[Haskell-cafe] Re: Suspected stupid Haskell Question
thomas.hartman at db.com
Thu Oct 18 10:59:01 EDT 2007
> But I would expect intTable to be faster,
But if I understand correctly, intTable can only deal with integer keys,
whereas BH's original question would have wanted string keys, and I can't
see a way to convert string to int and back.
"Chad Scherrer" <chad.scherrer at gmail.com>
10/17/2007 11:38 PM
Thomas Hartman/ext/dbcom at DBAmericas
haskell-cafe at haskell.org, haskell-cafe-bounces at haskell.org
Re: [Haskell-cafe] Re: Suspected stupid Haskell Question
Hmm, is insertWith' new? If I remember right, I think the stack overflows
were happening because Map.insertWith isn't strict enough. Otherwise I
think the code is the same. But I would expect intTable to be faster,
since it uses IntMap, and there's no IntMap.insertWith' as of 6.6.1
(though it may be easy enough to add one).
On 10/17/07, Thomas Hartman < thomas.hartman at db.com> wrote:
Since I'm interested in the stack overflow issue, and getting acquainted
with quickcheck, I thought I would take this opportunity to compare your
ordTable with some code Yitzchak Gale posted earlier, against Ham's
As far as I can tell, they're the same. They work on lists up to 100000
element lists of strings, but on 10^6 size lists I lose patience waiting
for them to finish.
Is there a more scientific way of figuring out if one version is better
than the other by using, say profiling tools?
Or by reasoning about the code?
import qualified Data.Map as M
Is there a library function to take a list of Strings and return a list of
ints showing how many times each String occurs in the list.
So for example:
["egg", "egg", "cheese"] would return [2,1]
testYitzGale n = do
l <- rgenBndStrRow (10,10) (10^n,10^n) -- 100000 strings, strings are
10 chars long, works. craps out on 10^6.
m <- return $ freqFold l
putStrLn $ "map items: " ++ ( show $ M.size m )
testCScherer n = do
l <- rgenBndStrRow (10,10) (10^n,10^n) -- same limitations as yitz gale
m <- return $ ordTable l
putStrLn $ "items: " ++ ( show $ length m )
-- slow for big lists
--freqArr = Prelude.map ( last &&& length ) . group . sort
-- yitz gale code. same as chad scherer code? it's simpler to understand,
but is it as fast?
freqFold :: [[Char]] -> M.Map [Char] Int
freqFold = foldl' g M.empty
where g accum x = M.insertWith' (+) x 1 accum
-- c scherer code. insists on ord. far as I can tell, same speed as yitz.
ordTable :: (Ord a) => [a] -> [(a,Int)]
ordTable xs = M.assocs $! foldl' f M.empty xs
where f m x = let m' = M.insertWith (+) x 1 m
Just v = M.lookup x m'
in v `seq` m'
l = ["egg","egg","cheese"]
-- other quickcheck stuff
--prop_unchanged_by_reverse = \l -> ( freqArr (l :: [[Char]]) ) == (
freqArr $ reverse l )
--prop_freqArr_eq_freqFold = \l -> ( freqArr (l :: [[Char]]) == (freqFold
--test1 = quickCheck prop_unchanged_by_reverse
--test2 = quickCheck prop_freqArr_eq_freqFold
--------------- generate test data:
genBndStrRow (minCols,maxCols) (minStrLen, maxStrLen) = rgen ( genBndLoL
(minStrLen, maxStrLen) (minCols,maxCols) )
gen gen = do
sg <- newStdGen
return $ generate 10000 sg gen
-- generator for a list with length between min and max
genBndList :: Arbitrary a => (Int, Int) -> Gen [a]
genBndList (min,max) = do
len <- choose (min,max)
-- lists of lists
--genBndLoL :: (Int, Int) -> (Int, Int) -> Gen [[a]]
genBndLoL (min1,max1) (min2,max2) = do
len1 <- choose (min1,max1)
len2 <- choose (min2,max2)
vec2 len1 len2
--vec2 :: Arbitrary a => Int -> Int -> Gen [[a]]
vec2 n m = sequence [ vector m | i <- [1..n] ]
This e-mail may contain confidential and/or privileged information. If you
are not the intended recipient (or have received this e-mail in error)
please notify the sender immediately and destroy this e-mail. Any
unauthorized copying, disclosure or distribution of the material in this
e-mail is strictly forbidden.
-------------- next part --------------
An HTML attachment was scrubbed...
More information about the Haskell-Cafe