用于生成层次结构的C#算法

Jud*_*ngo 19 c# algorithm hierarchy hierarchical-data

我有一个看起来像这样的文本文件:

{ Id = 1, ParentId = 0, Position = 0, Title = "root" }
{ Id = 2, ParentId = 1, Position = 0, Title = "child 1" }
{ Id = 3, ParentId = 1, Position = 1, Title = "child 2" }
{ Id = 4, ParentId = 1, Position = 2, Title = "child 3" }
{ Id = 5, ParentId = 4, Position = 0, Title = "grandchild 1" }
Run Code Online (Sandbox Code Playgroud)

我正在寻找一种通用的C#算法,它将从中创建一个对象层次结构.如果您愿意,可以使用"层次结构"功能将此数据转换为对象层次结构.

有任何想法吗?

编辑我已经将文件解析为.NET对象:

class Node
{
    public int Id { get; }
    public int ParentId { get; }
    public int Position { get; }
    public string Title { get; }
}
Run Code Online (Sandbox Code Playgroud)

现在我需要将对象实际排列到对象图中.

Jud*_*ngo 24

非常感谢Jon和mquander - 你们给了我足够的信息来帮助我以正确的,通用的方式解决这个问题.这是我的解决方案,一个将对象转换为层次结构形式的通用扩展方法:

public static IEnumerable<Node<T>> Hierarchize<T, TKey, TOrderKey>(
    this IEnumerable<T> elements, 
    TKey topMostKey, 
    Func<T, TKey> keySelector, 
    Func<T, TKey> parentKeySelector, 
    Func<T, TOrderKey> orderingKeySelector)
{
    var families = elements.ToLookup(parentKeySelector);
    var childrenFetcher = default(Func<TKey, IEnumerable<Node<T>>>);
    childrenFetcher = parentId => families[parentId]
        .OrderBy(orderingKeySelector)
        .Select(x => new Node<T>(x, childrenFetcher(keySelector(x))));

    return childrenFetcher(topMostKey);
}
Run Code Online (Sandbox Code Playgroud)

利用这个小节点类:

public class Node<T>
{
    public T Value { get; private set; }
    public IList<Node<T>> Children { get; private set; }

    public Node(T value, IEnumerable<Node<T>> children)
    {
        this.Value = value;
        this.Children = new List<Node<T>>(children);
    }
}
Run Code Online (Sandbox Code Playgroud)

它足够通用,可以解决各种问题,包括我的文本文件问题.漂亮!

****更新****:以下是你如何使用它:

// Given some example data:
var items = new[] 
{
   new Foo() 
   {
      Id = 1,
      ParentId = -1, // Indicates no parent
      Position = 0
   },
   new Foo() 
   {
      Id = 2,
      ParentId = 1,
      Position = 0
   },
   new Foo() 
   {
      Id = 3,
      ParentId = 1,
      Position = 1
   }
};

// Turn it into a hierarchy! 
// We'll get back a list of Node<T> containing the root nodes.
// Each node will have a list of child nodes.
var hierarchy = items.Hierarchize(
    -1, // The "root level" key. We're using -1 to indicate root level.
    f => f.Id, // The ID property on your object
    f => f.ParentId, // The property on your object that points to its parent
    f => f.Position, // The property on your object that specifies the order within its parent
    );
Run Code Online (Sandbox Code Playgroud)


Jon*_*eet 9

嗯......我不太明白这是怎么回事.2和5如何都有parent = 1,position = 0?5个父母应该有2个,3个还是4个?

好的,这个新版本经历了三次所有节点:

  • 加载所有节点并将它们放入地图中
  • 将每个节点与其父节点相关联
  • 按位置对每个节点的子节点进行排序

它没有很好的封装,很好的错误检查等 - 但它的工作原理.

using System;
using System.Collections.Generic;
using System.IO;

public class Node
{
    private static readonly char[] Braces = "{}".ToCharArray();
    private static readonly char[] StringTrim = "\" ".ToCharArray();

    public Node Parent { get; set; }
    public int ParentId { get; private set; }
    public int Id { get; private set; }
    public string Title { get; private set; }
    public int Position { get; private set; }
    private readonly List<Node> children = new List<Node>();
    public List<Node> Children { get { return children; } }

    public static Node FromLine(string line)
    {
        Node node = new Node();
        line = line.Trim(Braces);
        string[] bits = line.Split(',');
        foreach (string bit in bits)
        {
            string[] keyValue = bit.Split('=');
            string key = keyValue[0].Trim();
            string value = keyValue[1].Trim();
            switch (key)
            {
                case "Id":
                    node.Id = int.Parse(value);
                    break;
                case "ParentId":
                    node.ParentId = int.Parse(value);
                    break;
                case "Position":
                    node.Position = int.Parse(value);
                    break;
                case "Title":
                    node.Title = value.Trim(StringTrim);
                    break;
                default:
                    throw new ArgumentException("Bad line: " + line);
            }
        }
        return node;
    }

    public void Dump()
    {
        int depth = 0;
        Node node = this;
        while (node.Parent != null)
        {
            depth++;
            node = node.Parent;
        }
        Console.WriteLine(new string(' ', depth * 2) + Title);
        foreach (Node child in Children)
        {
            child.Dump();
        }
    }
}

class Test
{       
    static void Main(string[] args)
    {
        var dictionary = new Dictionary<int, Node>();

        using (TextReader reader = File.OpenText("test.txt"))
        {
            string line;
            while ((line = reader.ReadLine()) != null)
            {
                Node node = Node.FromLine(line);
                dictionary[node.Id] = node;
            }
        }
        foreach (Node node in dictionary.Values)
        {
            if (node.ParentId != 0)
            {
                node.Parent = dictionary[node.ParentId];
                node.Parent.Children.Add(node);
            }
        }

        foreach (Node node in dictionary.Values)
        {
            node.Children.Sort((n1, n2) =>
                               n1.Position.CompareTo(n2.Position));
        }

        Node root = dictionary[1];
        root.Dump();
    }
}
Run Code Online (Sandbox Code Playgroud)

示例文本文件:

{ Id = 5, ParentId = 4, Position = 0, Title = "grandchild 1" }
{ Id = 2, ParentId = 1, Position = 0, Title = "child 1" }
{ Id = 4, ParentId = 1, Position = 2, Title = "child 3" }
{ Id = 3, ParentId = 1, Position = 1, Title = "child 2" }
{ Id = 1, ParentId = 0, Position = 0, Title = "root" }
Run Code Online (Sandbox Code Playgroud)

输出:

root
  child 1
  child 2
  child 3
    grandchild 1
Run Code Online (Sandbox Code Playgroud)