Linq 不同 - 计数

Question

我希望对示例对象列表执行查询

Date     Username

01/01/2011 james
01/01/2011 jamie
01/01/2011 alex
01/01/2011 james
02/01/2011 matt
02/01/2011 jamie
02/01/2011 alex
02/01/2011 james
02/01/2011 james
02/01/2011 lucy
02/01/2011 alex
03/01/2011 james
03/01/2011 bob
03/01/2011 bob
03/01/2011 james
03/01/2011 james
04/01/2011 alex
04/01/2011 alex
04/01/2011 alex

我想使用 linq 查询具有唯一用户登录次数的日期列表。

例如：

01/01/2011 - 3
02/01/2011 - 5
03/01/2011 - 2
04/01/2011 - 1

我已经尝试测试了许多 linq 语句，但这些语句都没有给我想要的结果。我得到的最接近的是给我不同的日期，但包含所有用户的计数。

任何帮助将不胜感激。

Answer 1

logins
  .GroupBy(l => l.Date)
  .Select(g => new
  {
    Date = g.Key,
    Count = g.Select(l => l.Login).Distinct().Count()
  });

Answer 2

我意识到这是一个古老的问题，但我遇到了它并看到了关于想要方法语法的评论，并且无法帮助自己回答它......我可能有编码障碍。

在查询语法中，它看起来像这样...请注意，

Distinct

和

Count

没有查询语法

from l in logins
group l by l.Date into g
select new
{
    Date = g.Key,
    Count = (from l in g select l.Login).Distinct().Count() 
};

要与原始方法语法（我个人更喜欢）进行并排比较，请看这里...

logins
  .GroupBy(l => l.Date)
  .Select(g => new
  {
    Date = g.Key,
    Count = g.Select(l => l.Login).Distinct().Count()
  });

Answer 3

可以在单个 GroupBy 调用中完成，

  var Query = list.GroupBy(
                 (item => item.DateTime),
                 (key, elements) => new  { 
                                          key = key,
                                          count = elements
                                                  .Distinct()
                                                  .Count()
                                         }
                 );

Answer 4

也许是这样的？

var list = new List<MyClass>(new[] {
        new MyClass { Date = DateTime.Parse("01/01/2011"), Username = "james" },
        new MyClass { Date = DateTime.Parse("01/01/2011"), Username = "james" },
        new MyClass { Date = DateTime.Parse("01/01/2011"), Username = "alex" },
        new MyClass { Date = DateTime.Parse("01/01/2011"), Username = "james" },
        new MyClass { Date = DateTime.Parse("02/01/2011"), Username = "matt" },
        new MyClass { Date = DateTime.Parse("02/01/2011"), Username = "jamie" },
        new MyClass { Date = DateTime.Parse("02/01/2011"), Username = "alex" },
        new MyClass { Date = DateTime.Parse("02/01/2011"), Username = "james" },
        new MyClass { Date = DateTime.Parse("02/01/2011"), Username = "james" },
        new MyClass { Date = DateTime.Parse("02/01/2011"), Username = "lucy" },
        new MyClass { Date = DateTime.Parse("02/01/2011"), Username = "alex" },
        new MyClass { Date = DateTime.Parse("03/01/2011"), Username = "james" },
        new MyClass { Date = DateTime.Parse("03/01/2011"), Username = "bob" },
        new MyClass { Date = DateTime.Parse("03/01/2011"), Username = "bob" },
        new MyClass { Date = DateTime.Parse("03/01/2011"), Username = "james" },
        new MyClass { Date = DateTime.Parse("03/01/2011"), Username = "james" },
        new MyClass { Date = DateTime.Parse("04/01/2011"), Username = "alex" },
        new MyClass { Date = DateTime.Parse("04/01/2011"), Username = "alex" },
        new MyClass { Date = DateTime.Parse("04/01/2011"), Username = "alex" }
    });

list.GroupBy(l => l.Date, l => l.Username)
    .Select(g => new { 
                Date = g.Key, 
                Count = g.Distinct().Count() 
            });

Answer 5

解决这个问题的另一种方法是分组两次，检查样本

          var dist = listLogins.GroupBy(d => d.date + d.Username)
              .Select(x => x.First())
              .GroupBy(d => d.date).Select(y => new { date = y.Key, count = y.Count() }).ToList();

Answer 6

我认为奥廖尔的回答中不需要独特的计数。分组会自动选择不同的键，并且可以作为 g.key

进行访问

logins
  .GroupBy(l => l.Date)
  .Select(g => new
  {
    Date = g.Key,
    Count = g.Select(l => l.Login).Count()
  });

Answer 7

如果你真的想避免使用 Distinct 方法...

from login in logins
group login by login.Date into dateGroup
select new { 
     Date = dateGroup.Key, 
     UniqueLogins = (from login in dateGroup
                     group login by login.Name into uniqueNames
                     select uniqueNames).Count()
};