C# linq group by question - group by remainder

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Hello,

I am trying to Group elements on the remainder of an integer when they are divided by 5.

C#

nt[] numbers = { 5, 4, 1, 3, 9, 8, 6, 7, 2, 0 };

Such that, for instance, numbers with a remainder of 0 when divided by 5 are -- > 5 and 0 (0, new [] {5, 0})

However, I am receiving the following error:

I understand that it expects the Enumerable return type instead of IGrouping, but I cannot figure out how to fix it. Any help would be appreciated. Thank you.

Error	CS0266	Cannot implicitly convert type 'System.Collections.Generic.IEnumerable<<anonymous type: int Key, System.Linq.IGrouping<int, int> g>>'  
           
  to
 'System.Collections.Generic.IEnumerable<(int remainder, System.Collections.Generic.IEnumerable<int> numbers)>'

What I have tried:

C#

public static IEnumerable<(int remainder, IEnumerable<int> numbers)> Grouping()
        {
            int[] numbers = { 5, 4, 1, 3, 9, 8, 6, 7, 2, 0 };

            var numberGroups = from n in numbers
                               group n by n % 5 into g
                               select new { g.Key, g };

            return numberGroups;
        }

Posted 3-Mar-22 9:40am

LuckyChloe

Updated 4-Mar-22 10:32am

Sean Ewington

v2

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Maciej Los · Accepted Answer · 2022-03-03T11:15:00

Solution 1

Try this:

C#

void Main()
{
	int[] numbers = { 5, 4, 1, 3, 9, 8, 6, 7, 2, 0 };
	
	var result = Grouping(numbers);
	result.Dump();
}

// Define other methods and classes here
public static IEnumerable<(int remainder, IEnumerable<int> numbers)> Grouping(int[] input)
{
    return input
		.GroupBy(n => n % 5)
        .Select(grp=> (grp.Key, grp.Select(x=> x)));
}

Posted 3-Mar-22 11:15am

Maciej Los

Comments

LuckyChloe 3-Mar-22 17:39pm

💖

Maciej Los 4-Mar-22 1:52am

:D

Richard Deeming 4-Mar-22 3:49am

Or with the query syntax:

var numberGroups = from n in numbers
                   group n by n % 5 into g
                   select (g.Key, g.AsEnumerable());

Maciej Los 4-Mar-22 5:04am

Oh, yeah! I do like a lambda expression version ;)

LuckyChloe 4-Mar-22 7:27am

It was very useful thank you. I now know what AsEnumerable() is used for.