Get sum of two columns in one LINQ query

问题:

let's say that I have a table called Items (ID int, Done int, Total int)

I can do it by two queries:

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

But I'd like to do it in one query, something like this:

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

Surely there is a way to call aggregate functions from LINQ syntax...?

回答1:

This will do the trick:

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) 
};

回答2:

To sum the table, group by a constant:

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)
}

回答3:

How about

   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 });

回答4:

Figuring out where to extract the sums or other aggregate in the rest of my code confused me, until I remembered that the variable I constructed was an Iqueryable. Suppose we have a table in our database composed of Orders, and we want to produce a summary for the ABC company:

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) )
};

Now the fun part -- tempPrice and tempQty aren't declared anywhere but they must be part of myResult, no? Access them as follows:

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

A number of other Queryable methods could be used as well.

回答5:

With a helper tuple class, either your own or—in .NET 4—the standard ones you can do this:

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

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

And res.Item1 is the total of the Total column and res.Item2 of the Done column.

回答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)

回答7:

This has been answered already, but the other answers will still do multiple iterations over the collection (multiple calls to Sum) or create lots of intermediate objects/Tuples which may be fine, but if it isn't, then you can create an extension method (or multiple) that does it the old-fashioned way but fits well in a LINQ expression.

Such an extension method would look like this:

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);
}

And you can use it like this:

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);

回答8:

When you use group by Linq creates a new collection of items so you have two collections of items.

Here's a solution to both problems:

1. summing any amount of members in one iteration and
2. avoid duplicating your item's collection

Code:

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;
    }
  }
}

Here's the way you have to do the summation:

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

Here's a general example:

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);

as you can see

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