- Beginners - Haskell.org

Applying a function to two lists
by Matt Williams 23 Apr '16

23 Apr '16

Dear List, I am stuck. I have a function that needs to apply each item of one list to every element of the second list in turn. So far, I have this function: checkNum :: Int -> [Int] -> (Int,[Int]) checkNum a b = (a,filter (check a) $ b) which implements what I need, but I now need to apply it to every element of the first list. I am looking for something like: list1 = [1,2,3,4,5,6] list2 = [1,2,3,4,5,6] map checkNum list1 list2 to return: [(1,[1]),(2[3,4,5]),(6,[3]) (I have tried to simplify this a little, so my apologies if it looks pointless - the real function is useful) Any help would be appreciated. Matt

2 3

i have questions about Haskell
by Eunsu Kim 23 Apr '16

23 Apr '16

Hi when outputting the polynomial value, actually write out the polynomial, but: - skipping any missing monomials - not including any extraneous signs -not showing the constant term for the above example, the final line would be: The value of 1.0 x^3 - 2.0 x^2 + 10.0 evaluated at -1.0 is 7.0 how can I do this??? i have no idea now…. here is my code: evalpoly = do putStr "What is the degree of polynomial: " degree <- getLine coeffs <- (funcOfCoeff ((read degree::Int)+1) [] ) putStr "What value do you want to evaluate at: " value <- getLine putStr "The value of the polynomial is: " putStrLn (show (getResult (coeffs) (read value :: Float) )) --function loop to get coefficient-- funcOfCoeff 0 coeffs = do --to check the degree of 0 return coeffs --return list of coefficient funcOfCoeff degree coeffs = do putStr ("What is the x^" ++ show(degree-1)) putStr " coefficient: " coeff <- getLine loop <- funcOfCoeff (degree-1) ((read coeff :: Float) : coeffs) return loop getResult (coeffs) x = sum(map(\(a,b) -> a*x^b).zip coeffs.iterate (+1)$0) this is my output so far: > evalpoly What is the degree of the polynomial: 3 What is the x^3 coefficient: 1.0 What is the x^2 coefficient: - 2.0 What is the x^1 coefficient: 0 What is the x^0 coefficient: 10.0 What value do you want to evaluate at: -1.0 The value of the polynomial is 7.0

2 1

I have simple question about Haskell
by Eunsu Kim 23 Apr '16

23 Apr '16

Hi THIS IS MY CODE: evalpoly = do putStr "What is the degree of polynomial: " degree <- getLine coeffs <- (funcOfCoeff ((read degree::Int)+1) [] ) putStr "What value do you want to evaluate at: " value <- getLine putStr "the value of " putStr (printChar coeffs) putStr (" evaluated at "++ value ++" is ") putStrLn (show (getResult (coeffs) (read value :: Float) )) printChar coeffs =parser([x | x <-reverse (zip coeffs (iterate (+1) 0)), fst x /= 0]) parser []="" parser (a:as)=show(fst a) ++ "x^" ++show(snd a) ++ " " ++(parser as) —PROBLEM IS HERE!!! ---function loop to get coefficient-- funcOfCoeff 0 coeffs = do --to check the degree of 0 return coeffs --return list of coefficient funcOfCoeff degree coeffs = do putStr ("What is the x^" ++ show(degree-1)) putStr " coefficient: " coeff <- getLine loop <- funcOfCoeff (degree-1) ((read coeff :: Float) : coeffs) return loop getResult (coeffs) x = sum(map(\(a,b) -> a*x^b).zip coeffs.iterate (+1)$0) --evaluate polynomial with value HERE IS MY OUTPUT: *Main> evalpoly What is the degree of polynomial: 2 What is the x^2 coefficient: 3 What is the x^1 coefficient: 2 What is the x^0 coefficient: 1 What value do you want to evaluate at: 7 the value of 3.0x^2 2.0x^1 1.0x^0 evaluated at 7 is 162.0 in output, on the very bottom part, there should be + or - sign there like this: 3.0x^2 + 2.0x^1 - 1.0x^0 what should I do? I have no idea now….. thanks

1 0

Haskell triangular loop (correct use of (++))
by rohan sumant 23 Apr '16

23 Apr '16

Suppose I have a list of distinct integers and I wish to generate all possible unordered pairs (a,b) where a/=b. Ex: [1,2,3,0] --> [(1,2),(1,3),(1,0),(2,3),(2,0),(3,0)] The approach I am following is this :- mkpairs [] = [] mkpairs (x:xs) = (map (fn x) xs) ++ (mkpairs xs) fn x y = (x,y) It is generating the desired output but I am a bit unsure about the time complexity of the function mkpairs. In an imperative language a nested triangular for loop would do the trick in O(n^2) or more precisely (n*(n-1)/2) operations. Does my code follow the same strategy? I am particularly worried about the (++) operator. I think that (++) wouldn't add to the time complexity since the initial code fragment (map (fn x) xs) is to be computed anyway. Am I wrong here? Is this implementation running O(n^2)? If not, could you please show me how to write a nested triangular loop in Haskell? Rohan Sumant

