C# 用列表属性展平对象列表

Question

给定以下对象：

public class Person
{
    public string Name {get; set;}

    public string Age {get; set;}

    public list<string> Interests {get; set;}
}

是否有一种很好的单行 linq 方法来展平它（我对扩展方法持开放态度），例如，如果我们有

var People = new List<Person>(){
    new Person(){
        Name = "Bill",
        Age  = 40,
        Interests = new List<string>(){"Football", "Reading"}
    },
    new Person = new List<Person>(){
        Name = "John",
        Age = 32,
        Interests = new List<string>(){ "Gameshows", "Reading"}
    },
    new Person = new List<Person>(){
        Name = "Bill",
        Age = 40,
        Interests = new List<string>(){ "Golf"} 
    }
}

我们可以得到一个结果（即如果其他属性匹配，AddRange 到 Interests 列表属性）：

{
    {
        Name = "Bill",
        Age  = 40,
        Interests = {"Football", "Reading", "Golf"}
    },
    {
        Name = "John",
        Age = 32,
        Interests = { "Gameshows", "Reading"}
    }
}

Answer 1

我们可以尝试 GroupBy 和 SelectMany:

List<Person> People = ...

var result = People
  .GroupBy(person => new {
     person.Name,
     person.Age 
   })
  .Select(chunk => new Person() {
     Name      = chunk.Key.Name,
     Age       = chunk.Key.Age,
     Interests = chunk
       .SelectMany(item => item.Interests)
       .Distinct()
       .ToList()
   })
  .ToList(); // if we want List<People>

Answer 2

您可以使用GroupBy和SelectMany来串联兴趣

People.GroupBy(c => new
        {
            c.Name,
            c.Age
        })
       .Select(g => new Person() 
                  { Name = g.Key.Name, 
                    Age = g.Key.Age,
                    Interests = g.SelectMany(r => r. Interests).ToList()})

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