November 2010 - Beginners

Pointfree expeiments
by Paul Sargent 25 Nov '10

25 Nov '10

I'd been playing with pointfree, looking at what certain patterns transformed into, and I came across this: pauls@Mudskipper> pointfree -v "\x -> x + x" ~ Transformed to pointfree style: ap (+) id Optimized expression: ap (+) id join (+) Basically I wanted to see what happened when you used a parameter multiple times within a function. Intrigued by this, I did the following: Prelude Control.Monad> :t ap (+) id ap (+) id :: (Num a, Monad ((->) a)) => a -> a I'm aware that 'ap' is related the liftM, but what is this monad, and why are we working in a monad at all? ...and surely this isn't the same as the original code as we now have a different type signature?

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Re: Problem "grouping" a list
by Hein Hundal 25 Nov '10

25 Nov '10

> From: Jo?o Paulo Pizani Flor <joaopizani(a)gmail.com> > OK, so my specific problem goes like this: I have a list of > tuples > :t myList > [ (Float, Integer) ] > > And I want to "group" this list into a nested list > groupAtoms :: [ (Float,Integer) ] -> [ > [(Float,Integer)] ] > > Of course, I want that the concatenation of the groups > yield me the original > list, i.e, (foldl (++) [] . groupAtoms == id), and > the criterion that > defines a group is that: > "The sum of the first elements in the tuples comprising the > list must be > greater than or equal to 1.0". Jo?o, I am a beginner also and I liked Daniel Fischer's solution. Here is another (less elegant) solution: -- Start Code Here -- foldfun takes an existing list of lists of pairs and decides -- whether to append p to the current list, pl, or start a new -- list of pairs. foldfun done p (pl:pls) = if done pl then [p]:pl:pls else (p:pl):pls foldfun done p [] = [[p]] -- mygroup2 folds list of pairs, pl, using the foldfun, and then -- puts all the lists in the correct order using reverse twice. mygroup2 pl = map reverse $ reverse $ foldr (foldfun done) [] (reverse pl) where done pl = sum (map fst pl) >= 1.0 -- End of Code Cheers, Hein

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Re: [Haskell-beginners] Problem "grouping" a list
by João Paulo Pizani Flor 24 Nov '10

