[Haskell-cafe] Arrays performance
Paolo Veronelli
paolo.veronelli at gmail.com
Thu Jan 4 08:25:16 EST 2007
Quoting Paolo Veronelli <paolo.veronelli at gmail.com>:
> I paste new version in case you care give me some moe suggestion.
import Data.Maybe
import Data.List
import Data.Array.Diff
import System.Environment
import Control.Arrow
import Control.Monad
import Random
inc l i = l // [(i,l!i + 1)]
switch l i = l // [(i,not (l!i))]
constArray n v = listArray (0,n-1) (repeat v)
data Folding = Folding
{clusters :: [(Int,Int)], remi :: Int, colsCount :: DiffArray Int Int ,rowsCheck :: DiffArray Int Bool}
result (Folding cs _ _ _) = cs
rcluster ls d s = let
devil s@(Folding cs r hs fs) l@(row,col) = let
ns = s { clusters = (l:cs), rowsCheck = switch fs row, colsCount = inc hs col }
rowtest | c < d = ns
| (c == d) && (r > 0) = ns { remi = r - 1 }
| otherwise = s
where c = hs ! col
in if (not (fs ! row)) then rowtest else s
in foldl devil s ls
mcluster :: (Int,Int) -> [(Int,Int)] -> [(Int,[Int])]
mcluster (lr,lc) ls = let
(k,r) = divMod lr lc
start = Folding{clusters = [],remi = r,colsCount = constArray lc 0,rowsCheck = constArray lr False }
cs = result $ rcluster ls k start
in map collapse . groupBy (comp fst (==)) . sort . map swap $ cs
where
comp f g x y = (f x) `g` (f y)
swap = snd &&& fst
collapse = (head &&& unzip) >>> (fst *** snd)
cluster :: (Ord b) => (a -> a -> b) -> [a] -> [a] -> [(a,[a])]
cluster fxy xs ys = let
mkArray (l,xs) = (listArray :: (Int,Int) -> [e] -> DiffArray Int e) (0,l-1) xs
xls = mkArray (lc,xs)
yls = mkArray (rc,ys)
(lc,rc) = (length xs,length ys)
in
map ((yls !) *** map (xls !)) (mcluster (lc,rc) (snd.unzip.sort $ delta))
where
delta = [(fxy x y,(n,m))|(n,x) <- zip [0..] xs, (m,y) <- zip [0..] ys]
-- call it with 2 args, the number ov values and the number of clusters
-- <prog> 101 10 will cluster 101 values in 10 clusters
points m n = do gen <- getStdGen
return $ splitAt n (take (m + n) (randomRs (0,100::Int) gen))
test1 = do args <- getArgs
return $ map read args :: IO [Int]
main = do
[m,n] <- test1
--let [m,n] = [10,3200]
(ps,bs) <- points m n
print $ cluster (\x y -> abs (x - y)) ps bs
More information about the Haskell-Cafe
mailing list