Question

让我说我有一张叫做的桌子 Items（ID int，Done int，Total int）

我可以通过两个查询来完成：

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

但是我想在一个查询中执行此操作，如下所示：

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

当然有一种方法可以从LINQ语法调用聚合函数......？

Answer 1

这样可以解决问题：

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

要对表进行求和，请按常量分组：

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

怎么样

   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码中提取总和或其他聚合的位置使我感到困惑，直到我记得我构造的变量是一个Iqueryable。假设我们的数据库中有一个由Orders组成的表，我们想为ABC公司生成一个摘要：

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) )
};

现在有趣的部分 - tempPrice和tempQty未在任何地方声明，但它们必须是myResult的一部分，不是吗？按如下方式访问它们：

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

也可以使用许多其他可查询方法。

Answer 5

使用帮助器元组类，您可以使用自己的or-in .NET 4 - 标准的那些：

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Total列的res.Item2列和Done的总和。

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

已经回答了这个问题，但是其他答案仍会对集合进行多次迭代（多次调用Sum）或者创建大量的中间对象/元组可能没问题，但如果不是，那么你可以创建一个扩展方法（或多个），以旧式方式执行，但非常适合LINQ表达式。

这样的扩展方法如下所示：

private WidgetViewComponent(IWidgetService widgetService)
{
    _WidgetService = widgetService;
}

你可以像这样使用它：

public WidgetViewComponent(IWidgetService widgetService)
{
    _WidgetService = widgetService;
}

Answer 8

使用C# 7.0中引入的对元组的语言支持，您可以使用以下LINQ表达式解决此问题：

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

完整代码示例：

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

当您使用分组时，Linq会创建一个新的项目集合，因此您有两个项目集合。

以下是两个问题的解决方案：

在一次迭代中总结任意数量的成员
避免重复您的项目集合

代码：

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;
    }
  }
}

以下是你必须进行总结的方法：

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

这是一个一般的例子：

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);

你可以看到

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

在一个LINQ查询中获取两列的总和

9 个答案: