Question

转换扁平结构的最简单有效的方法是什么：

object[][] rawData = new object[][] 
{ 
  { "A1", "B1", "C1" }, 
  { "A1", "B1", "C2" },
  { "A2", "B2", "C3" }, 
  { "A2", "B2", "C4" }
  // .. more 
};

进入分层结构：

class X
{
  public X () 
  {
    Cs = new List<string>();
  }
  public string A { get; set; }
  public string B { get; set; }
  public List<string> Cs { get; private set; }
}

结果应如下所示

// pseudo code which describes structure:
result =
{
  new X() { A = "A1", B = "B1", Cs = { "C1", "C2" } },
  new X() { A = "A2", B = "B2", Cs = { "C3", "C4" } }
}

最好使用Linq扩展方法。目标类X可以更改（例如，列表的公共设置者），但前提是不可能/有用。

Answer 1

针对这种特殊情况：

   .GroupBy( x => new { a = x[0], b = x[1] } )
   .Select( x => new { A = x.Key.a, B = x.Key.b, C = x.Select( c => c[2] ) })

Answer 2

如果您的层次结构的深度有限，那么这样的事情应该有效（如您的示例中只有三个级别A，B和C）。我简化了你的X：

class X {
    public string A { get; set; }
    public string B { get; set; }
    public List<string> Cs { get; set; }
}

然后您可以根据需要多次使用嵌套GroupBy（取决于层次结构的深度）。将它重写为递归方法（这适用于任意深层次结构）也相对容易：

// Group by 'A'
rawData.GroupBy(aels => aels[0]).Select(a => 
  // Group by 'B'
  a.GroupBy(bels => bels[1]).Select(b =>
    // Generate result of type 'X' for the current grouping
    new X { A = a.Key, B = b.Key, 
            // Take the third element 
            Cs = b.Select(c => c[2]).ToList() }));

这比其他解决方案更明确，但也许它更具可读性，因为它更直接地编码了这个想法......

Answer 3

如果X成员是字符串而Cs是私有集，而rawData是对象数组的数组，我会向X public X(string a, string b, List<string> cs)添加构造函数然后执行此代码

var query = from row in rawData
            group row by new { A = row[0], B = row[1] } into rowgroup
            select new X((string)rowgroup.Key.A, (string)rowgroup.Key.B, rowgroup.Select(r => (string)r[2]).ToList());

这是关于以下原始数据

object[][] rawData = new object[][]  
    {  
        new object[] { "A1", "B1", "C1" },  
        new object[] { "A1", "B1", "C2" }, 
        new object[] { "A2", "B2", "C3" },  
        new object[] { "A2", "B2", "C4" } 
        // .. more  
    };

Answer 4

我想看看我是否可以在没有匿名实例的情况下编写此代码。这还不错：

IEnumerable<X> myList =
  from raw0 in rawData
  group raw0 by raw0[0] into g0
  let g1s =
  (
    from raw1 in g0
    group raw1 by raw1[1]
  )
  from g1 in g1s
  select new X()
  {
    A = g0.Key,
    B = g1.Key,
    C = g1.Select(raw2 => raw2[2]).ToList()
  }

Linq：将扁平结构转换为分层结构

4 个答案: