GroupBy - Simple 1

Query

public void Linq40()
{
    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 { Remainder = g.Key, Numbers = g };
      
    foreach (var g in numberGroups)
    {
        Console.WriteLine("Numbers with a remainder of {0} when divided by 5:", g.Remainder);
        foreach (var n in g.Numbers)
        {
            Console.WriteLine(n);
        }
    }
}

Lambda Expression

public void DataSetLinq40()
{
    var numbers = testDS.Tables["Numbers"].AsEnumerable();
 
    var numberGroups = numbers.GroupBy(n => n.Field("number")%5).Select(g => new {Remainder = g.Key, Numbers = g});
 
    foreach (var g in numberGroups)
    {
        Console.WriteLine("Numbers with a remainder of {0} when divided by 5:", g.Remainder);
        foreach (var n in g.Numbers)
        {
            Console.WriteLine(n.Field("number"));
        }
    }
}

Output

A backwards list of the digits with a second character of 'i':
nine
eight
six
five