Linq - SelectMany Confusion

Question

From what I understand from the documentation of SelectMany, one could use it to produce a (flattened) sequence of a 1-many relationship.

I have following classes

<

Answer 1

Here is another option using SelectMany

var customerOrders = customers.SelectMany(
  c => orders.Where(o => o.CustomerId == c.Id)
    .Select(p => new {CustomerId = c.Id, OrderDescription = p.Description}));

If you use the Entity Framework or LINQ to Sql and you have an association (relationship) between the entities, then you can do so:

var customerOrders = customers.SelectMany(
  c => c.orders
   .Select(p => new {CustomerId = c.Id, OrderDescription = p.Description}));

Answer 2

SelectMany() works like Select, but with that extra feature of flattening a collection that is selected. It should be used whenever you want a projection of elements of sub-collections, and don't care about the sub-collection's containing element.

For example, let's say your domain looked like this:

public class Customer
  {
    public int Id { get; set; }
    public string Name { get; set; }
    public List<Order> Orders { get; set; }
  }

  class Order
  {
    public int Id { get; set; }
    public Customer Customer { get; set; }
    public string Description { get; set; }
  }

To get the same list you wanted, your Linq would look something like this:

var customerOrders = Customers
                        .SelectMany(c=>c.Orders)
                        .Select(o=> new { CustomerId = o.Customer.Id, 
                                           OrderDescription = o.Description });

... which will produce the same result without needing the flat collection of Orders. The SelectMany takes each Customer's Orders collection and iterates through that to produce an IEnumerable<Order> from an IEnumerable<Customer>.

Answer 3

Though this is an old question, I thought I would improve the excellent answers a little:

SelectMany returns a list (which may be empty) for each element of the controlling list. Each element in these result lists are enumerated into the expressions' output sequence and so are concatenated into the result. Hence, a' list -> b' list[] -> concatenate -> b' list.

using System;
using Microsoft.VisualStudio.TestTools.UnitTesting;
using System.Linq;
using System.Diagnostics;
namespace Nop.Plugin.Misc.WebServices.Test
{
    [TestClass]
    public class TestBase
    {
        [TestMethod]
        public void TestMethod1()
        {  //See result in TestExplorer - test output 
            var a = new int[]{7,8};
            var b = new int[]
                    {12,23,343,6464,232,75676,213,1232,544,86,97867,43};
            Func<int, int, bool> numberHasDigit = 
                    (number
                     , digit) => 
                         ( number.ToString().Contains(digit.ToString()) );

            Debug.WriteLine("Unfiltered: All elements of 'b' for each element of 'a'");
            foreach(var l in a.SelectMany(aa => b))
                Debug.WriteLine(l);
            Debug.WriteLine(string.Empty);
            Debug.WriteLine("Filtered:" +  
            "All elements of 'b' for each element of 'a' filtered by the 'a' element");
            foreach(var l in a.SelectMany(aa => b.Where(bb => numberHasDigit(bb, aa))))
                Debug.WriteLine(l);
        }
    }
}

Answer 4

Here is your query using SelectMany, modeled exactly after your example. Same output!

        var customerOrders2 = customers.SelectMany(
            c => orders.Where(o => o.CustomerId == c.Id),
            (c, o) => new { CustomerId = c.Id, OrderDescription = o.Description });

The first argument maps each customer to a collection of orders (completely analagous to the 'where' clause you already have).

The second argument transforms each matched pair {(c1, o1), (c1, o2) .. (c3, o9)} into a new type, which I've made the same as your example.

So:

• arg1 maps each element in the base collection to another collection.
• arg2 (optional) transforms each pair into a new type

The resulting collection is flat like you'd expect in your original example.

If you were to omit the second argument, you would end up with a collection of all orders the match up to a customer. It'd be just that, a flat collection of Order objects.

Using it takes a lot of getting used to, I still have trouble wrapping my head around it sometimes. :(