使用 Haskell 中的 Lenses 遍历并将元素添加到 Data.Tree

Question

我开始使用镜头，直到现在我还无法在我正在编写的代码库的具体部分使用它们。我的 objective 是通过在现有节点之一中添加一个新节点来更新玫瑰树结构，例如 Data.Tree 中的结构。为此，我认为用唯一 id 标识每个节点是有意义的，所以它看起来像这样：

type MyTree = Tree Id
type Path = [Id]

addToTree :: MyTree -> MyTree -> Path -> MyTree
addToTree originalTree newNode path = undefined

函数 addToTree 必须按照 id 的路径遍历 originalTree 并在该级别添加 newNode ，返回整个更新的树。我没有为此制作 getter 的问题，但我无法找到合适的镜头来执行操作。

这就是我到目前为止所得到的：

import           Control.Lens
import           Data.Tree
import           Data.Tree.Lens

addToTree :: MyTree -> Path -> MyTree -> MyTree
addToTree tree path branch = tree & (traversalPath path) . branches %~ (branch:)

traversalPath :: (Foldable t, Applicative f, Contravariant f) => t Id -> (MyTree -> f MyTree) -> MyTree -> f MyTree
traversalPath = foldl (\acc id-> acc . childTraversal id) id

childTraversal :: (Indexable Int p, Applicative f) => Id -> p MyTree (f MyTree) -> MyTree -> f MyTree
childTraversal id = branches . traversed . withId id

withId :: (Choice p, Applicative f) => Id -> Optic' p f MyTree MyTree
withId id = filtered (\x -> rootLabel x == id)

但编译失败：

• No instance for (Contravariant Identity)
    arising from a use of ‘traversalPath’
• In the first argument of ‘(.)’, namely ‘(traversalPath path)’
  In the first argument of ‘(%~)’, namely
    ‘(traversalPath path) . branches’
  In the second argument of ‘(&)’, namely
    ‘(traversalPath path) . branches %~ (branch :)’

谢谢！

Answer 1

这不是特别优雅，但应该可以解决问题：

import Control.Lens
import Data.Monoid (Endo(..)) -- A tidier idiom for 'foldr (.) id'.
import Data.List.NonEmpty (NonEmpty(..)) -- You don't want an empty path.
import qualified Data.List.NonEmpty as N
import Data.Tree
import Data.Tree.Lens -- That's where I got 'branches' from.

addToTree :: Eq a => NonEmpty a -> Tree a -> Tree a -> Tree a
addToTree path newNode oldTree = head $ over pathForests (newNode :) [oldTree]
    where
    pathForests = appEndo $ foldMap (Endo . goDown) path 
    goDown x = traverse . filtered ((x ==) . rootLabel) . branches

（特别是，我从来不喜欢使用 head，即使在这种不可能失败的情况下也是如此。请随意将其替换为您最喜欢的绕口令。）

演示：

GHCi> addToTree (1 :| []) (Node 2 []) (Node 1 [])
Node {rootLabel = 1, subForest = [Node {rootLabel = 2, subForest = []}]}
GHCi> addToTree (4 :| []) (Node 2 []) (Node 1 [])
Node {rootLabel = 1, subForest = []}
GHCi> addToTree (1 :| [5]) (Node 2 []) (Node 1 [Node 5 [], Node 6 []])
Node {rootLabel = 1, subForest = [Node {rootLabel = 5, subForest = [Node {rootLabel = 2, subForest = []}]},Node {rootLabel = 6, subForest = []}]}
GHCi> addToTree (1 :| [7]) (Node 2 []) (Node 1 [Node 5 [], Node 6 []])
Node {rootLabel = 1, subForest = [Node {rootLabel = 5, subForest = []},Node {rootLabel = 6, subForest = []}]}
GHCi> addToTree (1 :| [5,3]) (Node 2 []) (Node 1 [Node 5 [], Node 6 []])
Node {rootLabel = 1, subForest = [Node {rootLabel = 5, subForest = []},Node {rootLabel = 6, subForest = []}]}

请注意，我们处理的是遍历，而不是镜头——不能保证或期望路径的目标存在或唯一。

这是一个更程式化的变体，没有 head 并使用 alaf 来处理 Endo 包装。

addToTree :: Eq a => NonEmpty a -> Tree a -> Tree a -> Tree a
addToTree (desiredRoot :| path) newNode oldTree@(Node x ts)
    | x == desiredRoot = Node x (over pathForests (newNode :) ts)
    | otherwise = oldTree
    where
    pathForests = alaf Endo foldMap goDown path 
    goDown x = traverse . filtered ((x ==) . rootLabel) . branches

使用 Haskell 中的 Lenses 遍历并将元素添加到 Data.Tree

Traversing and adding elements to a Data.Tree using Lenses in Haskell

tree

haskell

lenses

haskell-lens