Using LINQ, how would you filter out all but one item of a particular criteria from a list?

Question

I realize my title probably isn't very clear so here's an example:

I have a list of objects with two properties, A and B.

public class Item
{
    public int A { get; set; }
    public int B { get; set; }
}

var list = new List<Item>
{
    new Item() { A = 0, B = 0 },
    new Item() { A = 0, B = 1 },
    new Item() { A = 1, B = 0 },
    new Item() { A = 2, B = 0 },
    new Item() { A = 2, B = 1 },
    new Item() { A = 2, B = 2 },
    new Item() { A = 3, B = 0 },
    new Item() { A = 3, B = 1 },
}

Using LINQ, what's the most elegant way to collapse all the A = 2 items into the first A = 2 item and return along with all the other items? This would be the expected result.

var list = new List<Item>
{
    new Item() { A = 0, B = 0 },
    new Item() { A = 0, B = 1 },
    new Item() { A = 1, B = 0 },
    new Item() { A = 2, B = 0 },
    new Item() { A = 3, B = 0 },
    new Item() { A = 3, B = 1 },
}

I'm not a LINQ expert and already have a "manual" solution but I really like the expressiveness of LINQ and was curious to see if it could be done better.

Answer 1

How about:

var collapsed = list.GroupBy(i => i.A)
                    .SelectMany(g => g.Key == 2 ? g.Take(1) : g);

The idea is to first group them by A and then select those again (flattening it with .SelectMany) but in the case of the Key being the one we want to collapse, we just take the first entry with Take(1).

Answer 2

One way you can accomplish this is with GroupBy. Group the items by A, and use a SelectMany to project each group into a flat list again. In the SelectMany, check if A is 2 and if so Take(1), otherwise return all results for that group. We're using Take instead of First because the result has to be IEnumerable.

var grouped = list.GroupBy(g => g.A);
var collapsed = grouped.SelectMany(g =>
{
     if (g.Key == 2)
     {
         return g.Take(1);
     }
     return g;
});