c# - 嵌套分组策略/算法 c#

Question

不确定我是否必须为此类问题更正标签，但您对以下问题的通用解决方案有任何想法吗？

给定发票集合:

var invoices = new List<Invoice>()
{
new Invoice() { Id = 1, Customer = "a", Date = DateTime.Parse("1/1/2009") },
new Invoice() { Id = 2, Customer = "a", Date = DateTime.Parse("1/2/2009") },
new Invoice() { Id = 3, Customer = "a", Date = DateTime.Parse("1/2/2009") },
new Invoice() { Id = 4, Customer = "b", Date = DateTime.Parse("1/1/2009") },
new Invoice() { Id = 5, Customer = "b", Date = DateTime.Parse("1/1/2009") },
new Invoice() { Id = 6, Customer = "b", Date = DateTime.Parse("1/2/2009") }
}

我可以在指定类似内容的地方使用什么技术:

var tree = invoices.ToHeirarchy(t => {
    t.GroupBy(x => x.Date);
    t.GroupBy(x => x.Customer);
})

结果如下图:

Date "1/1/2009"
  Customer a
    Invoice 1
  Customer b
    Invoice 4
    Invoice 5
Date "1/2/2009"
  Customer a
    Invoice 2
    Invoice 3
  Customer b
    Invoice 6

并且还允许以下通过(允许对任何级别的发票进行计算)？

Assert.AreEqual(3, tree.Node[0].Items.Count)
Assert.AreEqual(DateTime.Parse("1/1/2009"), tree.Node[0].Key)

Assert.AreEqual(3, tree.Node[1].Items.Count)
Assert.AreEqual(DateTime.Parse("1/2/2009"), tree.Node[1].Key)

Assert.AreEqual("a", tree.Node[0].Node[0].Key)
Assert.AreEqual(1, tree.Node[0].Node[0].Items.Count)

Assert.AreEqual("b", tree.Node[0].Node[1].Key)
Assert.AreEqual(2, tree.Node[0].Node[1].Items.Count)

Answer 1

您要查找的术语是“嵌套的 groupby”。

http://msdn.microsoft.com/en-us/vcsharp/aa336754.aspx#nested