Question

I have a class like this:

class MyClass<T> {
    public string value1 { get; set; }
    public T objT { get; set; }
}

and a list of this class. I would like to use .net 3.5 lambda or linq to get a list of MyClass by distinct value1. I guess this is possible and much simpler than the way in .net 2.0 to cache a list like this:

List<MyClass<T>> list; 
...
List<MyClass<T>> listDistinct = new List<MyClass<T>>();
foreach (MyClass<T> instance in list)
{
    // some code to check if listDistinct does contain obj with intance.Value1
    // then listDistinct.Add(instance);
}

What is the lambda or LINQ way to do it?

Answer 1

Both Marc‘s and dahlbyk‘s answers seem to work very well. I have a much simpler solution though. Instead of using Distinct, you can use GroupBy. It goes like this:

var listDistinct
    = list.GroupBy(
        i => i.value1,
        (key, group) => group.First()
    ).ToArray();

Notice that I’ve passed two functions to the GroupBy(). The first is a key selector. The second gets only one item from each group. From your question, I assumed First() was the right one. You can write a different one, if you want to. You can try Last() to see what I mean.

I ran a test with the following input:

var list = new [] {
    new { value1 = "ABC", objT = 0 },
    new { value1 = "ABC", objT = 1 },
    new { value1 = "123", objT = 2 },
    new { value1 = "123", objT = 3 },
    new { value1 = "FOO", objT = 4 },
    new { value1 = "BAR", objT = 5 },
    new { value1 = "BAR", objT = 6 },
    new { value1 = "BAR", objT = 7 },
    new { value1 = "UGH", objT = 8 },
};

The result was:

//{ value1 = ABC, objT = 0 }
//{ value1 = 123, objT = 2 }
//{ value1 = FOO, objT = 4 }
//{ value1 = BAR, objT = 5 }
//{ value1 = UGH, objT = 8 }

I haven’t tested it for performance. I believe that this solution is probably a little bit slower than one that uses Distinct. Despite this disadvantage, there are two great advantages: simplicity and flexibility. Usually, it’s better to favor simplicity over optimization, but it really depends on the problem you’re trying to solve.