- Beginners - Haskell.org

parsec and source material with random order lines
by Emmanuel Touzery 26 Dec '12

26 Dec '12

Hi, I'm trying to parse ical files but the source material doesn't matter much. First, I know there is an icalendar library on hackage, but I'm trying to learn as well through this. Now the format is really quite simple and actually I'm parsing it, it works, but I don't like the code I'm writing, it feels wrong and I'm sure there is a better way. Actually for now I'm parsing it to an array of arrays, but I want to fill a proper "data" structure. For my purpose the file contains a bunch of records like this: BEGIN:VEVENT DTSTART:20121218T103000Z DTEND:20121218T120000Z [..] DESCRIPTION: [..] END:VEVENT There are a bunch of records I don't care about and also I want to parse no matter what is the order of directives (so, i want to parse also if DTEND appears before DTSTART for instance, and so on). That last part is my one problem. I can't do: parseBegin start <- parseStart end <- parseEnd skipRows desc <- parseDesc skipRows end <- parseEnd return Event { eventStart = start, eventEnd = end ...} my current working code is: parseEvent = do parseBegin contents <- many1 $ (try startDate) <|> (try endDate) <|> (try description) <|> unknownCalendarInfo parseEnd return contents But then contents of course returns an array, while I want to return only one element here. SOMEHOW what I would like is: parseEvent = do parseBegin contents <- many1 $ (start <- T.try startDate) <|> (end <- T.try endDate) <|> (desc <- T.try description) <|> unknownCalendarInfo parseEnd return Event { eventStart = start, eventEnd = end ...} But obviously as far as Parsec is concerned startDate could occur several times and also it's just not valid Haskell syntax. So, any hint about this problem? Parsing multi-line records with Parsec, when I don't know the order in which the lines will appear? I mean sure I can convert my array to the proper data structure... I find which element in the array contains the start date and then which contains the end date... and build my data structure.. But I'm sure something much nicer can be done... I just can't find how. I see the author of iCalendar fixed the problem but I can't completely understand his source, it's too many things at the same time for me, I need to take this one step at a time. Thank you! Emmanuel

