假设我有一张名为 项目（ID int、Done int、Total int）
我可以通过两个查询来做到这一点：
int total = m.Items.Sum(p=>p.Total)
int done = m.Items.Sum(p=>p.Done)
但我想在一个查询中完成，如下所示：
var x = from p in m.Items select new { Sum(p.Total), Sum(p.Done)};
肯定有一种方法可以从 LINQ 语法中调用聚合函数...？
【问题讨论】：
标签： c# .net linq linq-to-sql sum
使用C# 7.0 中引入的元组语言支持，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}");
【讨论】：
这已经得到回答，但其他答案仍会在集合上进行多次迭代（多次调用 Sum）或创建许多中间对象/元组，这可能很好，但如果不是，那么你可以创建一个（或多个）扩展方法，以老式方式执行此操作，但非常适合 LINQ 表达式。
这样的扩展方法如下所示：
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);
}
你可以这样使用它：
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);
【讨论】：
这样就可以了：
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)
};
【讨论】：
group p by p into g。
要对表格求和，请按常数分组：
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)
}
【讨论】：
当您使用 group by 时，Linq 会创建一个新的项目集合，因此您有两个项目集合。
以下是解决这两个问题的方法：
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;
}
}
}
这是你必须进行求和的方式：
double[] result = m.Items.SumMany(p => p.Total, q => q.Done);
这是一个通用示例：
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);
如你所见
result[0] = 7
result[1] = 11
【讨论】：
//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)
【讨论】：
弄清楚在我的其余代码中从何处提取总和或其他聚合使我感到困惑，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 没有在任何地方声明，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 Queryable 方法。
【讨论】：
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 });
【讨论】：
使用辅助元组类，无论是您自己的还是（在 .NET 4 中）标准的，您都可以这样做：
var init = Tuple.Create(0, 0);
var res = m.Items.Aggregate(init, (t,v) => Tuple.Create(t.Item1 + v.Total, t.Item2 + v.Done));
而res.Item1 是Total 列和Done 列的res.Item2 的总和。
【讨论】：