我正在为我编写的 LAN 派对网站构建一个插件,该插件将允许使用循环锦标赛。
一切进展顺利,但我对根据两个标准进行排名的最有效方法有一些疑问。
基本上,我想要以下排名布局:
Rank Wins TotalScore
PersonE 1 5 50
PersonD 2 3.5 37
PersonA 2 3.5 37
PersonC 4 2.5 26
PersonB 5 2.5 24
PersonF 6 0 12
在 SQL Server 中,我会使用:
SELECT
[Person],
RANK() OVER (ORDER BY Wins DESC, TotalScore DESC) [Rank],
[Wins],
[TotalScore]
现在,我只有 List、Dictionary 等可以使用
具体来说:
Dictionary<TournamentTeam, double> wins = new Dictionary<TournamentTeam, double>();
Dictionary<TournamentTeam, double> score = new Dictionary<TournamentTeam, double>();
有没有办法用 LINQ 进行这种排名方式?
如果没有,是否有一种可扩展的方式可以让我以后考虑赢-输-平,而不是如果我选择只赢?
编辑:
我对 TheSoftwareJedi 的回答的改编:
private class RRWinRecord : IComparable
{
public int Wins { get; set; }
public int Losses { get; set; }
public int Draws { get; set; }
public double OverallScore { get; set; }
public double WinRecord
{
get
{
return this.Wins * 1.0 + this.Draws * 0.5 + this.Losses * 0.0;
}
}
public int CompareTo(object obj) { ... }
public override bool Equals(object obj) { ... }
public override int GetHashCode() { ... }
public static bool operator ==(RRWinRecord lhs, RRWinRecord rhs) { ... }
public static bool operator !=(RRWinRecord lhs, RRWinRecord rhs) { ... }
public static bool operator >(RRWinRecord lhs, RRWinRecord rhs) { ... }
public static bool operator <(RRWinRecord lhs, RRWinRecord rhs) { ... }
public static bool operator >=(RRWinRecord lhs, RRWinRecord rhs) { ... }
public static bool operator <=(RRWinRecord lhs, RRWinRecord rhs) { ... }
}
...
int r = 1, lastRank = 1;
RRWinRecord lastRecord = null;
var ranks = from team in records.Keys
let teamRecord = records[team]
orderby teamRecord descending
select new RRRank() { Team = team, Rank = r++, Record = teamRecord };
foreach (var rank in ranks)
{
if (rank.Record != null && lastRecord == rank.Record)
{
rank.Rank = lastRank;
}
lastRecord = rank.Record;
lastRank = rank.Rank;
string scoreDescription = String.Format("{0}-{1}-{2}", rank.Record.Wins, rank.Record.Losses, rank.Record.Draws);
yield return new TournamentRanking(rank.Team, rank.Rank, scoreDescription);
}
yield break;