CODEWITHSUNDEEP
c#.netlinq
Mike
Mike

Reputation: 5770

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

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();

Upvotes: 181

Views: 84594

Answers (22)

Justin Lessard
Justin Lessard

Reputation: 11901

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://learn.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.

Upvotes: 107

Emiliano Ruiz
Emiliano Ruiz

Reputation: 482

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.

Upvotes: 22

John Stevens
John Stevens

Reputation: 1

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();
    }
}

Upvotes: -3

Tarik
Tarik

Reputation: 11209

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();
            }
        }
    }
}

Upvotes: 1

einord
einord

Reputation: 2325

If speed is a requirement, this old school way should be the fastest, even though the code doesn't look as smooth as linq could make it.

int[] newSequence = int[sequence.Length - 1];
for (int x = 0; x < sequence.Length - 1; x++)
{
    newSequence[x] = sequence[x];
}

This requires that the sequence is an array since it has a fixed length and indexed items.

Upvotes: -3

Chibueze Opata
Chibueze Opata

Reputation: 10054

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;
    }
}

Upvotes: -1

Noldorin
Noldorin

Reputation: 147290

Nothing in the BCL (or MoreLinq I believe), but you could create your own extension method.

public static IEnumerable<T> TakeAllButLast<T>(this IEnumerable<T> source)
{
    using (var enumerator = source.GetEnumerator())
        bool first = true;
        T prev;
        while(enumerator.MoveNext())
        {
            if (!first)
                yield return prev;
            first = false;
            prev = enumerator.Current;
        }
    }
}

Upvotes: 12

Robert Schmidt
Robert Schmidt

Reputation: 699

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.

Upvotes: 1

RoJaIt
RoJaIt

Reputation: 461

You could write:

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

Upvotes: -1

Matthew Layton
Matthew Layton

Reputation: 42260

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);

Upvotes: 3

ddur
ddur

Reputation: 75

    public static IEnumerable<T> NoLast<T> (this IEnumerable<T> items) {
        if (items != null) {
            var e = items.GetEnumerator();
            if (e.MoveNext ()) {
                T head = e.Current;
                while (e.MoveNext ()) {
                    yield return head; ;
                    head = e.Current;
                }
            }
        }
    }

Upvotes: 1

Adhemar
Adhemar

Reputation: 151

The solution that I use for this problem is slightly more elaborate.

My util static class contains an extension method MarkEnd which converts the T-items in EndMarkedItem<T>-items. Each element is marked with an extra int, which is either 0; or (in case one is particularly interested in the last 3 items) -3, -2, or -1 for the last 3 items.

This could be useful on its own, e.g. when you want to create a list in a simple foreach-loop with commas after each element except the last 2, with the second-to-last item followed by a conjunction word (such as “and” or “or”), and the last element followed by a point.

For generating the entire list without the last n items, the extension method ButLast simply iterates over the EndMarkedItem<T>s while EndMark == 0.

If you don’t specify tailLength, only the last item is marked (in MarkEnd()) or dropped (in ButLast()).

Like the other solutions, this works by buffering.

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

namespace Adhemar.Util.Linq {

    public struct EndMarkedItem<T> {
        public T Item { get; private set; }
        public int EndMark { get; private set; }

        public EndMarkedItem(T item, int endMark) : this() {
            Item = item;
            EndMark = endMark;
        }
    }

    public static class TailEnumerables {

        public static IEnumerable<T> ButLast<T>(this IEnumerable<T> ts) {
            return ts.ButLast(1);
        }

        public static IEnumerable<T> ButLast<T>(this IEnumerable<T> ts, int tailLength) {
            return ts.MarkEnd(tailLength).TakeWhile(te => te.EndMark == 0).Select(te => te.Item);
        }

        public static IEnumerable<EndMarkedItem<T>> MarkEnd<T>(this IEnumerable<T> ts) {
            return ts.MarkEnd(1);
        }

        public static IEnumerable<EndMarkedItem<T>> MarkEnd<T>(this IEnumerable<T> ts, int tailLength) {
            if (tailLength < 0) {
                throw new ArgumentOutOfRangeException("tailLength");
            }
            else if (tailLength == 0) {
                foreach (var t in ts) {
                    yield return new EndMarkedItem<T>(t, 0);
                }
            }
            else {
                var buffer = new T[tailLength];
                var index = -buffer.Length;
                foreach (var t in ts) {
                    if (index < 0) {
                        buffer[buffer.Length + index] = t;
                        index++;
                    }
                    else {
                        yield return new EndMarkedItem<T>(buffer[index], 0);
                        buffer[index] = t;
                        index++;
                        if (index == buffer.Length) {
                            index = 0;
                        }
                    }
                }
                if (index >= 0) {
                    for (var i = index; i < buffer.Length; i++) {
                        yield return new EndMarkedItem<T>(buffer[i], i - buffer.Length - index);
                    }
                    for (var j = 0; j < index; j++) {
                        yield return new EndMarkedItem<T>(buffer[j], j - index);
                    }
                }
                else {
                    for (var k = 0; k < buffer.Length + index; k++) {
                        yield return new EndMarkedItem<T>(buffer[k], k - buffer.Length - index);
                    }
                }
            }    
        }
    }
}

Upvotes: 1

Joren
Joren

Reputation: 14746

Because I'm not a fan of explicitly using an Enumerator, here's an alternative. Note that the wrapper methods are needed to let invalid arguments throw early, rather than deferring the checks until the sequence is actually enumerated.

