C# - 将多对多对象列表转换为两个对象列表的不同组合列表

Question

我似乎无法掌握如何做到这一点，也找不到简单的方法来解释它......所以我希望这个简化的示例有意义。

被给予一个对象列表<>，如下:

public class FlatManyToMany
{
    public string BookTitle { get; set; }
    public int BookPages { get; set; }
    public string ReaderName { get; set; }
    public int ReaderAge { get; set; }
}

var flatManyToMany = new List<FlatManyToMany>();

flatManyToMany.Add(new FlatManyToMany { BookTitle = "How to Do This Double List", BookPages = 105, ReaderName = "Kyle", ReaderAge = 29 });
flatManyToMany.Add(new FlatManyToMany { BookTitle = "How to Do This Double List", BookPages = 105, ReaderName = "Bob", ReaderAge = 34 });
flatManyToMany.Add(new FlatManyToMany { BookTitle = "Gone With Jon Skeet", BookPages = 192, ReaderName = "Kyle", ReaderAge = 29 });
flatManyToMany.Add(new FlatManyToMany { BookTitle = "Gone With Jon Skeet", BookPages = 192, ReaderName = "James", ReaderAge = 45 });
flatManyToMany.Add(new FlatManyToMany { BookTitle = "Gone With Jon Skeet", BookPages = 192, ReaderName = "Brian", ReaderAge = 15 });
flatManyToMany.Add(new FlatManyToMany { BookTitle = "Why Is This So Hard?", BookPages = 56, ReaderName = "Kyle", ReaderAge = 29 });
flatManyToMany.Add(new FlatManyToMany { BookTitle = "Why Is This So Hard?", BookPages = 56, ReaderName = "James", ReaderAge = 45 });
flatManyToMany.Add(new FlatManyToMany { BookTitle = "Why Is This So Hard?", BookPages = 56, ReaderName = "Brian", ReaderAge = 15 });
flatManyToMany.Add(new FlatManyToMany { BookTitle = "Impostor Syndrome", BookPages = 454, ReaderName = "Kyle", ReaderAge = 29 });
flatManyToMany.Add(new FlatManyToMany { BookTitle = "Self Doubt and You", BookPages = 999, ReaderName = "Kyle", ReaderAge = 29 });

我需要的结果是两个对象列表的列表，如下所示:

public class ResultDoubleList
{
    public List<Book> Books { get; set; } = new List<Book>();
    public List<Reader> Readers { get; set; } = new List<Reader>();
}

public class Book
{
    public string Title { get; set; }
    public int Pages { get; set; }
}

public class Reader
{
    public string Name { get; set; }
    public int Age { get; set; }
}

这本书在最终结果中应该只出现一次，但读者可以出现多次。如果同一读者阅读过多本书，则可以将它们放在一起。

这就是我需要结果的方式:

List<ResultDoubleList> results = new List<ResultDoubleList>();

result(1):
Books
    How to Do This Double List  105
Readers
    Kyle    29
    Bob     34

result(2):
Books
    Gone With Jon Skeet     192
    Why Is This So Hard?    56
Readers
    Kyle    29
    James   45
    Brian   15

result(3):
Books
    Impostor Syndrome   454
    Self Doubt and You  999
Readers
    Kyle    29

最终结果是图书列表和读者列表的不同组合。这本书只出现一次，但读者可以出现多次。具有完全相同读者列表的书籍将被分组在一起。

即使有人能告诉我这种最终结果叫什么，我也会很感激。

Answer 1

您可以使用这个冗长的 LINQ 查询来完成此操作:

var result = flatManyToMany
    .GroupBy(f1 => (f1.BookTitle, f1.BookPages))
    .Select(g1 => (bookInfo: g1.Key,
                    readers:
                        g1.Select(f2 => new Reader { Name= f2.ReaderName, Age= f2.ReaderAge }),
                    readerKey:
                        String.Join("|", g1.Select(f3 => $"{f3.ReaderName}{f3.ReaderAge}"))))
    .GroupBy(a1 => a1.readerKey)
    .Select(g2 => new ResultDoubleList {
        Books = g2.Select(a2 => new Book {
                    Title = a2.bookInfo.BookTitle,
                    Pages = a2.bookInfo.BookPages
                }
            ).ToList(),
        Readers = g2.First().readers.ToList() // Any will do, since they have the same readers
    })
    .ToList();

这个想法是分组两次。每本书一次，每个读者组一次。

首先，我们按 ValueTuple 进行分组(f1.BookTitle, f1.BookPages) 。与创建 Book 相比的优势对象是ValueTuple自动覆盖Equals和GetHashCode 。对于用作字典或查找中的键的类型，这是必需的，如 GroupBy做。或者，您可以覆盖 Book 中的这些方法。类(class)和分组 Book对象。如果您有唯一的图书 ID，请使用此 ID。

然后我们用 Select 创建一个临时结果。我们再次创建一个包含 3 个字段的元组。包含书籍信息的元组，可枚举 Reader对象，最后，我们创建一个包含所有读者的字符串作为键，稍后我们将使用该字符串按唯一的读者组进行分组。如果您有唯一的读者 ID，请使用此 ID，而不是姓名和年龄。

到目前为止，我们已经有了

IEnumerable<(
    (string BookTitle, int BookPages) bookInfo,
    IEnumerable<Reader> readers,
    string readerKey
)>

现在我们按readerKey分组然后创建 ResultDoubleList 的列表对象。

如果您难以理解详细信息，请将 LINQ 查询分解为多个查询。通过使用“Make Explicit”重构，您可以看到得到的结果类型。 (这就是我从上面得到复杂的 IEnumerable<T> 的方式。)这还允许您检查调试器中的中间结果。

---

本次测试...

int resultNo = 1;
foreach (ResultDoubleList item in result) {
    Console.WriteLine($"\r\nresult({resultNo++}):");
    Console.WriteLine("Books");
    foreach (var book in item.Books) {
        Console.WriteLine($"    {book.Title,-28} {book.Pages,3}");
    }
    Console.WriteLine("Readers");
    foreach (var reader in item.Readers) {
        Console.WriteLine($"    {reader.Name,-8} {reader.Age,2}");
    }
}
Console.ReadKey();

...产量:

result(1):
Books
    How to Do This Double List   105
Readers
    Kyle     29
    Bob      34

result(2):
Books
    Gone With Jon Skeet          192
    Why Is This So Hard?          56
Readers
    Kyle     29
    James    45
    Brian    15

result(3):
Books
    Impostor Syndrome            454
    Self Doubt and You           999
Readers
    Kyle     29