F＃将列表转换为树 [英] F# transform list to a tree

问题描述

我有一个元组int * string列表，其中int是级别，字符串是名称

I have a list of tuples int*string where int is level and string is name

let src = [
        (0, "root");
            (1, "a");
                (2, "a1");
                (2, "a2");
            (1, "b");
                (2, "b1");
                    (3, "b11");
                (2, "b2");
        ]

我需要将其转换为以下内容

and i need to transform it to following

let expectingTree = 
    Branch("root", 
    [
        Branch("a",
            [
                Leaf("a1");
                Leaf("a2")
            ]);
        Branch("b",
            [
                Branch("b1", [Leaf("b11")]);
                Leaf("b2")
            ]);
    ]);

下面是我的操作方式，但是任何人都可以提出更好的建议。
我是F＃的新手，执行相同操作的C＃代码会更短一些，所以我想我做错了。

Below is the way how i did it, but could anybody advice with a better way to achieve that. I'm new to F#, and C# code to do the same thing would be some shorter, so i guess i'm going it wrong.

type Node = 
    | Branch of (string * Node list)
    | Leaf of string

let src = [
            (0, "root");
                (1, "a");
                    (2, "a1");
                    (2, "a2");
                (1, "b");
                    (2, "b1");
                        (3, "b11");
                    (2, "b2");
            ]

let rec setParents (level:int) (parents:list<int>) (lst:list<int*int*string>) : list<int*int*string> =
    //skip n items and return the rest
    let rec skip n xs = 
        match (n, xs) with
        | n, _ when n <= 0 -> xs
        | _, [] -> []
        | n, _::xs -> skip (n-1) xs

    //get parent id for given level
    let parentId (level) = 
        let n = List.length parents - (level + 1)
        skip n parents |> List.head 

    //create new parent list and append new id to begin
    let newParents level id =
        let n = List.length parents - (level + 1)
        id :: skip n parents

    match lst with
    | (id, l, n) :: tail -> 
                        if l = level then (id, parentId(l), n) :: setParents l (newParents l id) tail
                        elif l <= level + 1 then setParents l parents lst
                        else [] //items should be in order, e.g. there shouldnt be item with level 5 after item with level 3
    | _ -> []


let rec getTree (root:int) (lst: list<int*int*string>) =

    let getChildren parent = 
        List.filter (fun (_, p, _) -> p = parent) lst

    let rec getTreeNode (id:int) (name:string) =
        let children = getChildren id
        match List.length children with
        | 0 -> Leaf(name)
        | _ -> Branch(name, 
                        children
                        |> List.map (fun (_id, _, _name) -> getTreeNode _id _name))

    match getChildren root with
    | (id, _, n) :: _ -> getTreeNode id n
    | _ -> Leaf("")

let rec printTree (ident:string) (tree:Node) = 
    match tree with
    | Leaf(name) -> 
        printfn "%s%s" ident name
    | Branch(name, children) -> 
        printfn "%s%s" ident name
        List.iter (fun (node) -> printTree ("   " + ident) node) children

let tree = 
    src
    |> List.mapi (fun i (l, n) -> (i+1, l, n)) //set unique id to each item
    |> setParents 0 [0] //set parentId to each item
    |> getTree 0


printTree "" tree

Console.ReadKey() |> ignore

推荐答案

首先，您并不需要如果您的分支包含子树列表，则 Leaf 有一个特殊的情况，因为leaf只是一个没有子树的分支。因此，我将使用以下树类型：

First of all, you do not really need to have a distinguished case for Leaf if your branch is containing a list of sub-trees, because leaf is just a branch with no sub-trees. So, I'm going to use the following tree type:

type Tree = 
  | Branch of string * list<Tree>

使用显式递归列表处理可能更容易实现将列表转为树的核心功能。您可以一次性完成-只要找到嵌套索引就可以遍历元素并开始一个新分支（或者当索引较小时从适当数量的递归调用返回）。这是我的尝试：

The core function for turning list to a tree is probably easier to implement using explicit recursive list processing. You can do it in one pass - just go over the elements and start a new branch whenever you find nested index (or return from an appropriate number of recursive calls when you get a smaller index). This is my attempt:

/// Build a tree from elements of 'list' that have larger index than 'offset'. As soon
/// as it finds element below or equal to 'offset', it returns trees found so far
/// together with unprocessed elements.
let rec buildTree offset trees list = 
  match list with
  | [] -> trees, [] // No more elements, return trees collected so far
  | (x, _)::xs when x <= offset -> 
      trees, list // The node is below the offset, so we return unprocessed elements
  | (x, n)::xs ->
      /// Collect all subtrees from 'xs' that have index larger than 'x'
      /// (repeatedly call 'buildTree' to find all of them)
      let rec collectSubTrees xs trees = 
        match buildTree x [] xs with
        | [], rest -> trees, rest
        | newtrees, rest -> collectSubTrees rest (trees @ newtrees)
      let sub, rest = collectSubTrees xs []
      [Branch(n, sub)], rest

该函数获取初始偏移量，并收集到目前为止的树。 trees 参数始终将为 [] ，并且您需要一些初始偏移值。结果是低于给定级别的树木列表和其余元素的列表：

The function takes initial offset and trees collected so far. The trees parameter is always going to be [] and you need some value for initial offset. The result is a list of trees below the given level and a list of remaining elements:

let res = buildTrees -1 [] src

假设根在-1以上，您可以忽略元组的第二部分（应该为空列表）并获得第一棵树（应该只有一棵树）：

Assuming root is above -1, you can just ignore the second part of the tuple (it should be empty list) and get the first tree (there should be only one):

/// A helper that nicely prints a tree
let rec print depth (Branch(n, sub)) =
  printfn "%s%s" depth n
  for s in sub do print (depth + "  ") s

res |> fst |> Seq.head |> print ""

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