Question

我有以下代码

class Person
{
    public string FirstName { get; set; }
    public string LastName { get; set; }
}

class Child
{
    public string Name { get; set; }
    public Person Owner { get; set; }
}
public class JoinTest
{
    public static void LeftOuterJoinExample()
    {
        Person magnus = new Person { FirstName = "Magnus", LastName = "Hedlund" };
        Person terry = new Person { FirstName = "Terry", LastName = "Adams" };
        Person charlotte = new Person { FirstName = "Charlotte", LastName = "Weiss" };
        Person arlene = new Person { FirstName = "Arlene", LastName = "Huff" };

        Child barley = new Child { Name = "Barley", Owner = terry };
        Child boots = new Child { Name = "Boots", Owner = terry };
        Child whiskers = new Child { Name = "Whiskers", Owner = charlotte };
        Child bluemoon = new Child { Name = "Blue Moon", Owner = terry };
        Child daisy = new Child { Name = "Daisy", Owner = magnus };

        // Create two lists.
        List<Person> people = new List<Person> { magnus, terry, charlotte, arlene };
        List<Child> childs = new List<Child> { barley, boots, whiskers, bluemoon, daisy };

        var query = from person in people
                    join child in childs
                    on person equals child.Owner into gj
                    from subpet in gj.DefaultIfEmpty()
                    select new
                    {
                        person.FirstName,
                        ChildName = subpet!=null? subpet.Name:"No Child"
                    };
                       // PetName = subpet?.Name ?? String.Empty };

        foreach (var v in query)
        {
            Console.WriteLine($"{v.FirstName + ":",-25}{v.ChildName}");
        }
    }

    // This code produces the following output:
    //
    // Magnus:        Daisy
    // Terry:         Barley
    // Terry:         Boots
    // Terry:         Blue Moon
    // Charlotte:     Whiskers
    // Arlene:        No Child

我只需要使用val。我不知道如何在不使用var的情况下完成循环。（这是我给出的总和要求）。

该代码试图从投资组合（字符串列表）和给定的年份范围中获取股票代码的所有第一个价格。内部列表用于股票代码和年份的外部列表。

如何将给定代码转换为不使用任何可变数据结构的代码？

由于

Answer 1

您应该使用.map。我不知道get_first_price()知道什么，我们无法帮助你，但在你的函数中做这样的事情可能就是你想要的：

years.map(y => {
  portfolio.map(p => {
    getFirstPrice(p, y)  
  })
})

将使用可变数据的代码转换为不可变数据

1 个答案: