基于关联计算表中行之间相似度的最佳方法是什么?

这可能不是最有效的,但相对简单:

WITH SimlarBookPrefs(person_id, similar_person_id, booksInCommon) AS
(
 Select p1.person_id, p2.person_id AS simlar_person_id,   
 /* Find the number of books p1 and p2 have in common */
   (SELECT COUNT(*) FROM PersonBook pb1, PersonBook pb2 
     JOIN pb1=book_id=pb2.book_id
   WHERE pb1.person_id=p1.person_id AND pb2.person_id=p2.person_id) As BooksInCommon
   FROM Person p1 CROSS JOIN Person p2
)

这将给你每个人,其他人的名单和共同的号码簿。

要获得最相似的人,请添加(在同一查询中)

SELECT TOP 1 similar_person_id FROM SimilarBookPrefs 
   WHERE person_id = <person_to_match>
   ORDER By booksInCommon DESC;

第一部分不必是CTE(即带有…),它可以是视图,甚至是派生表。为了简洁起见,这里有一个CTE。

如果我在C#中这样做,我可能会这样处理它

var query = from personBook in personBooks
            where personBook.PersonId != basePersonId // ID of person to match
            join bookbase in personBooks
            on personBook.BookId equals bookbase.BookId
            where bookbase.PersonId == basePersonId // ID of person to match
            join person in persons 
            on personBook.PersonId equals person.Id 
            group person by person into bookgroup
            select new
            {
                Person = bookgroup.Key, 
                BooksInCommon = bookgroup.Count()
            };

这可以通过实体框架或linqtosql来完成,也可以直接转换成SQL。

class CommonBooks
{
    static void Main()
    {
        List<Person> persons = new List<Person>()
        {
            new Person(1, "Jane"), new Person(2, "Joan"), new Person(3, "Jim"), new Person(4, "John"), new Person(5, "Jill")
        };

        List<Book> books = new List<Book>()
        {
            new Book(1), new Book(2), new Book(3), new Book(4), new Book(5)
        };

        List<PersonBook> personBooks = new List<PersonBook>()
        {
            new PersonBook(1,1), new PersonBook(1,2), new PersonBook(1,3), new PersonBook(1,4), new PersonBook(1,5), 
            new PersonBook(2,2), new PersonBook(2,3), new PersonBook(2,5), 
            new PersonBook(3,2), new PersonBook(3,4), new PersonBook(3,5), 
            new PersonBook(4,1), new PersonBook(4,4),
            new PersonBook(5,1), new PersonBook(5,3), new PersonBook(5,5)
        };

        int basePersonId = 4; // person to match likeness

        var query = from personBook in personBooks
                    where personBook.PersonId != basePersonId
                    join bookbase in personBooks
                    on personBook.BookId equals bookbase.BookId
                    where bookbase.PersonId == basePersonId
                    join person in persons
                    on personBook.PersonId equals person.Id
                    group person by person into bookgroup
                    select new
                    {
                        Person = bookgroup.Key,
                        BooksInCommon = bookgroup.Count()
                    };

        foreach (var item in query)
        {
            Console.WriteLine("{0}\t{1}", item.Person.Name, item.BooksInCommon);
        }

        Console.Read();
    }
}

class Person
{
    public int Id { get; set; }
    public string Name { get; set; }
    public Person(int id, string name) { Id = id; Name = name; }
}

class Book
{
    public int Id { get; set; }
    public Book(int id) { Id = id; }
}

class PersonBook
{
    public int PersonId { get; set; }
    public int BookId { get; set; }
    public PersonBook(int personId, int bookId) { PersonId = personId; BookId = bookId; }
}

您描述的问题通常被称为“协作过滤”,并使用“推荐系统”解决。在谷歌上搜索这两个词中的任何一个都会让你获得大量有用的信息。