獲取兩列的總和一個LINQ查詢

Question

讓我們說，我有一個表稱爲 項目（ID INT，完成INT，總INT）

我可以用兩個查詢做到這一點：

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

但我想這樣做在一個查詢中，這樣的事情：

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

當然還有一種方法叫聚合函數從LINQ語法...？

Answer 1

這將這樣的伎倆：

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

在一名助手的元組類，無論是你自己或—在.NET 4級—通用的標準，你可以這樣做：

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

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

而且res.Item1是總的Total柱和res.Item2的Done列。

Answer 3

如何

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

總結表，組由一個常數：

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 5

搞清楚提取的款項或者在我的其他代碼中的其他集合混淆了我，直到我記得我構建的變量是一個Iqueryable。假設我們有我們的訂單組成的數據庫中的表，我們要生產的ABC公司的一個總結：

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)) 
}; 

現在最有趣的部分 - tempPrice和tempQty沒有在任何聲明，但他們必須是一部分myResult的，不是？如下訪問它們：

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

也可以使用許多其他的Queryable方法。

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

當您使用組由，所以你有一個項目的兩個集合的LINQ創建項目的新的集合。

下面就以這兩個問題的解決方案：在一個迭代和

避免重複你的項目的集合

1. 總結成員的任何量

代碼：

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; 
    } 
    } 
} 

這裏的你必須做的總結方式：

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

下面是一個普通的例子：

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 

Answer 8

這已經被回答了，但其他的答案仍然會做在收集多次迭代（多個呼叫或者創建大量可能很好的中間對象/元組，但是如果不是這樣，那麼您可以創建一個擴展方法（或多個），它以傳統方式進行，但很適合LINQ表達式。

這樣的擴展方法是這樣的：

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); 
} 

而且你可以使用它像這樣：

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);