foldr和foldl进一步的解释和示例
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08-10-2019 - |
解决方案
使用
foldr f z [] = z
foldr f z (x:xs) = x `f` foldr f z xs
和
k y ys = ys ++ [y]
让我们打开包装:
foldr k [] [1,2,3]
= k 1 (foldr k [] [2,3]
= k 1 (k 2 (foldr k [] [3]))
= k 1 (k 2 (k 3 (foldr k [] [])))
= (k 2 (k 3 (foldr k [] []))) ++ [1]
= ((k 3 (foldr k [] [])) ++ [2]) ++ [1]
= (((foldr k [] []) ++ [3]) ++ [2]) ++ [1]
= ((([]) ++ [3]) ++ [2]) ++ [1]
= (([3]) ++ [2]) ++ [1]
= ([3,2]) ++ [1]
= [3,2,1]
其他提示
FOLDR是一件容易的事:
foldr :: (a->b->b) -> b -> [a] -> b
它采用的函数在某种程度上类似于(:),
(:) :: a -> [a] -> [a]
和一个类似于空列表[]的值
[] :: [a]
并替换每个:[和]在某些列表中。
看起来像这样:
foldr f e (1:2:3:[]) = 1 `f` (2 `f` (3 `f` e))
您也可以想象FOLDR也是某些州机器词汇器:
f是过渡,
f :: input -> state -> state
E是开始状态。
e :: state
foldr(foldright)在输入列表上以过渡f和启动状态E运行状态机,从右端开始。想象一下f在infix符号中的f,因为pacman从右翼出现。
foldl(foldleft)从左左右执行相同的操作,但是用infix符号编写的过渡函数从正确地获取其输入参数。因此,机器从左端开始消耗列表。 Pacman由于嘴(B-> a-> B)而不是(A-> B-> B)而以左右左右左右消耗清单。
foldl :: (b->a->b) -> b -> [a] -> b
为了清楚这一点,请想象功能 (-)
作为过渡:
foldl (-) 100 [1] = 99 = ((100)-1)
foldl (-) 100 [1,2] = 97 = (( 99)-2) = (((100)-1)-2)
foldl (-) 100 [1,2,3] = 94 = (( 97)-3)
foldl (-) 100 [1,2,3,4] = 90 = (( 94)-4)
foldl (-) 100 [1,2,3,4,5] = 85 = (( 90)-5)
foldr (-) 100 [1] = -99 = (1-(100))
foldr (-) 100 [2,1] = 101 = (2-(-99)) = (2-(1-(100)))
foldr (-) 100 [3,2,1] = -98 = (3-(101))
foldr (-) 100 [4,3,2,1] = 102 = (4-(-98))
foldr (-) 100 [5,4,3,2,1] = -97 = (5-(102))
您可能想在列表可以是无限的情况下使用foldr,并且评估应该懒惰:
foldr (either (\l ~(ls,rs)->(l:ls,rs))
(\r ~(ls,rs)->(ls,r:rs))
) ([],[]) :: [Either l r]->([l],[r])
而且,当您消耗整个列表以产生其输出时,您可能想使用严格的foldl版本。它可能会表现更好,并且可能会阻止您由于非常长的列表与懒惰评估的极端列表,因此无法使用堆栈堆叠或存储外例外(取决于编译器):
foldl' (+) 0 [1..100000000] = 5000000050000000
foldl (+) 0 [1..100000000] = error "stack overflow or out of memory" -- dont try in ghci
foldr (+) 0 [1..100000000] = error "stack overflow or out of memory" -- dont try in ghci
第一个 - 步骤逐步创建列表的一个条目,对其进行评估并消耗它。
第二个是首先创建一个很长的公式,将内存浪费((((((((((((((((0+1)'''')),内存))))之后都会对其进行评估。
第三个就像第二个,但有另一个公式。
定义 foldr
是:
foldr f z [] = z
foldr f z (x:xs) = f x (foldr f z xs)
因此,这是逐步减少您的示例:
foldr (\y ys -> ys ++ [y]) [] [1,2,3]
= (\y ys -> ys ++ [y]) 1 (foldr (\y ys -> ys ++ [y]) [] [2,3])
= (foldr (\y ys -> ys ++ [y]) [] [2,3]) ++ [1]
= (\y ys -> ys ++ [y]) 2 (foldr (\y ys -> ys ++ [y]) [] [3]) ++ [1]
= (foldr (\y ys -> ys ++ [y]) [] [3]) ++ [2] ++ [1]
= (\y ys -> ys ++ [y]) 3 (foldr (\y ys -> ys ++ [y]) [] []) ++ [2] ++ [1]
= (foldr (\y ys -> ys ++ [y]) [] []) ++ [3] ++ [2] ++ [1]
= [] ++ [3] ++ [2] ++ [1]
= [3,2,1]
infix符号在这里可能会更清楚。
让我们从定义开始:
foldr f z [] = z
foldr f z (x:xs) = x `f` (foldr f z xs)
为了简洁起见,让我们写信 g
代替 (\y ys -> ys ++ [y])
. 。以下行是等效的:
foldr g [] [1,2,3]
1 `g` (foldr g [] [2,3])
1 `g` (2 `g` (foldr g [] [3]))
1 `g` (2 `g` (3 `g` (foldr g [] [])))
1 `g` (2 `g` (3 `g` []))
(2 `g` (3 `g` [])) ++ [1]
(3 `g` []) ++ [2] ++ [1]
[3] ++ [2] ++ [1]
[3,2,1]
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