在一個LINQ查詢中獲取兩列的總和

Question

假設我有一個名為Items的表（ID int，Done int，Total int）

我可以通過兩個查詢來完成：

int total = m.Items.Sum(p=>p.Total)
int done = m.Items.Sum(p=>p.Done)

但我想在一個查詢中執行此操作，如下所示：

var x = from p in m.Items select new { Sum(p.Total), Sum(p.Done)};

當然有一種方法可以從LINQ語法中調用聚合函數......？

Answer 1

這樣就可以了：

from p in m.Items
group p by 1 into g
select new
{
    SumTotal = g.Sum(x => x.Total), 
    SumDone = g.Sum(x => x.Done) 
};

Answer 2

要對表進行求和，請按常量分組：

from p in m.Items
group p by 1 into g
select new {
    SumTotal = g.Sum(x => x.Total),
    SumDone = g.Sum(x => x.Done)
}

Answer 3

怎么樣

   m.Items.Select(item => new { Total = item.Total, Done = item.Done })
          .Aggregate((t1, t2) => new { Total = t1.Total + t2.Total, Done = t1.Done + t2.Done });

Answer 4

弄清楚在我的其余代碼中提取總和或其他聚合的位置使我感到困惑，直到我記得我構造的變量是一個Iqueryable。 假設我們的數據庫中有一個由Orders組成的表，我們想為ABC公司生成一個摘要：

var myResult = from g in dbcontext.Ordertable
               group p by (p.CUSTNAME == "ABC") into q  // i.e., all of ABC company at once
               select new
{
    tempPrice = q.Sum( x => (x.PRICE ?? 0m) ),  // (?? makes sure we don't get back a nullable)
    tempQty = q.Sum( x => (x.QTY ?? 0m) )
};

現在有趣的部分 - tempPrice和tempQty沒有在任何地方聲明，但它們必須是myResult的一部分，不是嗎？ 訪問它們如下：

Console.Writeline(string.Format("You ordered {0} for a total price of {1:C}",
                                 myResult.Single().tempQty,
                                 myResult.Single().tempPrice ));

還可以使用許多其他可查詢方法。

Answer 5

使用幫助器元組類，您可以使用自己的或者.NET 4 - 標准的那些：

var init = Tuple.Create(0, 0);

var res = m.Items.Aggregate(init, (t,v) => Tuple.Create(t.Item1 + v.Total, t.Item2 + v.Done));

res.Item1是Done列的Total列和res.Item2的Total 。

Answer 6

//Calculate the total in list field values
//Use the header file: 

Using System.Linq;
int i = Total.Sum(G => G.First);

//By using LINQ to calculate the total in a list field,

var T = (from t in Total group t by Total into g select g.Sum(t => t.First)).ToList();

//Here Total is a List and First is the one of the integer field in list(Total)

Answer 7

這已經得到了回答，但是其他答案仍然會對集合進行多次迭代（多次調用Sum）或者創建大量的中間對象/元組可能沒問題，但如果不是，那么你可以創建一個擴展方法（或多個）以舊式方式執行，但非常適合LINQ表達式。

這樣的擴展方法如下所示：

public static Tuple<int, int> Sum<T>(this IEnumerable<T> collection, Func<T, int> selector1, Func<T, int> selector2)
{
    int a = 0;
    int b = 0;

    foreach(var i in collection)
    {
        a += selector1(i);
        b += selector2(i);
    }

    return Tuple.Create(a, b);
}

你可以像這樣使用它：

public class Stuff
{
    public int X;
    public int Y;
}

//...

var stuffs = new List<Stuff>()
{
    new Stuff { X = 1, Y = 10 }, 
    new Stuff { X = 1, Y = 10 }
};

var sums = stuffs.Sum(s => s.X, s => s.Y);

Answer 8

使用C＃7.0中引入的元組語言支持，您可以使用以下LINQ表達式解決此問題：

var itemSums = m.Items.Aggregate((Total: 0, Done: 0), (sums, item) => (sums.Total + item.Total, sums.Done + item.Done));

完整代碼示例：

var m = new
{
    Items = new[]
    {
        new { Total = 10, Done = 1 },
        new { Total = 10, Done = 1 },
        new { Total = 10, Done = 1 },
        new { Total = 10, Done = 1 },
        new { Total = 10, Done = 1 },
    },
};

var itemSums = m.Items.Aggregate((Total: 0, Done: 0), (sums, item) => (sums.Total + item.Total, sums.Done + item.Done));

Console.WriteLine($"Sum of Total: {itemSums.Total}, Sum of Done: {itemSums.Done}");

Answer 9

當你使用group by Linq創建一個新的項目集合時，你有兩個項目集合。

這是兩個問題的解決方案：

1. 在一次迭代中總結任意數量的成員
2. 避免重復項目的集合

碼：

public static class LinqExtensions
{
  /// <summary>
  /// Computes the sum of the sequence of System.Double values that are obtained 
  /// by invoking one or more transform functions on each element of the input sequence.
  /// </summary>
  /// <param name="source">A sequence of values that are used to calculate a sum.</param>
  /// <param name="selectors">The transform functions to apply to each element.</param>    
  public static double[] SumMany<TSource>(this IEnumerable<TSource> source, params Func<TSource, double>[] selectors)
  {
    if (selectors.Length == 0)
    {
      return null;
    }
    else
    {
      double[] result = new double[selectors.Length];

      foreach (var item in source)
      {
        for (int i = 0; i < selectors.Length; i++)
        {
          result[i] += selectors[i](item);
        }
      }

      return result;
    }
  }

  /// <summary>
  /// Computes the sum of the sequence of System.Decimal values that are obtained 
  /// by invoking one or more transform functions on each element of the input sequence.
  /// </summary>
  /// <param name="source">A sequence of values that are used to calculate a sum.</param>
  /// <param name="selectors">The transform functions to apply to each element.</param>
  public static double?[] SumMany<TSource>(this IEnumerable<TSource> source, params Func<TSource, double?>[] selectors) 
  { 
    if (selectors.Length == 0)
    {
      return null;
    }
    else
    {
      double?[] result = new double?[selectors.Length];

      for (int i = 0; i < selectors.Length; i++)
      {
        result[i] = 0;
      }

      foreach (var item in source)
      {
        for (int i = 0; i < selectors.Length; i++)
        {
          double? value = selectors[i](item);

          if (value != null)
          {
            result[i] += value;
          }
        }
      }

      return result;
    }
  }
}

這是你必須做總結的方式：

double[] result = m.Items.SumMany(p => p.Total, q => q.Done);

這是一個一般的例子：

struct MyStruct
{
  public double x;
  public double y;
}

MyStruct[] ms = new MyStruct[2];

ms[0] = new MyStruct() { x = 3, y = 5 };
ms[1] = new MyStruct() { x = 4, y = 6 };

// sum both x and y members in one iteration without duplicating the array "ms" by GROUPing it
double[] result = ms.SumMany(a => a.x, b => b.y);

如你看到的

result[0] = 7 
result[1] = 11