foldr 版本比 foldl 版本快:
文件夹版本:
cartProdN9 :: [[a]] -> [[a]]
cartProdN9 xss =
foldr h1 [[]] xss where
h1 xs yss = foldr g [] xs where
g x zss = foldr f zss yss where
f xs yss = (x:xs):yss
折叠版
cartProdN11 :: [[a]] -> [[a]]
cartProdN11 xss =
foldl h1 [[]] xss where
h1 yss xs = foldl g [] xs where
g zss x = foldl f zss yss where
f yss xs = (x:xs):yss
过程cartProdN9 [[1,2]| i <- [1 .. 1000]]
没问题。但是cartProdN11 [[1,2]| i <- [1 .. 1000]]
不行。
严格的版本fold'
仍然不行:
foldl' f z [] = z
foldl' f z (x:xs) = let z' = z `f` x
in z' `seq` foldl' f z' xs
即使使用https://www.fpcomplete.com/haskell/tutorial/all-about-strictness/中的技术
{-# LANGUAGE BangPatterns #-}
module D where
data StrictList a = Cons !a !(StrictList a) | Nil
strictMap :: (a -> b) -> StrictList a -> StrictList b
strictMap _ Nil = Nil
strictMap f (Cons a list) =
let !b = f a
!list' = strictMap f list
in b `seq` list' `seq` Cons b list'
strictEnum :: Int -> Int -> StrictList Int
strictEnum low high =
go low
where
go !x
| x == high = Cons x Nil
| otherwise = Cons x (go $! x + 1)
list :: Int -> StrictList Int
list !x = Cons x (Cons x Nil)
foldlS = \f z l ->
case l of
Nil -> z
Cons !x !xs -> let !z' = z `f` x
in z' `seq` foldlS f z' xs
listlist :: StrictList (StrictList Int)
listlist = strictMap list $! strictEnum 1 10
cartProdN12 :: StrictList (StrictList a) -> StrictList (StrictList a)
cartProdN12 xss =
foldlS h1 (Cons Nil Nil) xss where
h1 !yss !xs = foldlS g Nil xs where
g !zss !x = foldlS f zss yss where
f !yss !xs = Cons (Cons x xs ) yss
myhead :: StrictList a -> a
myhead = \l ->
case l of
Cons x xs -> x
r = cartProdN12 listlist
hr :: Int
hr = myhead( myhead r)
listlist = strictMap list $! strictEnum 1 100
仍然太慢而无法计算。
所以我的问题是:如何使foldl
版本计算与版本一样快foldr
?有可能的?