3 4

Cartesian Product in Standard Haskell Libraries
by Jay Sulzberger 26 Dec '12

26 Dec '12

I need the usual Cartesian product of a list of lists. In the posts http://stackoverflow.com/questions/4119730/cartesian-product http://stackoverflow.com/questions/3387359/calculate-n-ary-cartesian-product it is recommended to use the sequence procedure, which is in the standard Prelude, as a "Cartesian product of a list of lists" function. That is we do have that sequence [["a", "b"], ["c", "d"]] evaluates to [["a","c"],["a","d"],["b","c"],["b","d"]] and often, sequence seems to give the expected answer. But we have also that, and here we quote from a GHCi session, > sequence [] [] it :: [()] This looks to me to be a violation of the rule that the Cartesian product of an empty list of lists is a list with one element in it. It looks to be a violation because "[]" looks like a name for an empty list. But we also have > length (sequence []) 1 it :: Int which almost reassures me. Appended are a Scheme session and a GHCi session showing some of what puzzles me. I remember when I wrote my now standard Cartesian product Scheme procedure I was delighted that, we get > (cprd '()) (()) which gives '() as the single element of the Cartesian product of an empty list of lists. ad Old Types vs New: Indeed every particular application of the Scheme procedure cprd should be to (some instance of) some "type" of lists of lists, and thus the one element in the output, in case the list of lists is the empty list, should be of the type of the elements, that is, the inner lists, of the input list of lists. In Haskell I do not know enough to say, but, certainly the behavior of sequence-plus-GHCi-plus-printing-conventions was surprising to me. Oi. Perhaps an/the object of type () just does not get printed clearly. So a list with only one object, and that one object of type (), gets printed on the page as [] which is a text name for the list, which is of length one, but this text name looks on the page just like the text name of an empty list. oo--JS. <blockquote what="Scheme session showing my standard version of Cartesian product" date="Monday 24 December 2012 01:30:37 -0500"> SCM version 5e5, Copyright (C) 1990-2006 Free Software Foundation. SCM comes with ABSOLUTELY NO WARRANTY; for details type `(terms)'. This is free software, and you are welcome to redistribute it under certain conditions; type `(terms)' for details. ;loading /usr/share/slib/require ;done loading /usr/share/slib/require.scm ;loading /usr/share/slib/require ;done loading /usr/share/slib/require.scm ;loading /usr/lib/scm/Link ;done loading /usr/lib/scm/Link.scm ;loading /usr/lib/scm/Transcen ;done loading /usr/lib/scm/Transcen.scm > (define cartesian:multiply-by-one-list (lambda (multiplier old-product) (if (equal? multiplier '()) '() (if (equal? (cdr multiplier) '()) (map (lambda (x) (cons (car multiplier) x)) old-product) (append (map (lambda (x) (cons (car multiplier) x)) old-product) (cartesian:multiply-by-one-list (cdr multiplier) old-product)))))) #<unspecified> > (define cartesian:product (lambda (l) (if (equal? l '()) (list '()) (cartesian:multiply-by-one-list (car l) (cartesian:product (cdr l)))))) #<unspecified> > (define cprd cartesian:product) #<unspecified> > (cprd '()) (()) > (cprd '(())) () > (cprd '((a))) ((a)) > (cprd '((a b))) ((a) (b)) > (cprd '(() (a))) () > (cprd '((a) ())) () > (cprd '((a) (b))) ((a b)) > (cprd '((a b) (c d))) ((a c) (a d) (b c) (b d)) > (cprd '((a b) (c d) (e) (f g h))) ((a c e f) (a c e g) (a c e h) (a d e f) (a d e g) (a d e h) (b c e f) (b c e g) (b c e h) (b d e f) (b d e g) (b d e h)) > (cprd '(() ())) () > ;; OK, this looks right. On lists of lists of symbols ;; cprd seems to be a version of Cartesian product. (quit) ;EXIT Process scheme finished </blockquote> <blockquote what="GHCi session showing sequence and an untested version of a Haskell Cartesian product function" point="To me sequence's behavior on the list of lists [] is puzzling. sequence seems not to obey the rule that the Cartesian product of an empty list of lists is a list with one element in it, whose name/manifestation, in general, might be hard to