[Haskell-cafe] Is this haskelly enough?

[Thomas Hartman]

hartthoma at linuxpt:~/ProjectRepos/learning$ ghc -fglasgow-exts -e 'main' 
maxSubArrays.hs
should be [2,5,-1,3]:
[2,5,-1,3]
hartthoma at linuxpt:~/ProjectRepos/learning$ cat maxSubArrays.hs
import Data.List
-- maximum sub-array:  [2, 5, -1, 3]
main = do putStrLn $ "should be " ++ show [2, 5, -1, 3] ++ ":"
          putStrLn $ show $ maxsubarray [-1, 2, 5, -1, 3, -2, 1]

maxsubarray :: forall a. (Ord [a], Ord a, Num a) => [a] -> [a]
maxsubarray a = head $ reverse $ sortBy comparelists $ sublists a

comparelists l1 l2 = compare (sum l1) (sum l2)
sublists a = nub $ sort $ concat $ map inits $ tails a
hartthoma at linuxpt:~/ProjectRepos/learning$

cheers :)

t.




James Hunt <james at j-hunt.co.uk> 
Sent by: haskell-cafe-bounces at haskell.org
07/17/2007 04:26 PM

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Subject
[Haskell-cafe] Is this haskelly enough?






Hi,

As a struggling newbie, I've started to try various exercises in order 
to improve. I decided to try the latest Ruby Quiz 
(http://www.rubyquiz.com/quiz131.html) in Haskell. Would someone be kind 
enough to cast their eye over my code? I get the feeling there's a 
better way of doing it!

subarrays :: [a] -> [[a]]
subarrays [] = [[]]
subarrays xs = (sa xs) ++ subarrays (tail xs)
  where sa xs = [ys | n <- [1..length xs], ys <- [(take n xs)]]

maxsubarrays :: [Integer] -> [Integer]
maxsubarrays xs = msa [] (subarrays xs)
  where
    msa m [] = m
    msa m (x:xs)
      | sum x > sum m = msa x xs
      | otherwise     = msa m xs

--for testing: should return [2, 5, -1, 3]
main = maxsubarrays [-1, 2, 5, -1, 3, -2, 1]

I've read tutorials about the syntax of Haskell, but I can't seem to 
find any that teach you how to really "think" in a Haskell way. Is there 
anything (books, online tutorials, exercises) that anyone could recommend?

Thanks,
James
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