I would like to get ALL elements in enumerable whose property is the maximum :
IEnumerable<Person> allOldest = people.GroupBy(p => p.Age) .OrderByDescending(i=>i.Key).First().Select(_=>_); .NET 6 introduce MaxBy which return THE max item:
Person uniqueOldest = people.MaxBy(person => person.Age); MaxBy can't help me ? Is there another more elegant solution than first example?
3 Answers
Using Max is a simple plain way of doing this:
var maxAge = items.Max(y => y.Age); var maxItems = items.Where(x => x.Age == maxAge); If you are OK with adding an external dependency to your project, you could install the respectable MoreLinq package (originally developed by the one and only Jon Skeet) and do this:
IEnumerable<Person> allOldest = MoreLinq.MoreEnumerable.MaxBy(people, p => p.Age); The signature of the MoreEnumerable.MaxBy method:
public static IExtremaEnumerable<TSource> MaxBy<TSource, TKey>( this IEnumerable<TSource> source, Func<TSource, TKey> selector); You could also try to use it as an extension method, by adding this using directive at the top:
using static MoreLinq.Extensions.MaxByExtension; ...but in .NET 6 this will result (most likely) in a name resolution conflict.
This is one of those times that linq isn't necessarily the best option. You could create a linq style extension to do it but really the most direct solution is just to write a function. Here's a simple example. I've used a value tuple as a stand in for your person object.
static void Main(string[] _) { var source = new (int key, int value)[] { (0, 0), (1, 5), (2, 1), (3, 5), (4, 0), (5, 5) }; var allMax = new List<(int, int)>(); int currentMax = int.MinValue; foreach (var (key, value) in source) { if (currentMax < value) { allMax.Clear(); allMax.Add((key, value)); currentMax = value; } else if (currentMax == value) { allMax.Add((key, value)); } } Console.WriteLine($"{allMax.Count} Max valued items found"); foreach (var (key, value) in allMax) Console.WriteLine($"({key}, {value})"); } ncG1vNJzZmirpJawrLvVnqmfpJ%2Bse6S7zGiorp2jqbawutJobXJwZmh%2BdX2OpaCnqV2irrmu2GauoqyYYq6tuIyeo56llaPBtA%3D%3D