May 2014 - Beginners - Haskell.org

Re: [Haskell-beginners] Haskell type Array a b
by Kevin Van Horn 27 May '14

27 May '14

On May 26, Brandon Allbery <allbery.b(a)gmail.com> wrote: > On Mon, May 26, 2014 at 5:12 PM, Kevin Van Horn <kevin(a)ksvanhorn.com> wrote: > I'm looking on the Haskell 98 Report, and the definition of (==) for values of type Array a b looks wrong to me. Here is the definition given: > > a == a' = (assocs a == assocs a') > > But this fails to account for the array length in each dimension; hence, if > > Have you tested this? Yes. For the arrays I defined I had assocs a = [ ], because the length in one dimension was equal to 0. So I had x == y but bounds x != bounds y. > I think if you're looking for anything more precise, you need to switch to something that can't be represented in H'98 or even H'2010, namely arrays with the bounds in the type. Why would you need that? Array equality could have been defined to be a1 == a2 = (bounds a1 == bounds a2) && (assocs a1 == assocs a2) and I don't understand why this is not the definition.

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Haskell type Array a b
by Kevin Van Horn 27 May '14

27 May '14

I'm looking on the Haskell 98 Report, and the definition of (==) for values of type Array a b looks wrong to me. Here is the definition given: a == a' = (assocs a == assocs a') But this fails to account for the array length in each dimension; hence, if x is a 0 x 2 array, y is a 2 x 0, and z is a 0 x 0 array, then x == y x == z, and y == z are all True, even though bounds x != bounds y, bounds x != bounds z, and bounds y != bounds z. This violates the most fundamental property of equality: that if a == b, then you can substitute a for b in an expression without changing the value of the expression. I can't possibly be the first person to have noticed this. So why is equality for Arrays defined this way?

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Evaluate function
by Derek McLoughlin 27 May '14

27 May '14

Reading "Parallel and Concurrent Programming in Haskell", at the bottom of page 27 the author wants to force the evaluation of a list of Strings: evaluate ( length puzzles ) Why not just evaluate puzzles ? https://github.com/simonmar/parconc-examples/blob/master/sudoku4.hs

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apply a function to all params of another function
by Kees Bleijenberg 27 May '14

27 May '14

In have a lot of functions in a program with params that are alle Int's. f :: Int -> Int -> Int g :: Int -> Int -> Int h :: Int -> Int -> Int -> Int ... and convertParam :: Int -> Int -- say (+1) I want to call functions like f, g and h, but apply convertParam to all params first. f2 a b c = f (convertParam a) (convertParam b) (convertParam c) I tried: run1p f conv = \a -> f (conv a) -- for functions with one param run2p f conv = \a b -> f (conv a) (conv b) -- for functions with two params . Can you do this in a more generalized way (using currying?) Any ideas? Kees

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A better way to "integrate"
by Dimitri DeFigueiredo 23 May '14

23 May '14

Awesome haskellers, I am coding up a little function that aggregates "ask orders" in a currency exchange. Another way to look at it, is that the function takes as input a histogram or fdf (in list format) and outputs the cumulative distribution cdf (also in list format). So we are kind of "integrating" the input list. When given a list of asks in order of increasing price, the function returns a list of points in the graph of the total supply curve. Here's an example: asks: returned list: [ (Price 42, Volume 0.5), [ (Price 21, Volume 0.5), (Price 50, Volume 1 ), (Price 21+50=71, Volume 1.5), (Price 55, Volume 0.2)] (Price 21+50+11=82,Volume 1.7)] the returned list gives us the total supply curve (price = y-axis, quantity/volume = x-axis, so the order is flipped) Summarizing * We're adding up the volumes. The last volume on the list is the total volume available for sale. * We calculate the total amount to be paid to buy the current volume (for each item in the list). I have written up a simple function to do this: aggregate :: Num a => [(a,a)] -> [(a,a)] aggregate xs = aggregate' 0 0 xs aggregate' :: Num a => a -> a -> [(a,a)] -> [(a,a)] aggregate' _ _ [] = [] aggregate' accX accY ((x,y):ls) = let accX' = accX + x * y accY' = accY + y in (accX',accY') : aggregate' accX' accY' ls main = print $ aggregate [(42,0.5),(50,1),(55,0.2)] However, this does not look very good to me and it feels like I'm reinventing the wheel. Question: Is there a better Haskell way to do this? I'm really anal about making it easy to read. Thanks! Dimitri

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using withAnnotations from Graphics.Rendering.Plot.Figure
by Albert Cardona 23 May '14

