Thanks Bulat.
FWIW, i take it that
http://www.haskell.org/haskellwiki/Shootout/Knucleotide
is what Edward was referring to, with the shootouts. It seems that a
lot of progress has been made but not much has been migrated back to
hackage.
Going back to my original question, I am now looking for a dead simple
motivating example for showing the example of using a (good) hashtable
over Data.Map, with a tangible demo of O(n) over O(n log n) running
times. I mean, something where running an input of (10^4) size versus
(10^6) size shows a noticeably laggier run when using Set versus
hashtable.
I don't think maybe my original example quite qualifies because I
think in practice the computation is dominated by space complexity.
However, I haven't yet ported it over to a hashtable version, so not
sure.
(And the shootout example doesn't satisfy my sense of "dead simple.")
2009/7/18 Bulat Ziganshin
Hello Thomas,
Saturday, July 18, 2009, 2:24:21 AM, you wrote:
Further, is there a hashtable implementation for haskell that doesn't live in IO? Maybe in ST or something?
import Prelude hiding (lookup) import qualified Data.HashTable import Data.Array import qualified Data.List as List
data HT a b = HT (a->Int) (Array Int [(a,b)])
-- size is the size of array (we implement a closed hash) -- hash is the hash function (a->Int) -- list is assoclist of items to put in hash create size hash list = HT hashfunc (accumArray (flip (:)) [] (0, arrsize-1) (map (\(a,b) -> (hashfunc a,(a,b))) list) )
where arrsize = head$ filter (>size)$ iterate (\x->3*x+1) 1 hashfunc a = hash a `mod` arrsize
lookup a (HT hash arr) = List.lookup a (arr!hash a)
main = do let assoclist = [("one", 1), ("two", 2), ("three", 3)] hash = create 10 (fromEnum . Data.HashTable.hashString) assoclist print (lookup "one" hash) print (lookup "zero" hash)
-- Best regards, Bulat mailto:Bulat.Ziganshin@gmail.com