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

user_5198 0
Viewed 68720

Let's say I have a sequence.

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

Getting the sequence is not cheap and is dynamically generated, and I want to iterate through it once only.

I want to get 0 - 999999 (i.e. everything but the last element)

I recognize that I could do something like:

sequence.Take(sequence.Count() - 1);

but that results in two enumerations over the big sequence.

Is there a LINQ construct that lets me do:

sequence.TakeAllButTheLastElement();
linq .net c#
22 Answers

The Enumerable.SkipLast(IEnumerable<TSource>, Int32) method was added in .NET Standard 2.1. It does exactly what you want.

IEnumerable<int> sequence = GetSequenceFromExpensiveSource();

var allExceptLast = sequence.SkipLast(1);

From https://docs.microsoft.com/en-us/dotnet/api/system.linq.enumerable.skiplast

Returns a new enumerable collection that contains the elements from source with the last count elements of the source collection omitted.

user_21884070

With C# 8.0 you can use Ranges and indices for that.

var allButLast = sequence[..^1];

By default C# 8.0 requires .NET Core 3.0 or .NET Standard 2.1 (or above). Check this thread to use with older implementations.

user_119177640

I don't think it can get more succinct than this - also ensuring to Dispose the IEnumerator<T>:

public static IEnumerable<T> SkipLast<T>(this IEnumerable<T> source)
{
    using (var it = source.GetEnumerator())
    {
        if (it.MoveNext())
        {
            var item = it.Current;
            while (it.MoveNext())
            {
                yield return item;
                item = it.Current;
            }
        }
    }
}

Edit: technically identical to this answer.

user_7917060

My traditional IEnumerable approach:

/// <summary>
/// Skips first element of an IEnumerable
/// </summary>
/// <typeparam name="U">Enumerable type</typeparam>
/// <param name="models">The enumerable</param>
/// <returns>IEnumerable of type skipping first element</returns>
private IEnumerable<U> SkipFirstEnumerable<U>(IEnumerable<U> models)
{
    using (var e = models.GetEnumerator())
    {
        if (!e.MoveNext()) return;
        for (;e.MoveNext();) yield return e.Current;
        yield return e.Current;
    }
}

/// <summary>
/// Skips last element of an IEnumerable
/// </summary>
/// <typeparam name="U">Enumerable type</typeparam>
/// <param name="models">The enumerable</param>
/// <returns>IEnumerable of type skipping last element</returns>
private IEnumerable<U> SkipLastEnumerable<U>(IEnumerable<U> models)
{
    using (var e = models.GetEnumerator())
    {
        if (!e.MoveNext()) return;
        yield return e.Current;
        for (;e.MoveNext();) yield return e.Current;
    }
}
user_6127170

A simple way would be to just convert to a queue and dequeue until only the number of items you want to skip is left.

public static IEnumerable<T> SkipLast<T>(this IEnumerable<T> source, int n)
{
    var queue = new Queue<T>(source);

    while (queue.Count() > n)
    {
        yield return queue.Dequeue();
    }
}
user_33207560

Could be:

var allBuLast = sequence.TakeWhile(e => e != sequence.Last());

I guess it should be like de "Where" but preserving the order(?).

user_11908620
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