24 Nov '10

Thank you all for the helpful responses! The solution from Daniel Fischer worked like a charm, the solution from Aai kinda worked, but I can't use because I don't want to modify the order of the elements in the list... And yes, my "criterion" was a bit ill-defined, so the very clever function from Ozgur satisfied it :P The real goal is to maximize the number of groups though. I'll test the solutions more hardly today and try to make the functions from Daniel even more clearer and elegant. I post the results back here! Thanks a lot! João Paulo Pizani Flor joaopizani(a)gmail.com Federal University of Santa Catarina 2010/11/23 Daniel Fischer <daniel.is.fischer(a)web.de> > On Tuesday 23 November 2010 18:23:28, João Paulo Pizani Flor wrote: > > Hello dear Haskellers! > > > > I've been a user and big fan of Haskell for a while, but only now I'm > > writing my first "big" project in Haskell (some thousands of lines of > > code perhaps). It's an interpreter for a programming language, the > > source code is music! Yes, sheet music! :D > > > > OK, so my specific problem goes like this: I have a list of tuples > > > > :t myList > > > > [ (Float, Integer) ] > > > > And I want to "group" this list into a nested list > > groupAtoms :: [ (Float,Integer) ] -> [ [(Float,Integer)] ] > > > > Of course, I want that the concatenation of the groups yield me the > > original list, i.e, (foldl (++) [] . groupAtoms == id), > > Here, foldr is the better choice. (++) can start delivering its result > before inspecting its second argument, hence > > foldr (++) [] (list1 : list2 : list3 : ...) > = list1 ++ (foldr (++) [] (list2 : list3 : ...)) > = list1 ++ (list2 ++ (foldr (++) [] (list3 : ...))) > > can be consumed as it is constructed, so it can run in constant space, and > it can stop early if not the entire result is needed. And it can deal with > infinite lists. > > The prelude function concat does this. > > foldl on the other hand can't deliver anything before the whole input list > has been traversed, in particular it doesn't terminate on infinite input, > it needs at least O(length result) space and it can't stop early. > Additionally, foldl (++) [] constructs left-nested (++)-applications, which > are bad for performance. > > > and the criterion that defines a group is that: > > "The sum of the first elements in the tuples comprising the list must be > > greater than or equal to 1.0". That is, given a list of tuples, the > > boolean predicate deciding whether this list is a PROPER group (True) or > > TOO SMALL (False) is: > > \g -> sum (map fst g) >= 1.0 > > > > > > Although the criterion is very clear, I've tried hard until now and > > couldn't come up with a function for producing the nested list based on > > this grouping criterion. > > Split the task into two parts. > > 1. collect the first group and the remainder of the list as a pair of lists > 2. cons the first group to what recursion on the remainder delivers > > Very much like the definition of `lines'. > > For your use-case, the condition when a group is complete doesn't only > depend on the list element but also on the previous, concretely the running > sum of the first components. > For a general such function, you need > - the stopping condition > - an accumulation parameter > - an update function for the accumulating parameter > as arguments. > For example (stupid names, sorry) > > > breakAcc :: (b -> Bool) -> (a -> b -> b) -> b -> [a] -> ([a],[a]) > breakAcc _ _ _ [] = ([],[]) > breakAcc cond comb acc xxs@(x:xs) > | cond acc = ([],xxs) > | otherwise = (x:ys,zs) > where > (ys, zs) = breakAcc cond comb (comb x acc) xs > > solves part 1, then > > > groupsAcc :: (b -> Bool) -> (a -> b -> b) -> b -> [a] -> [[a]] > groupsAcc _ _ _ [] = [] > groupsAcc cond comb acc xs = let (g,tl) = breakAcc cond comb acc xs > in g : groupsAcc cond comb acc tl > > gives you part 2, and your particular function is > > > groupAtoms :: [(Float,Integer)] -> [[(Float,Integer)]] > groupAtoms = groupsAcc (>= 1.0) ((+) . fst) 0 > > > Of course the last group need not be proper. > If there are long groups, the above can cause a space leak (as lines had in > GHC < 7), that can be avoided by sacrificing a little non-strictness and > replacing the let (g,tl) = ... in with a > > case ... of > (g,tl) -> g : groupsAcc ... > > or, if you need the non-strictness, with a slightly convoluted method as > used for lines in GHC 7. > > > I am sure that the Haskell Hierarchical > > Libraries have something to help me, but only now I see I'm still a big > > noob :P > > > > Could someone please help me writing this function? > > > > > > My best regards from Brazil, > > > > João Paulo Pizani Flor > > joaopizani(a)gmail.com > > Federal University of Santa Catarina > >

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Problem "grouping" a list
by João Paulo Pizani Flor 23 Nov '10

23 Nov '10

Hello dear Haskellers! I've been a user and big fan of Haskell for a while, but only now I'm writing my first "big" project in Haskell (some thousands of lines of code perhaps). It's an interpreter for a programming language, the source code is music! Yes, sheet music! :D OK, so my specific problem goes like this: I have a list of tuples :t myList [ (Float, Integer) ] And I want to "group" this list into a nested list groupAtoms :: [ (Float,Integer) ] -> [ [(Float,Integer)] ] Of course, I want that the concatenation of the groups yield me the original list, i.e, (foldl (++) [] . groupAtoms == id), and the criterion that defines a group is that: "The sum of the first elements in the tuples comprising the list must be greater than or equal to 1.0". That is, given a list of tuples, the boolean predicate deciding whether this list is a PROPER group (True) or TOO SMALL (False) is: \g -> sum (map fst g) >= 1.0 Although the criterion is very clear, I've tried hard until now and couldn't come up with a function for producing the nested list based on this grouping criterion. I am sure that the Haskell Hierarchical Libraries have something to help me, but only now I see I'm still a big noob :P Could someone please help me writing this function? My best regards from Brazil, João Paulo Pizani Flor joaopizani(a)gmail.com Federal University of Santa Catarina

