Question

So the distinct is based on unique Month/Year, not just one distinct month (so I would want January of 2011 and January of 2012 to be distinct)

 // Test set of data
        List<DateTime> CompleteListOfDates = new List<DateTime>();
        CompleteListOfDates.Add(new DateTime(2011, 1, 1));
        CompleteListOfDates.Add(new DateTime(2011, 1, 5));
        CompleteListOfDates.Add(new DateTime(2011, 3, 1));
        CompleteListOfDates.Add(new DateTime(2011, 5, 1));
        CompleteListOfDates.Add(new DateTime(2011, 5, 1));
        CompleteListOfDates.Add(new DateTime(2012, 1, 1));
        CompleteListOfDates.Add(new DateTime(2012, 2, 1));

        List<DateTime> UniqueMonthYears = new List<DateTime>();

        /* need distinct list of DateTimes that are distinct by Month and Year and should be in UniqueMonthYears
         For example:
        new DateTime(2011, 1, 1)
        new DateTime(2011, 3, 1)
        new DateTime(2011, 5, 1)
        new DateTime(2012, 1, 1)
        new DateTime(2012, 2, 1)
        */

Answer 1

List<DateTime> result = source
  .Select(d => new DateTime(d.Year, d.Month, 1))
  .Distinct()
  .ToList();

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Also useful:

ILookup<DateTime, Order> ordersByMonth = ordersSource
  .ToLookup(order => new DateTime(order.OrderDate.Year, order.OrderDate.Month, 1));