guess." date="Monday 24 December 2012 01:32:56 -0500"> GHCi, version 7.4.1: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer-gmp ... linking ... done. Loading package base ... linking ... done. Prelude> :set +t Prelude> :set prompt "> " > sequence [] [] it :: [()] > sequence [[]] [] it :: [[a]] > sequence [["a"]] [["a"]] it :: [[[Char]]] > sequence [["a", "b"]] [["a"],["b"]] it :: [[[Char]]] > sequence [[], ["a"]] [] it :: [[[Char]]] > sequence [["a"], []] [] it :: [[[Char]]] > sequence [["a"], ["b"]] [["a","b"]] it :: [[[Char]]] > sequence [["a", "b"], ["c", "d"]] [["a","c"],["a","d"],["b","c"],["b","d"]] it :: [[[Char]]] > sequence [["a", "b"], ["c", "d"], ["e"], ["f", "g", "h"]] [["a","c","e","f"],["a","c","e","g"],["a","c","e","h"],["a","d","e","f"],["a","d","e","g"],["a","d","e","h"],["b","c","e","f"],["b","c","e","g"],["b","c","e","h"],["b","d","e","f"],["b","d","e","g"],["b","d","e","h"]] it :: [[[Char]]] > length it 12 it :: Int > 2 * 2 * 1 * 3 12 it :: Integer > sequence [[], []] [] it :: [[a]] > -- OK, it looks as though 'sequence' as suggested in the > -- Stackoverflow post > -- http://stackoverflow.com/questions/4119730/cartesian-product > -- in Answer 10, does most of the time compute a version > -- of Cartesian product very much as does the Scheme version. > -- Let us do the example in the answer: > sequence [[1,2,3],[4,5,6]] [[1,4],[1,5],[1,6],[2,4],[2,5],[2,6],[3,4],[3,5],[3,6]] it :: [[Integer]] > length it 9 it :: Int > 3 * 3 9 it :: Integer > -- But this seems to violate the convention that > -- Cartesian product, when handed an empty list of lists > -- returns a list with one element. This convention > -- makes the 'length' function a homomorphism from the monoid > -- of lists, with cartesian product as multiplication, and the empty list > -- as the identity element, to the monoid of non-negative integers > -- with '*' as multiplication, and '1' as the identity element. > -- Let us have a look again as what 'sequence' does: > sequence [] [] it :: [()] > length it 0 it :: Int > sequence [[]] [] it :: [[a]] > length it 0 it :: Int > length (sequence []) 1 it :: Int > :t (sequence []) (sequence []) :: Monad m => m [a] > :t (sequence [[]]) (sequence [[]]) :: [[a]] > -- Let us try a version of our Scheme procedure 'cprd'. > let hmbol multiplier oldproduct = [ x:xs | x <- multiplier, xs <- oldproduct] hmbol :: [a] -> [[a]] -> [[a]] > hmbol [] [["a"]] [] it :: [[[Char]]] > hmbol [] [[]] [] it :: [[a]] > hmbol ["a"] [[]] [["a"]] it :: [[[Char]]] > hmbol ["a", "b"] [[]] [["a"],["b"]] it :: [[[Char]]] > hmbol ["a", "b"] [] [] it :: [[[Char]]] > hmbol ["a", "b"] [["c"]] [["a","c"],["b","c"]] it :: [[[Char]]] > hmbol ["a", "b"] [["c"], ["d", "e"]] [["a","c"],["a","d","e"],["b","c"],["b","d","e"]] it :: [[[Char]]] > hmbol ["a", "b"] [["c"], [], ["d", "e"]] [["a","c"],["a"],["a","d","e"],["b","c"],["b"],["b","d","e"]] it :: [[[Char]]] > let hcprod = foldr hmbol [[]] hcprod :: [[a]] -> [[a]] > hcprod [] [[]] it :: [[a]] > hcprod [[]] [] it :: [[a]] > hcprod [["a", "b"], ["c"], [], ["d", "e"]] [] it :: [[[Char]]] > -- Let us repeat all the tests of 'cprod' in the Scheme session: > hcprod [] [[]] it :: [[a]] > hcprod [[]] [] it :: [[a]] > hcprod [["a"]] [["a"]] it :: [[[Char]]] > hcprod [["a", "b"]] [["a"],["b"]] it :: [[[Char]]] > hcprod [[], ["a"]] [] it :: [[[Char]]] > hcprod [["a"], []] [] it :: [[[Char]]] > hcprod [["a"], ["b"]] [["a","b"]] it :: [[[Char]]] > hcprod [["a", "b"], ["c", "d"]] [["a","c"],["a","d"],["b","c"],["b","d"]] it :: [[[Char]]] > hcprod [["a", "b"], ["c", "d"], ["e"], ["f", "g", "h"]] [["a","c","e","f"],["a","c","e","g"],["a","c","e","h"],["a","d","e","f"],["a","d","e","g"],["a","d","e","h"],["b","c","e","f"],["b","c","e","g"],["b","c","e","h"],["b","d","e","f"],["b","d","e","g"],["b","d","e","h"]] it :: [[[Char]]] > length it 12 it :: Int > 2 * 2 * 1 * 3 12 it :: Integer > hcprod [[], []] [] it :: [[a]] > -- My two puzzles are: > -- Why does > -- sequence [] > -- evaluate to > -- [] > -- and not a list with one element? > -- Why is the type of the thing that > -- sequence [] > -- evaluates to > -- [()] > -- ? > :q Leaving GHCi. Process haskell finished </blockquote>