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Portability of a Haskell program
by Ondřej Garncarz 22 Nov '10

22 Nov '10

I want to run my Haskell program which I've compiled on other computer, so I try: $ ./eternitypar ./eternitypar: error while loading shared libraries: libgmp.so.10: cannot open shared object file: No such file or directory $ LD_PRELOAD=./libgmp.so.10 ./eternitypar ./eternitypar: /lib/libc.so.6: version `GLIBC_2.7' not found (required by ./libgmp.so.10) $ LD_PRELOAD=./libc.so:./libgmp.so.10 ./eternitypar ERROR: ld.so: object './libc.so' from LD_PRELOAD cannot be preloaded: ignored. ./eternitypar: /lib/libc.so.6: version `GLIBC_2.7' not found (required by ./libgmp.so.10) $ LD_PRELOAD=./ld-linux.so.2:./libc.so:./libgmp.so.10 ./eternitypar ERROR: ld.so: object './ld-linux.so.2' from LD_PRELOAD cannot be preloaded: ignored. ERROR: ld.so: object './libc.so' from LD_PRELOAD cannot be preloaded: ignored. ./eternitypar: /lib/libc.so.6: version `GLIBC_2.7' not found (required by ./libgmp.so.10) So I download a package from http://new-www.haskell.org/ghc/download_ghc_7_0_1#x86_64linux and I try: $ ./configure checking for path to top of build tree... utils/ghc-pwd/ghc-pwd: error while loading shared libraries: libgmp.so.3: cannot open shared object file: No such file or directory configure: error: cannot determine current directory Why is it looking for such an old version, when 7.0.1 should be up-to-date? But fine, I can replace ghc-pwd with symlinked /bin/pwd, but I have a problem with some pgcc installed on the machine, so I run ./configure --with-gcc=/usr/bin/gcc: (/usr/bin/gcc exists and works) checking for path to top of build tree... /u/gar061/tmp/ghc-7.0.1 checking for perl... /usr/bin/perl checking if your perl works in shell scripts... yes checking for a BSD-compatible install... /usr/bin/install -c checking whether ln -s works... yes checking for ar... /usr/bin/ar checking whether /usr/bin/ar is GNU ar... yes checking for ar arguments... q checking for gcc... /tools/pgi725/linux86-64/7.2-5/bin/pgcc checking for C compiler default output file name... a.out checking whether the C compiler works... yes checking whether we are cross compiling... no checking for suffix of executables... checking for suffix of object files... o checking whether we are using the GNU C compiler... no checking whether /tools/pgi725/linux86-64/7.2-5/bin/pgcc accepts -g... yes checking for /tools/pgi725/linux86-64/7.2-5/bin/pgcc option to accept ISO C89... none needed checking whether ranlib is needed... no checking for sed... /bin/sed configure: error: gcc is required It's ignoring my choice of GCC. Installation of 6.12.3 runs the same way. Then I move to compilation from source codes, but here - http://new-www.haskell.org/ghc/download_ghc_7_0_1#sources - I read: "The source distribution needs an installed GHC (version 6.10 at least)." Oh, so I need the application to build the application? What's the reason? Is there any other way to make it work? Thanks, Ondřej Garncarz

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Parsing an integer
by Tim Baumgartner 22 Nov '10

22 Nov '10

Hi folks, since I always get great answers here, I have another question. I'm currently parsing a (positive) integer using data Token = Number Int -- ... number :: Parser Token number = many1 digit ↠ (read >>> Number >>> return) -- = many1 digit ↠ return∘Number∘read But I wonder what an advanced haskeller would code instead. Particularly, I'd like to get rid of the 'return'. Thanks in advance Tim

