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，但要计算所有用户。

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，无法帮助自己回答......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() 
};

为了与原始方法语法（我个人更喜欢）进行并排比较，你去......

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 调用内完成，

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

Answer 4

也许像这样的东西？

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

我认为在 Oryol 的回答中不需要 Distinct 计数。 groupup 会自动选择不同的键，它可以作为 g.key 访问

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