[Haskell-cafe] efficient combination of foldl' and foldr -> foldl'r
lemming at henning-thielemann.de
Fri Dec 5 10:04:22 EST 2008
I want to do a foldl' and a foldr in parallel on a list. I assumed it
would be no good idea to run foldl' and foldr separately, because then the
input list must be stored completely between the calls of foldl' and
foldr. I wanted to be clever and implemented a routine which does foldl'
and foldr in one go. But surprisingly, at least in GHCi, my clever routine
is less efficient than the naive one.
Is foldl'rNaive better than I expect, or is foldl'r worse than I hope?
module FoldLR where
import Data.List (foldl', )
import Control.Arrow (first, second, (***), )
foldl'r, foldl'rNaive ::
(b -> a -> b) -> b -> (c -> d -> d) -> d -> [(a,c)] -> (b,d)
foldl'r f b0 g d0 =
first ($b0) .
foldr (\(a,c) ~(k,d) -> (\b -> k $! f b a, g c d)) (id,d0)
foldl'rNaive f b g d xs =
(foldl' f b *** foldr g d) $ unzip xs
test, testNaive :: (Integer, Char)
second last $ foldl'r (+) 0 (:) "" $ replicate 1000000 (1,'a')
(2.65 secs, 237509960 bytes)
second last $ foldl'rNaive (+) 0 (:) "" $ replicate 1000000 (1,'a')
(0.50 secs, 141034352 bytes)
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