Does having 'zipped' at the top level mean that the program is keeping the entire 100,000-element list in memory?
I don't know, but I tested with zipped at the top, in the where, and it appears to make no performance or memory difference.
Also, would performance improve if you used Map.fromList? How about Map.fromAscList?
HUGE IMPROVEMENT. The old code had maximum residency of 11Mb and running time of 41s. The code using Map.fromList has max. residency of 6.4Mb and running time of 5.2s!
I guess foldlStrict is HUGELY more efficient than than foldr. (Why isn't it in the prelude?) But even using foldrStrict (fromList), we are still using 60 bytes per item and have ~30 bytes unaccounted for. Any hints?
import qualified Map
main = print $ length $ Map.keys theMap
where
zipped =zip [1..] [1..100000]::[(Int,Int)]
theMap = Map.fromList zipped-Alex- ______________________________________________________________ S. Alexander Jacobson tel:917-770-6565 http://alexjacobson.com
S. Alexander Jacobson writes:
After actually running the correct test, I am still getting semi-ridiculous space behavior (6k/pair)!
import qualified Map zipped =zip [1..] [1..100000]::[(Int,Int)] untup f (x,y) = f x y produce = foldr (untup Map.insert) Map.empty zipped fm = length $ Map.keys produce main = print $ fm
Two questions I'm currently too lazy to investigate myself:
-- David Menendez <[EMAIL PROTECTED]> | "In this house, we obey the laws <http://www.eyrie.org/~zednenem> | of thermodynamics!"
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