in the message. This should be äquivalent to (length msg), correct? Now I send an udp message (by connecting over the forwarded port with enet), and get this output: Received: 52 = 41 than I run it again, and get this output: Received: 52 = 43 Why is the length of msg not equal to 52? And why is it different from run to run? enet is unable to connect over this forward, by the way. Thanks! Nathan

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