How to find consecutive same values items as a Linq group

Question

var schedules = new List{
    new Item { Id=1, Name = \"S\" },
    new Item { Id=2, Name = \"P\" },
    new Item { Id=3, Name = \"X\" },
    new Item { Id=4,         


        

Answer 1

What you're looking for here is a GroupWhile method.

Credit to user L.B for the solution. Go give his original answer an UpDoot https://stackoverflow.com/a/20469961/30155

    var schedules = new List{
        new Item { Id=1, Name = "S" },
        new Item { Id=2, Name = "P" },
        new Item { Id=3, Name = "X" },
        new Item { Id=4, Name = "X" },
        new Item { Id=5, Name = "P" },
        new Item { Id=6, Name = "P" },
        new Item { Id=7, Name = "P" },
        new Item { Id=8, Name = "S" }
    };

    var results = schedules
        .GroupWhile((preceding, next) => preceding.Name == next.Name) 
        //Group items, while the next is equal to the preceding one
        .Where(s => s.Count() > 1)
        //Only include results where the generated sublist have more than 1 element.
        .ToList();

    foreach (var sublist in results)
    {
        foreach (Item i in sublist)
        {
            Console.WriteLine($"{i.Name} - {i.Id}");
        }
        Console.WriteLine("");
    }

    Console.ReadLine();

You can add the implementation as an Extension Method to all IEnumerable like so.

public static class Extensions
{
    public static IEnumerable> GroupWhile(this IEnumerable seq, Func condition)
    {
        T prev = seq.First();
        List list = new List() { prev };

        foreach (T item in seq.Skip(1))
        {
            if (condition(prev, item) == false)
            {
                yield return list;
                list = new List();
            }
            list.Add(item);
            prev = item;
        }

        yield return list;
    }
}