public static IEnumerable<T> DropLast<T>(this IEnumerable<T> source)
{
    if (source == null)
        throw new ArgumentNullException("source");

    return InternalDropLast(source);
}

private static IEnumerable<T> InternalDropLast<T>(IEnumerable<T> source)
{
    T buffer = default(T);
    bool buffered = false;

    foreach (T x in source)
    {
        if (buffered)
            yield return buffer;

        buffer = x;
        buffered = true;
    }
}

As per Eric Lippert's suggestion, it easily generalizes to n items:

public static IEnumerable<T> DropLast<T>(this IEnumerable<T> source, int n)
{
    if (source == null)
        throw new ArgumentNullException("source");

    if (n < 0)
        throw new ArgumentOutOfRangeException("n", 
            "Argument n should be non-negative.");

    return InternalDropLast(source, n);
}

private static IEnumerable<T> InternalDropLast<T>(IEnumerable<T> source, int n)
{
    Queue<T> buffer = new Queue<T>(n + 1);

    foreach (T x in source)
    {
        buffer.Enqueue(x);

        if (buffer.Count == n + 1)
            yield return buffer.Dequeue();
    }
}

Where I now buffer before yielding instead of after yielding, so that the n == 0 case does not need special handling.

Upvotes: 44

jrr
jrr

Reputation: 366

A slight variation on the accepted answer, which (for my tastes) is a bit simpler:

    public static IEnumerable<T> AllButLast<T>(this IEnumerable<T> enumerable, int n = 1)
    {
        // for efficiency, handle degenerate n == 0 case separately 
        if (n == 0)
        {
            foreach (var item in enumerable)
                yield return item;
            yield break;
        }

        var queue = new Queue<T>(n);
        foreach (var item in enumerable)
        {
            if (queue.Count == n)
                yield return queue.Dequeue();

            queue.Enqueue(item);
        }
    }

Upvotes: 2

Brady Moritz
Brady Moritz

Reputation: 8903

Why not just .ToList<type>() on the sequence, then call count and take like you did originally..but since it's been pulled into a list, it shouldnt do an expensive enumeration twice. Right?

Upvotes: 1

SmallBizGuy
SmallBizGuy

Reputation: 89

if you don't have time to roll out your own extension, here's a quicker way:

var next = sequence.First();
sequence.Skip(1)
    .Select(s => 
    { 
        var selected = next;
        next = s;
        return selected;
    });

Upvotes: 4

Guillermo Ares
Guillermo Ares

Reputation: 197

Could be:

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

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

Upvotes: -3

einord
einord

Reputation: 2325

I would probably do something like this:

sequence.Where(x => x != sequence.LastOrDefault())

This is one iteration with a check that it isn't the last one for each time though.

Upvotes: -6

Alex Aza
Alex Aza

Reputation: 78457

It would be helpful if .NET Framework was shipped with extension method like this.

public static IEnumerable<T> SkipLast<T>(this IEnumerable<T> source, int count)
{
    var enumerator = source.GetEnumerator();
    var queue = new Queue<T>(count + 1);

    while (true)
    {
        if (!enumerator.MoveNext())
            break;
        queue.Enqueue(enumerator.Current);
        if (queue.Count > count)
            yield return queue.Dequeue();
    }
}

Upvotes: 7

Silas Davis
Silas Davis

Reputation: 742

A slight expansion on Joren's elegant solution:

public static IEnumerable<T> Shrink<T>(this IEnumerable<T> source, int left, int right)
{
    int i = 0;
    var buffer = new Queue<T>(right + 1);

    foreach (T x in source)
    {
        if (i >= left) // Read past left many elements at the start
        {
            buffer.Enqueue(x);
            if (buffer.Count > right) // Build a buffer to drop right many elements at the end
                yield return buffer.Dequeue();    
        } 
        else i++;
    }
}
public static IEnumerable<T> WithoutLast<T>(this IEnumerable<T> source, int n = 1)
{
    return source.Shrink(0, n);
}
public static IEnumerable<T> WithoutFirst<T>(this IEnumerable<T> source, int n = 1)
{
    return source.Shrink(n, 0);
}

Where shrink implements a simple count forward to drop the first left many elements and the same discarded buffer to drop the last right many elements.

Upvotes: 3

Dario
Dario

Reputation: 49218

I don't know a Linq solution - But you can easily code the algorithm by yourself using generators (yield return). 

public static IEnumerable<T> TakeAllButLast<T>(this IEnumerable<T> source) {
    var it = source.GetEnumerator();
    bool hasRemainingItems = false;
    bool isFirst = true;
    T item = default(T);

    do {
        hasRemainingItems = it.MoveNext();
        if (hasRemainingItems) {
            if (!isFirst) yield return item;
            item = it.Current;
            isFirst = false;
        }
    } while (hasRemainingItems);
}

static void Main(string[] args) {
    var Seq = Enumerable.Range(1, 10);

    Console.WriteLine(string.Join(", ", Seq.Select(x => x.ToString()).ToArray()));
    Console.WriteLine(string.Join(", ", Seq.TakeAllButLast().Select(x => x.ToString()).ToArray()));
}

Or as a generalized solution discarding the last n items (using a queue like suggested in the comments):

public static IEnumerable<T> SkipLastN<T>(this IEnumerable<T> source, int n) {
    var  it = source.GetEnumerator();
    bool hasRemainingItems = false;
    var  cache = new Queue<T>(n + 1);

    do {
        if (hasRemainingItems = it.MoveNext()) {
            cache.Enqueue(it.Current);
            if (cache.Count > n)
                yield return cache.Dequeue();
        }
    } while (hasRemainingItems);
}

static void Main(string[] args) {
    var Seq = Enumerable.Range(1, 4);

    Console.WriteLine(string.Join(", ", Seq.Select(x => x.ToString()).ToArray()));
    Console.WriteLine(string.Join(", ", Seq.SkipLastN(3).Select(x => x.ToString()).ToArray()));
}

Upvotes: 69

Kamarey
Kamarey

Reputation: 11079

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);

Upvotes: 50

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