LINQ to Get Closest Value?

Question

I have a List, MyStuff has a property of Type Float.

There are objects with property values of 10,20,22,30.

I need to write a query that finds the objects cl

Answer 1

Try sorting them by the absolute value of the difference between the number and 21 and then take the first item:

float closest = MyStuff
    .Select (n => new { n, distance = Math.Abs (n - 21) })
    .OrderBy (p => p.distance)
    .First().n;

Or shorten it according to @Yuriy Faktorovich's comment:

float closest = MyStuff
    .OrderBy(n => Math.Abs(n - 21))
    .First();

Answer 2

List<float> numbers = new List<float>() { 10f, 20f, 22f, 30f };
float pivot = 21f;
var result = numbers.Where(x => x >= pivot).OrderBy(x => x).FirstOrDefault();

OR

var result = (from n in numbers
              where n>=pivot
              orderby n
              select n).FirstOrDefault();

and here comes an extension method:

public static T Closest<T,TKey>(this IEnumerable<T> source, Func<T, TKey> keySelector, TKey pivot) where TKey : IComparable<TKey>
{
    return source.Where(x => pivot.CompareTo(keySelector(x)) <= 0).OrderBy(keySelector).FirstOrDefault();
}

Usage:

var result = numbers.Closest(n => n, pivot);

Answer 3

Based on this post at the Microsoft Linq forums:

var numbers = new List<float> { 10f, 20f, 22f, 30f };
var target = 21f;

//gets single number which is closest
var closest = numbers.Select( n => new { n, distance = Math.Abs( n - target ) } )
  .OrderBy( p => p.distance )
  .First().n;

//get two closest
var take = 2;
var closests = numbers.Select( n => new { n, distance = Math.Abs( n - target ) } )
  .OrderBy( p => p.distance )
  .Select( p => p.n )
  .Take( take );       

//gets any that are within x of target
var within = 1;
var withins = numbers.Select( n => new { n, distance = Math.Abs( n - target ) } )
  .Where( p => p.distance <= within )
  .Select( p => p.n );

Answer 4

Here's a solution that satisfies the second query in linear time:

var pivot = 21f;
var closestBelow = pivot - numbers.Where(n => n <= pivot)
                                  .Min(n => pivot - n);

(Edited from 'above' to 'below' after clarification)

As for the first query, it would be easiest to use MoreLinq's MinBy extension:

var closest = numbers.MinBy(n => Math.Abs(pivot - n));

It's also possible to do it in standard LINQ in linear time, but with 2 passes of the source:

var minDistance = numbers.Min(n => Math.Abs(pivot - n));
var closest = numbers.First(n => Math.Abs(pivot - n) == minDistance);

If efficiency is not an issue, you could sort the sequence and pick the first value in O(n * log n) as others have posted.