How to use linq to find the minimum

Question

I have a class A { public float Score; ... } and an IEnumerable<A> items and would like to find the A which has minimal score.

Using items.Min(x => x.Score) gives the minimal score and not the instance with minimal score.

How can I get the instance by iterating only once through my data?

Edit: So long there are three main solutions:

• Writing an extension method (proposed by Svish). Pros: Easy to use and evaluates Score only once per item. Cons: Needs an extension method. (I choosed this solution for my application.)

• Using Aggregate (proposed by Daniel Renshaw). Pros: Uses a built-in LINQ method. Cons: Slightly obfuscated to the untrained eye and calls evaluator more than once.

• Implementing IComparable (proposed by cyberzed). Pros: Can use Linq.Min directly. Cons: Fixed to one comparer - can not freely choose comparer when performing the minimum computation.

Answer 1

Try items.OrderBy(s => s.Score).FirstOrDefault();

Answer 2

The quick way as I see it would be implementing IComparable for your class A (if possible ofcourse)

class A : IComparable<A>

It's a simple implementation where you write the CompareTo(A other) have a look at IEnumerable(of T).Min on MSDN for a reference guide

Answer 3

Jamie Penney got my vote. You can also say

items.OrderBy(s => s.Score).Take(1);

Which has the same effect. Use Take(5) to take the lowest 5 etc.

Answer 4

This can be solved with a little simple iteration:

float minScore = float.MaxValue;
A minItem = null;

foreach(A item in items)
{
   if(item.Score < minScore)
       minItem = item;
}

return minItem;

It's not a nice LINQ query, but it does avoid a sorting operation and only iterates the list once, as per the question's requirements.

Answer 5

Have a look at the MinBy extension method in morelinq by Jon Skeet (unless I'm mistaken).

Answer 6

Use Aggregate:

items.Aggregate((c, d) => c.Score < d.Score ? c : d)

Answer 7

A little fold should do:

var minItem = items.Aggregate((acc, c) => acc.Score < c.Score? acc : c);

Ah... too slow.