23 May '14

Hi all, given two Vector and a list of strings, I generate a plot of one vector as a function of the other [1]. At each x,y point, I am trying to place the corresponding string. For the purpose, I zip3 all three sequences, define an annotation function, and invoke it with mapM_ in the context of withAnnotations [2]: withPlot (1, 1) $ do ... let annotation (x, y, label) = text (x + 2, y) (setText label) in withAnnotations $ do mapM_ annotation $ zip3 (toList vx) (toList vy) labels The code above compiles and runs, the plot shows, but no text annotations appear at all. I understand that I am not using the Annote monad correctly: the mapM_ drops every returned value, so the state is not altered. What is the correct way to invoke withAnnotations over a list of (x, y, String) ? Needless to say, I am a beginner in Haskell. Apologies if the above is too trivial. Thanks very much in advance. Albert [1] https://github.com/acardona/haskell-lib/blob/master/CATMAID/Analysis/Graph/… [2] http://hackage.haskell.org/package/plot-0.2.3/docs/Graphics-Rendering-Plot-… PS: unfortunately I had sent this message to haskell-cafe; sorry! -- http://albert.rierol.net http://www.ini.uzh.ch/~acardona/

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Re: [Haskell-beginners] haskell on OpenBSD 5.5
by MJ Williams 22 May '14

22 May '14

Any reason for installing OpenBSD and not NetBSD or FreeBSD? I think you can run any Linux package including ghc on both of those. At 20:10 20/05/2014, you wrote: >Good day all, > >I installed OpenBSD 5.5 in order to learn haskell. >There are a few glitches however. I installed the haskell-platform >meta-package and the issues I noticed are: >- hs-vector wouldn't install >- ghci wouldn't start, the message: unable to load package `integer-gmp' > >Has anyone experienced the same issue? Is this happening on OpenBSD 5.4? >Is this OpenBSD specific? > >Thanks in advance. > > > >_______________________________________________ >Beginners mailing list >Beginners(a)haskell.org >http://www.haskell.org/mailman/listinfo/beginners

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Instance declaration needs more info?
by Dimitri DeFigueiredo 22 May '14

22 May '14

Hi All, I'm trying to write a simplified dimensional library where where quantities in meters, seconds and meters/second can all co-exist adjusting their respective units when multiplied and/or divided. Also, meter+meter is allowed, but meter+second should cause the type checker to complain. This is a bit like a *much* simplified version of the Units library. However, I am having trouble understanding why my instance declaration below appears to be under specified. Here's the code: ------------ module Dimensional where import qualified Prelude -- A Group allows you to add and subtract (but not multiply or divide) class Group a where (+) :: a -> a -> a (-) :: a -> a -> a x - y = x + negate y negate :: a -> a negate x = fromInteger 0 - x fromInteger :: Prelude.Integer -> a class Unit a where (*) :: (Unit a, Unit b, Unit c) => a -> b -> c (/) :: (Unit a, Unit b, Unit c) => a -> b -> c toDouble :: a -> Prelude.Double instance Unit Prelude.Double where (*) x y = (Prelude.*) x (toDouble y) -- <----- Error here (/) x y = (Prelude./) x (toDouble y) toDouble x = x ------------ GHC complains as follows: Could not deduce (c ~ Prelude.Double) from the context (Unit Prelude.Double, Unit b, Unit c) bound by the type signature for * :: (Unit Prelude.Double, Unit b, Unit c) => Prelude.Double -> b -> c at /code/haskell/dimensional.hs:25:5-43 `c' is a rigid type variable bound by the type signature for * :: (Unit Prelude.Double, Unit b, Unit c) => Prelude.Double -> b -> c at /code/haskell/dimensional.hs:25:5 In the first argument of `(Prelude.*)', namely `x' In the expression: (Prelude.*) x (toDouble y) In an equation for `*': * x y = (Prelude.*) x (toDouble y) Any pointers would be much appreciated! Thanks, Dimitri

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Haskell network sample
by yang.zhao 21 May '14

21 May '14

hi guys, I'm newbie, just begin to learn Haskell. now i write a very simple server and client . Server: import Control.Concurrent import System.IO import Network port :: Int port = 1234 main :: IO () main = withSocketsDo $ do s <- listenOn $ PortNumber $ fromIntegral port (h, _, _) <- accept s sline <- hGetLine h hPutStrLn h sline putStrLn $ "send "++sline threadDelay 1000000 hClose h sClose s Client : import System.IO import Network port :: Int port = 1234 main :: IO () main = withSocketsDo $ do h <- connectTo "localhost" $ PortNumber $ fromIntegral port hPutStrLn h "hello" bs <- hGetContents h putStrLn bs hClose h And, it doesn't work now . I run ./serv , then run ./cli , they will block all the time. but, when i run ./serv, and telnet localhost 1234 in another terminal, it works fine. so i don't know what's the wrong with my code. anybody can tell me about my problem? os is Debian 7, haskell-platform 2012.2.0.0 thanks a lot!!! -- K.I.S.S.

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haskell on OpenBSD 5.5
by Michael S 21 May '14

21 May '14

Good day all, I installed OpenBSD 5.5 in order to learn haskell. There are a few glitches however. I installed the haskell-platform meta-package and the issues I noticed are: - hs-vector wouldn't install - ghci wouldn't start, the message: unable to load package `integer-gmp' Has anyone experienced the same issue? Is this happening on OpenBSD 5.4? Is this OpenBSD specific? Thanks in advance.

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