How to take all but the last element in a sequence using LINQ?

Question

Let\'s say I have a sequence.

IEnumerable sequence = GetSequenceFromExpensiveSource();
// sequence now contains: 0,1,2,3,...,999999,1000000


        

Answer 1

This is a general and IMHO elegant solution that will handle all cases correctly:

using System;
using System.Collections.Generic;
using System.Linq;

public class Program
{
    public static void Main()
    {
        IEnumerable<int> r = Enumerable.Range(1, 20);
        foreach (int i in r.AllButLast(3))
            Console.WriteLine(i);

        Console.ReadKey();
    }
}

public static class LinqExt
{
    public static IEnumerable<T> AllButLast<T>(this IEnumerable<T> enumerable, int n = 1)
    {
        using (IEnumerator<T> enumerator = enumerable.GetEnumerator())
        {
            Queue<T> queue = new Queue<T>(n);

            for (int i = 0; i < n && enumerator.MoveNext(); i++)
                queue.Enqueue(enumerator.Current);

            while (enumerator.MoveNext())
            {
                queue.Enqueue(enumerator.Current);
                yield return queue.Dequeue();
            }
        }
    }
}

Answer 2

As an alternative to creating your own method and in a case the elements order is not important, the next will work:

var result = sequence.Reverse().Skip(1);

Answer 3

If you can get the Count or Length of an enumerable, which in most cases you can, then just Take(n - 1)

Example with arrays

int[] arr = new int[] { 1, 2, 3, 4, 5 };
int[] sub = arr.Take(arr.Length - 1).ToArray();

Example with IEnumerable<T>

IEnumerable<int> enu = Enumerable.Range(1, 100);
IEnumerable<int> sub = enu.Take(enu.Count() - 1);

Answer 4

You could write:

var list = xyz.Select(x=>x.Id).ToList();
list.RemoveAt(list.Count - 1);