Tree Functor和Foldable但有節點。 對它有任何概括嗎？

Question

data Tree t = Empty | Node t (Tree t) (Tree t)

我們可以創建Functor實例並使用

fmap :: (t -> a) -> Tree t -> Tree a

但是如果不是（t - > a）我想要（樹t - > a）那么我可以訪問整個（節點t）而不僅僅是t

treeMap :: (Tree t -> a) -> Tree t -> Tree a
treeMap f Empty = Empty
treeMap f n@(Node _ l r) = Node (f n) (treeMap f l) (treeMap f r)

與折疊相同

treeFold :: (Tree t -> a -> a) -> a -> Tree t -> a

對這些功能有任何概括嗎？

map :: (f t -> a) -> f t -> f a
fold ::  (f t -> a -> a) -> a -> f t -> a

Answer 1

你剛剛發現了comonads！ 好吧，差不多。

class Functor f => Comonad f where
  extract :: f a -> a
  duplicate :: f a -> f (f a)

instance Comonad Tree where
  extract (Node x _ _) = x   -- this one only works if your trees are guaranteed non-empty
  duplicate t@(Node n b1 b2) = Node t (duplicate b1) (duplicate b2)

使用duplicate您可以實現您的功能：

treeMap f = fmap f . duplicate
freeFold f i = foldr f i . duplicate

要正確執行此操作，您應該通過類型系統強制執行非空白：

type Tree' a = Maybe (Tree'' a)

data Tree'' t = Node' t (Tree' t) (Tree' t)
   deriving (Functor)

instance Comonad Tree'' where
  extract (Node' x _ _) = x
  duplicate t@(Node' _ b1 b2) = Node' t (fmap duplicate b1) (fmap duplicate b2)