【问题标题】:Need better way to sort one list against another list需要更好的方法来根据另一个列表对一个列表进行排序
【发布时间】:2014-07-12 08:16:12
【问题描述】:

我有两个列表 - 来宾列表和 VIP 列表。我需要对访客列表进行排序,以便如果它包含 VIP 列表中的第一个人,他们会进入列表顶部,依此类推。 VIP 名单用完后,其余的客人名单仍保持原始顺序。排序必须同时使用名字和姓氏。我已经使用 List 和 foreach 语句完成了这项工作,但似乎应该有更优雅的方式。

有没有更简单、更现代的方法来做这种排序?

class Guest 
{
    public int NumberInParty { get; set; }
    public string FirstName { get; set; }
    public string LastName { get; set; }
}

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

class TrackedGuest
{
    public Guest guest;
    public bool isTaken;

    public TrackedGuest(Guest g)
    {
        this.guest = g;
        isTaken = false;
    }
}
static void Main(string[] args)
{
    List<Guest> guests = new List<Guest>();

    guests.Add(new Guest { FirstName = "Rob", LastName = "Carson", NumberInParty = 5 });
    guests.Add(new Guest { FirstName = "George", LastName = "Waverly", NumberInParty = 3 });
    guests.Add(new Guest { FirstName = "Pete", LastName = "Spacely", NumberInParty = 2 });
    guests.Add(new Guest { FirstName = "George", LastName = "Jetson", NumberInParty = 6 });
    guests.Add(new Guest { FirstName = "Cosmo", LastName = "Spacely", NumberInParty = 2 });

    List<VIP> vips = new List<VIP>();
    vips.Add(new VIP { FirstName = "George", LastName = "Jetson" });
    vips.Add(new VIP { FirstName = "Cosmo", LastName = "Spacely" });

    List<TrackedGuest> TrackedGuests = new List<TrackedGuest>();

    foreach (Guest g in guests)
    {
        TrackedGuests.Add(new TrackedGuest(g));
    }

    List<Guest>SortedGuests = new List<Guest>();

    // Copy each guest on the VIP list in order
    foreach (VIP vip in vips)
    {
        foreach (TrackedGuest tGuest in TrackedGuests)
        {
            if (
                (tGuest.isTaken == false) &&
                (vip.FirstName == tGuest.guest.FirstName) &&
                (vip.LastName == tGuest.guest.LastName)
                )
            {
                SortedGuests.Add(tGuest.guest);
                tGuest.isTaken = true;
            }
        }        
    }

    // Process the rest of the guests
    if (SortedGuests.Count < guests.Count)
    {
        foreach (TrackedGuest tGuest in TrackedGuests)
        {
            if (tGuest.isTaken == false)
            {
                SortedGuests.Add(tGuest.guest);
                tGuest.isTaken = true;
            }
        }
    }

    foreach (Guest guest in SortedGuests)
    {
        Console.WriteLine(guest.FirstName + " " + guest.LastName + ": " + guest.NumberInParty + " in party.");

    }

    Console.ReadLine();
}

【问题讨论】:

  • 它可能会建议您将其发布到 Code Review 而不是 Stack Overflow。前者旨在对此进行同行评审;一段有效的代码,但可能会被改进。
  • 我有一段时间没写c#了,但我相信你可以使用一个哈希表作为VIP列表,并为你的Guest类使用一个值(可能称为VIP)来表明它是一个VIP。然后,当您构建您的访客列表时,在哈希表中查找名称,如果存在,请在您的访客类中设置值以指示它。然后,首先在 VIP 字段中对 Guest 类进行排序,并将名称作为辅助键。看起来这将是一种直接的方法,需要的代码要少得多。
  • 试试 linq;参考这个:stackoverflow.com/questions/3945935/…

标签: c# list sorting


【解决方案1】:
var sorted = new List<Guest>();
var guestvips = from g in guests
                from v in vips.Where(vip => vip.FirstName == g.FirstName && vip.LastName == g.LastName).DefaultIfEmpty()
                where v != null
                select g;
var guestsimple = from g in guests
                  from v in vips.Where(vip => vip.FirstName == g.FirstName && vip.LastName == g.LastName).DefaultIfEmpty()
                  where v == null
                  select g;

sorted.AddRange(guestvips.Concat(guestsimple));

此代码两次“离开加入”贵宾。在第一种情况下,它需要那些具有相同vip的客人,其次是那些没有相同vip的客人。第一种情况实际上可以用'join'关键字重写。

【讨论】:

  • 这是正确的答案,虽然它可以通过 1-2 句解释它的作用来改进。只是输入代码片段不是 SO 上首选的回答方法
  • 我认为这是一个近乎完美的答案。言归正传,不废话。
  • 这个答案最接近我正在寻找的答案类型,但我从所有答案和 cmets 中学到了很多东西。谢谢!
【解决方案2】:
// dictionary to easily get vips order
// uses anonymous types, to get value equality for free
var vipsOrder = vips.Select((v, i) => new { v, i })
                    .ToDictionary(x => new { x.v.FirstName, x.v.LastName },
                                  x => x.i);

// sort first by order taken from vipsOrder and then by name
var sortedGuests = (from g in guests
                    let info = new { g.FirstName, g.LastName }
                    let oorder
                     = vipsOrder.ContainsKey(info)
                         ? vipsOrder[info] : vips.Count
                    orderby oorder, info.FirstName, info.LastName
                    select g).ToList();

【讨论】:

    猜你喜欢
    • 1970-01-01
    • 2018-11-15
    • 1970-01-01
    • 2023-04-01
    • 2021-05-02
    • 2011-03-22
    • 1970-01-01
    • 2012-08-27
    相关资源
    最近更新 更多