5 6

I have question about Haskell
by Eunsu Kim 22 Apr '16

22 Apr '16

Hi i have a problem in my code! here is my code: -- Baic I/O and Loop (50 Points) evalpoly = do putStr "What is the degree of polynomial: " degree <- getLine coeffs <- (funcOfCoeff ((read degree::Int)+1) []) putStr "What value do you want to evaluate at: " value <- getLine putStr "The value of the polynomial is: " putStr (show (polyEvaluate (coeffs) (read value :: Float) )) putStr "\n" --function loop to get coefficient-- funcOfCoeff 0 coeffs = do --to check the degree of 0 return coeffs --return list of coefficient funcOfCoeff degree coeffs = do putStr ("What is the x^" ++ show(degree-1)) putStr " coefficient: " coeff <- getLine loop <- funcOfCoeff (degree-1) ((read coeff :: Int) : coeffs) return loop polyEvaluate (coeffs) x = do powers <- zip coeffs (iterate (+1) 0) result <- map (\(a,b)-> a+b) powers —PROBLEM IS HERE!!!! return result here is error message: in very bottom function (polyEvaluate), why is not working “result <- map (\(a,b) -> a+b) powers” ??? in Prelude, it is working Thanks!

3 2

ghci :load vs import
by Doug McIlroy 19 Apr '16

19 Apr '16

I have module Powser stored in haskell/Powser.hs. There is no file ./Powser*. This loads the module shellprompt> ghci -ihaskell Prelude> :load Powser But this can't find it shellprompt> ghci -ihaskell Prelude> import Powser What might cause the difference? (I am running ghci 7.8.4.) Doug

3 3

ghci :load vs import
by Doug McIlroy 19 Apr '16

19 Apr '16

I have module Powser stored in haskell/Powser.hs. There is no file ./Powser*. This loads the module shellprompt> ghci -ihaskell Prelude> :load Powser But this can't find it shellprompt> ghci -ihaskell Prelude> import Powser What might cause the difference? (I am running ghci 7.8.4.) Doug

1 0

Parsec and Parsing
by mike h 18 Apr '16

18 Apr '16

Hi, I have two types, Tag = String Val = String and I want to creat expressions based on equality, t = v (where t is Tag, v is Val) and containment ie t IN v1, v2, v3, v4 ie a tag equals a value or a tag is contained in a list of values. Furthyermore I would like to apply AND and OR, i.e. (t1 = v OR t1 IN v1, v2, v3) AND (t2 = v2) ect etc I want to use Parsec and I'm ok with basic combinators but I want to get as far as making this a simple DSL but I don't want to use buildExpressionParser - at least not initially. I want to do this from first principles, understand what I'm doing and then maybe use buildExpressionParser. Any help would be really appreciated. Thanks Mike

1 0

Trying to install "Craft3e" from Hackage using stack
by Albin Otterhäll 16 Apr '16

16 Apr '16

I'm currently reading "Haskell: the Craft of Functional Programming", by Simon Thompson and I'm trying to download and install the package books code. But when I run "stack build Craft3e" I get the following error message: > Run from outside a project, using implicit global project config > Using resolver: lts-5.12 from implicit global project's config file: > /home/dumbl3d0re/.stack/global-project/stack.yaml > While constructing the BuildPlan the following exceptions were encountered: > > -- Failure when adding dependencies: > HUnit: needed (==1.2.*), 1.3.1.1 found (latest applicable is 1.2.5.2) > mtl: needed (>=1.1 && <2.2), 2.2.1 found (latest applicable is 2.1.3.1) > needed for package Craft3e-0.1.0.10 How should I go about to install "Craft3e"? Thanks in advance! Regards, Albin

2 4

Unicode characters in function names: some don't work?
by Silent Leaf 14 Apr '16

14 Apr '16

Hi! say I wanna use "—" as new infix operator: it's the big dash used a bit like parenthesis, especially at the end of sentences —like this. in ghci directly: Prelude> let (—) a b = a + b No problem, is accepted and usable. Same in files. Now I try using (»), a French (amongst others) punctuation sign, typically replaces the quote-ends, « like this ». Doesn't work: <interactive>:2:6: lexical error at character '\187' I thought Haskell was Unicode-friendly? Why some symbols but not others? :'(

2 5