Distinct() with lambda?

Question

Right, so I have an enumerable and wish to get distinct values from it.

Using System.Linq, there\'s of course an extension method called Distinct<

Answer 1

All solutions I've seen here rely on selecting an already comparable field. If one needs to compare in a different way, though, this solution here seems to work generally, for something like:

somedoubles.Distinct(new LambdaComparer((x, y) => Math.Abs(x - y) < double.Epsilon)).Count()