Question

（我已经重写了这个问题，试图更好地提供我想要做的事情的例子。如果这仍是一个令人困惑的问题，我道歉...... ）

以什么方式可以找到包含为其他对象属性的n组对象的并集？

以下面的例子为例：

class Author
{
    public string FirstName;
    public string LastName;
}

class Magazine
{
    public string Title;
    public string ISBN;

    public List<Article> TOC;
}

class Article
{
    public string Title;
    public string Header;
    public string Body;

    public int PageNumber;

    public Magazine Source;
    public Author Author;
}

我获得List<Magazine>，其中包含List<Article>。
Article对象有一个杂志和作者对象作为属性（该文章可能存在于杂志的1：N - 作为副本或转载）。
事实上，我将每篇文章视为一个独立的对象。我不关心给我的List<Magazine>。我被要求提供List<Author>。
在获得此列表之前，我对可能的文章或作者有 NO 知识。

我提供的这些数据看起来像这样（在树形视图布局中）。

Magazine01
    Article01 (Title, Header, ... Magazine01, Author01)
    Article02 (Title, Header, ... Magazine01, Author02)
    Article03 (Title, Header, ... Magazine01, Author03)
Magazine02
    Article04 (Title, Header, ... Magazine02, Author04)
    Article05 (Title, Header, ... Magazine02, Author05)
    Article06 (Title, Header, ... Magazine02, Author04)
Magazine03
    Article07 (Title, Header, ... Magazine03, Author03)
    Article08 (Title, Header, ... Magazine03, Author02)
    Article09 (Title, Header, ... Magazine03, Author01)

我被要求提供的这些数据看起来像这样（在树视图布局中）。

Author01
    Article01 (Title, Header, ... Magazine01, Author01)
    Article09 (Title, Header, ... Magazine03, Author01)
Author02
    Article02 (Title, Header, ... Magazine01, Author02)
    Article08 (Title, Header, ... Magazine03, Author02)
Author03
    Article03 (Title, Header, ... Magazine01, Author03)
    Article07 (Title, Header, ... Magazine03, Author03)
Author04
    Article04 (Title, Header, ... Magazine02, Author04)
    Article06 (Title, Header, ... Magazine02, Author04)
Author05
    Article05 (Title, Header, ... Magazine02, Author05)

我给出的List<Magazine>没有特别的顺序。
我还没有尝试对输出进行排序，但这样会很好。
可能会有很多杂志，每个杂志都有很多文章。
因此，作者可能会有很多文章。
当我收到杂志对象时，我正在将这些文章聚合成List<Article>（但我不会继续这样做，它似乎好像是个好主意。。）

除了我的（可耻的）“减员减员”方法之外，我还能以什么方式减少通过文章列表的循环：

List<Magazine> magazineList; // Magazines contain a list of Articles which contain an Author.
List<Article> articleList; // As each Magazine is created, a summary list of the Articles is recorded here (I realize that this is probably unnecessary, but that is an aside to my problem)
Bag<Author> authorBag;

for (i = 0; i < (Magazines.Count - 1); i++)
{
    BuildAuthorBag();
}

ParseArticleList();

//...

BuildAuthorBag()
{
    // Finds occurrences of Author in the Article list of each Magazine.
    // Limits the inclusion of Author based on preferences ( i > 1, 2, etc. )
}

ParseArticleList()
{
    // I clone the articleList (i.e. tmpArticleList)
    // I create two lists of Article type.  One will be the new leaf nodes.  The second is a swap list.

    // Using the authorBag...

    // In a loop...
    // I create the new parent node from the current Author in the bag.
    // Then, find occurrences of the Author in the tmpArticleList adding that Article as a child node to the parent node.
    // or...
    // I place them into the second list to be swapped for the tmpArticleList before the next loop.
}

Answer 1

这会满足您的需求吗？

  public class Student
  {
    public List<Course> Courses { get; set; }
    public List<Grade> Grades { get; set; }

    public Dictionary<Course, Grade> CourseGrades { get; set; }
  }

  public class Course
  {
    public int Id { get; set; }
  }

  public class Grade
  {
    public double Value { get; set; }
  }

  class Program
  {
    static void Main(string[] args)
    {
      var c0 = new Course() { Id = 0 };
      var c1 = new Course() { Id = 1 };

      var students = new List<Student>()
      {
        new Student() { CourseGrades = new Dictionary<Course,Grade>()
        { 
          { c0, new Grade() { Value = 2 } },
          { c1 , new Grade() { Value = 1 } }
        }},
        new Student() { CourseGrades = new Dictionary<Course,Grade>()
        { 
          { c1 , new Grade() { Value = 3 } },
          { c0 , new Grade() { Value = 4 } }
        }},
      };


      Dictionary<Course, List<Grade>> courseGrades = SelectUnion(students.SelectMany(s => s.CourseGrades), cg => cg.Key, cg => cg.Value);
    }

    private static Dictionary<TKey, List<TValue>> SelectUnion<TSource, TKey, TValue>(IEnumerable<TSource> set, Func<TSource, TKey> keyGen, Func<TSource, TValue> valueGen)
    {
      var result = new Dictionary<TKey, List<TValue>>();

      foreach (var src in set)
      {
        var key = keyGen(src);
        if (!result.ContainsKey(key))
        {
          result[key] = new List<TValue>();
        }

        result[key].Add(valueGen(src));
      }

      return result;
    }

我知道这很可能只适合你给出的例子。因此，如果这还不够，也许您可以提供您想要的最终结果的示例结构，以便我们知道要拍摄的内容。

Answer 2

根据我的理解，你需要这样的东西。（我使用的是VS2010和C＃4，但任何支持LINQ的版本都应该这样做。）

using System;
using System.Collections.Generic;
using System.Linq;

namespace SO6299404
{
    class Author
    {
        public string FirstName;
        public string LastName;
    }

    class Magazine
    {
        public string Title;
        public string ISBN;

        public List<Article> TOC;