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Ignoring the result of a monadic computation
by Tim Baumgartner 21 Nov '10

21 Nov '10

Hi, while learning about monads, I had something like do line <- getLine something putStrLn line and I wondered if I could write it in one line, without naming of parameters. I finally came up with getLine >>= ignore something >>= putStrLn using ignore :: Monad m => m a -> b -> m b ignore m a = m >> return a I'm satisfied with this solution but searching hoogle I didn't find a standard function for my ignore. Am I missing something? Cheers, Tim -- GRATIS! Movie-FLAT mit über 300 Videos. Jetzt freischalten unter http://portal.gmx.net/de/go/maxdome

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Chain computations
by Tim Baumgartner 20 Nov '10

20 Nov '10

Hi, I really like fooling around with operators. In elementary school, we always had to make chain computations, e.g. teacher: 2 teacher: +1 teacher: *2 teacher: divided by 3 Again, I didn't find the following operator on hoogle (very similar to ($)): (->>) :: a -> (a -> b) -> b a ->> f = f a Using it, we can nicely write down the above computation: 2 ->> (+1) ->> (*2) ->> (`div` 3) Cheers Tim

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Rekommendation for select i GUI
by Anders Persson 19 Nov '10

19 Nov '10

Hi! i have installed Haskell Platform om a Windows Vista, I will also run Haskell on a OS/X machine. I have looked at a good GUI tool, tested wxHaskell, and hs2GTK but no one seems easy to be install, lot off errors, etc... Have i missed something and could i get some comments om a GUI toolkit to select. Best regards Anders -- med vänliga hälsningar Anders ______________________________________________ Anders Persson anders.u.persson(a)gmail.com

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Re: [Haskell-cafe] About "Fun with type functions" example
by Bastian Erdnüß 19 Nov '10

19 Nov '10

On Nov 19, 2010, at 1:07, Daniel Peebles wrote: > The best you can do with fromInt is something like Int -> (forall n. (Nat n) > => n -> r) -> r, since the type isn't known at compile time. Or you put it in the type class as fromInt :: Int -> n and do in Zero: fromInt _ = Zero and in the other: fromInt _ = Succ $ fromInt (undefined :: n). Shouldn't that work, too? > On Thu, Nov 18, 2010 at 2:52 PM, Arnaud Bailly <arnaud.oqube(a)gmail.com>wrote: > >> Thanks a lot, that works perfectly fine! >> Did not know this one... >> BTW, I would be interested in the fromInt too. >> >> Arnaud >> >> On Thu, Nov 18, 2010 at 8:22 PM, Erik Hesselink <hesselink(a)gmail.com> >> wrote: >>> On Thu, Nov 18, 2010 at 20:17, Arnaud Bailly <arnaud.oqube(a)gmail.com> >> wrote: >>>> Hello, >>>> I am trying to understand and use the Nat n type defined in the >>>> aforementioned article. Unfortunately, the given code does not compile >>>> properly: >>> >>> [snip] >>> >>>> instance (Nat n) => Nat (Succ n) where >>>> toInt _ = 1 + toInt (undefined :: n) >>> >>> [snip] >>> >>>> And here is the error: >>>> >>>> Naturals.hs:16:18: >>>> Ambiguous type variable `n' in the constraint: >>>> `Nat n' arising from a use of `toInt' at Naturals.hs:16:18-39 >>>> Probable fix: add a type signature that fixes these type variable(s) >>> >>> You need to turn on the ScopedTypeVariables extension (using {-# >>> LANGUAGE ScopedTypeVariables #-} at the top of your file, or >>> -XScopedTypeVariables at the command line). Otherwise, the 'n' in the >>> class declaration and in the function definition are different, and >>> you want them to be the same 'n'. >>> >>> Erik >>> >> _______________________________________________ >> Haskell-Cafe mailing list >> Haskell-Cafe(a)haskell.org >> http://www.haskell.org/mailman/listinfo/haskell-cafe >> > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe(a)haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe

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