3 10

Overlapping instances
by Nathan Hüsken 26 Dec '12

26 Dec '12

Hi, I have a test file: module Main where class A a where doSomething :: a -> IO () class B b where doMore :: b -> IO () instance B b => A b where doSomething = doMore instance A String where doSomething = putStrLn main = doSomething "Hello, World! So there is a class A, and a class B. If something is part of B it is automatically part of A (so A is kind of a superclass to B). But String is just part of A. I try to compile it with: ghc Test.hs -XFlexibleInstances -XUndecidableInstances I get: Test.hs:14:8: Overlapping instances for A [Char] arising from a use of `doSomething' Matching instances: instance B b => A b -- Defined at Test.hs:8:10 instance A String -- Defined at Test.hs:11:10 In the expression: doSomething "Hello, World!" In an equation for `main': main = doSomething "Hello, World!" Why are the instances overlapping? String is not part of B??? How can I do this? Thanks! Nathan

4 4

Fwd: deep seq and bang patterns
by Emmanuel Touzery 25 Dec '12

25 Dec '12

Hello, On Tue, Dec 25, 2012 at 3:39 PM, Daniel Fischer < daniel.is.fischer(a)googlemail.com> wrote: > > Defining TvShow with strict fields for title, startTime and summary makes > sure > these fields are evaluated (to WHNF, but in case of `Data.Text`, that means > fully evaluated) **when the TvShow value is evaluated to WHNF**. > how do I know what does "evaluated to WHNF" means for every possible type? For instance, what does that mean for "String"? What does that mean for "Maybe String"? If I want the field to be fully evaluated, is !(Maybe String) enough or should I even do !(Maybe !String) or something more? Any source of information where I can read about this in depth would be welcome (I have yet to read Real World Haskell, maybe it's covered there?). Thank you! Emmanuel

3 2

deep seq and bang patterns
by Emmanuel Touzery 25 Dec '12

25 Dec '12

Hello, laziness just gave me some headache now. Because some of my code was evaluated much later, well first the memory use exploded but also I got an error which was really difficult to understand because it manifested much later than I would intuitively expect (when I read the record field, not when I wrote it...). Now, if nothing else for space usage needs, I need to make this strict. At first I tried bang patterns on my record: {-# LANGUAGE BangPatterns #-} data TvShow = TvShow { channel :: Channel, title :: !T.Text, startTime :: !T.Text, summary :: !T.Text } deriving (Eq, Show) That did not help. Then I tried deepseq: instance NFData TvShow return $!! result That did not help. And here's the catch: doing BOTH helps... I mean I think I'll check it again because I find it hard to believe but right now it really seems it behaves like that... Is that possible? That I need to combine BOTH deepseq and bang patterns to actually get my code to evaluate when filling in the data? Or am I going crazy? Thank you! Emmanuel

3 6

Christmas
by Julien 24 Dec '12

24 Dec '12

Happy christmas (Joyeux Noël !) Sent from my iPhone

1 0

{..} syntax
by Emmanuel Touzery 24 Dec '12

24 Dec '12

Hello, I see the {..} syntax in haskell code and at this point I don't understand what it means.. Any hint? The best would maybe be a link to the relevant "real world haskell" or "learn you a haskell" pages, if they exist. Unfortunately as soon as you have such special characters, google is pretty useless... Thanks! Emmanuel

1 1

How does this function append the log to the beginning of the list?
by jugree＠lavabit.com 24 Dec '12

24 Dec '12

Hello. Could you explain this example(0)? Could you show its step by step execution? gcd' :: Int -> Int -> Writer (DiffList String) Int gcd' a b | b == 0 = do tell (toDiffList ["Finished with " ++ show a]) return a | otherwise = do result <- gcd' b (a `mod` b) tell (toDiffList [show a ++ " mod " ++ show b ++ " = " ++ show (a `mod` b)]) return result Why does the above append the log to the beginning of the list? What value will result have in the following? result <- gcd' 2 (3 `mod` 2) (0) http://learnyouahaskell.com/for-a-few-monads-more#writer

3 3

process with mutable state in haskell
by Alexander _ 24 Dec '12

24 Dec '12

Hello, Can i run process with some mutable state and communicate with it in haskell? Something like gen_server in erlang. Thank you.

3 2

programming novice :: chess engine interest - request for resource suggestions.
by B F 23 Dec '12

23 Dec '12

Dear All, So far, I have experienced nothing but a warm reception to the Haskell community. I am new to programming (a dabbler in Python), a mathematics grad. student and a chess enthusiast. As a personal goal, I would like to learn Haskell by exploring ideas in chess evaluation development from the statistical and pattern similarity side (rather than simply max ply search). As I will be learning principles of (functional) programming via Haskell - in the context of chess related concepts- I would like to request some Haskell resources that are explicative enough to suit the novice, while at the same time tailored to learning concepts related to pattern analysis, array methods, and matrix manipulation. Many Many Thanks - in advance- Diabellical

3 2