        // Code Until the end of the class is for testing only
        public void Dump()
        {
            Dump(0);
        }

        public void Dump(int indent)
        {
            Console.WriteLine("".PadLeft(indent) + Title);
            if (TOC != null)
            {
                foreach (Article article in TOC)
                {
                    article.Dump(indent + 2);
                }
            }
        }
    }

    class Article
    {
        public string Title;
        public string Header;
        public string Body;

        public int PageNumber;

        public Magazine Source;
        public Author Author;

        // Code Until the end of the class is for testing only
        public void Dump()
        {
            Dump(0);
        }

        public void Dump(int indent)
        {
            string author = Author == null ? "" : Author.FirstName;
            Console.WriteLine("".PadLeft(indent) + Title + ", " + author);
        }
    }

    // A foreach extension. Not strictly nescessary, but having this extension
    // makes it a bit easier to write foreach loops
    public static class EnumerableExtension
    {
        public static void Foreach<T>(this IEnumerable<T> enumerable, Action<T> action)
        {
            if (enumerable == null) throw new ArgumentNullException("enumerable");
            if (action == null) throw new ArgumentNullException("action");

            foreach (T value in enumerable)
                action(value);
        }
    }

    class Program
    {

        static void Main()
        {
            List<Magazine> magazines = SetupTestCase();

            // Let's print out our test case
            Console.WriteLine("==[Setup]===========================");
            magazines.Foreach(x => x.Dump());

            // So now we need to extraxct all Authors and list articles belonging to them
            // First we get Authorl; Article pair and then we group by Authors
            var temp = magazines.SelectMany(m => m.TOC, (a, b) => new {b.Author, Article = b}).GroupBy(x => x.Author);

            // Ok, we are done. Let's print out the results
            Console.WriteLine("==[REsult]===========================");
            temp.Foreach(x =>
            {
                Console.WriteLine(x.Key.FirstName);
                x.Foreach( y => y.Article.Dump(2));
            });

        }

        // The code from here until the end of the class is for generating a test case only

        private static List<Magazine> SetupTestCase()
        {
            // Let's set up a test case similar to the example in the question

            // We generate 5 Authors
            Author[] authors = Seed(1, 5).Select(x => GenerateTestAuthor(x.ToString())).ToArray();

            // And 9 Articles
            Article[] articles = Seed(1,9).Select(x => GenerateTestArticle(x.ToString())).ToArray();

            // This defines how articles are connected to authors
            int[] articleToAuthor = new[] {0,1,2,3,4,3,2,1,0};

            // Let's connect articles and authors as per definition abbove
            Seed(9).Foreach(x=> {articles[x].Author = authors[articleToAuthor[x]];});

            // Now 3 Magazines
            Magazine[] magazines = Seed(1,3).Select(x => GenerateTestMagazine(x.ToString())).ToArray();

            // This deines which articles go in which magazine
            int[] articleToMagazine = new[] {0,0,0,1,1,1,2,2,2};

            // Let's add artices to the Magazines
            Seed(9).Foreach(x=> magazines[articleToMagazine[x]].TOC.Add(articles[x]));

            // And now let us link back from articles to Magazines
            magazines.Foreach(x => x.TOC.Foreach(z => z.Source = x));
            return magazines.ToList();
        }

        static IEnumerable<int> Seed(int start, int count)
        {
            return Enumerable.Range(start, count);
        }

        static IEnumerable<int> Seed(int n)
        {
            return Seed(0, n);
        }

        static Article GenerateTestArticle(string id)
        {
            return new Article
            {
                Title = "Title" + id,
                Header = "Title" + id,
                Body = "Title" + id,
                PageNumber = 1,
            };
        }

        static Author GenerateTestAuthor(string id)
        {
            return new Author
            {
                FirstName = "Author" + id,
                LastName = "Author" + id,
            };
        }

        static Magazine GenerateTestMagazine(string id)
        {
            return new Magazine
            {
                Title = "Magazine" + id,
                ISBN = "Magazine" + id,
                TOC = new List<Article>()
            };
        }
    }
}

这就是我在屏幕上看到的，当我跑步时。

==[Setup]===========================
Magazine1
    Title1, Author1
    Title2, Author2
    Title3, Author3
Magazine2
    Title4, Author4
    Title5, Author5
    Title6, Author4
Magazine3
    Title7, Author3
    Title8, Author2
    Title9, Author1
==[REsult]===========================
Author1
    Title1, Author1
    Title9, Author1
Author2
    Title2, Author2
    Title8, Author2
Author3
    Title3, Author3
    Title7, Author3
Author4
    Title4, Author4
    Title6, Author4
Author5
    Title5, Author5

在列表中查找对象的联合（复合对象/嵌套列表）

2 个答案: