Re: Why is partition no good list producer?

what you observe is possibly related to this:

http://www.mail-archive.com/haskell@haskell.org/msg07508.html

Although I'm adressing purely the problem of runtime, you are probably right. At least the implementation of partition is still defined in terms of foldr (from Data.List in ghc 6.2): -- > partition p xs == (filter p xs, filter (not . p) xs) partition :: (a -> Bool) -> [a] -> ([a],[a]) {-# INLINE partition #-} partition p xs = foldr (select p) ([],[]) xs select p x (ts,fs) | p x = (x:ts,fs) | otherwise = (ts, x:fs) Is the above strict version of partition more suitable for other optimizations than my partition/foldr case?? Cheers, Jan