我的解决方案是基于c#版本8中引入的范围。

        public static IEnumerable<T> TakeLast<T>(this IEnumerable<T> source, int N)
        {
            return source.ToArray()[(source.Count()-N)..];
        }

在用大多数评价的解决方案(以及我谦卑地提出的解决方案)运行了一个基准测试后:

    public static class TakeLastExtension
    {
        public static IEnumerable<T> TakeLastMarkByers<T>(this IEnumerable<T> source, int takeCount)
        {
            if (source == null) { throw new ArgumentNullException("source"); }
            if (takeCount < 0) { throw new ArgumentOutOfRangeException("takeCount", "must not be negative"); }
            if (takeCount == 0) { yield break; }

            T[] result = new T[takeCount];
            int i = 0;

            int sourceCount = 0;
            foreach (T element in source)
            {
                result[i] = element;
                i = (i + 1) % takeCount;
                sourceCount++;
            }

            if (sourceCount < takeCount)
            {
                takeCount = sourceCount;
                i = 0;
            }

            for (int j = 0; j < takeCount; ++j)
            {
                yield return result[(i + j) % takeCount];
            }
        }

        public static IEnumerable<T> TakeLastKbrimington<T>(this IEnumerable<T> source, int N)
        {
            return source.Skip(Math.Max(0, source.Count() - N));
        }

        public static IEnumerable<T> TakeLastJamesCurran<T>(this IEnumerable<T> source, int N)
        {
            return source.Reverse().Take(N).Reverse();
        }

        public static IEnumerable<T> TakeLastAlex<T>(this IEnumerable<T> source, int N)
        {
            return source.ToArray()[(source.Count()-N)..];
        }
    }

Test

    [MemoryDiagnoser]
    public class TakeLastBenchmark
    {
        [Params(10000)]
        public int N;

        private readonly List<string> l = new();

        [GlobalSetup]
        public void Setup()
        {
            for (var i = 0; i < this.N; i++)
            {
                this.l.Add($"i");
            }
        }

        [Benchmark]
        public void Benchmark1_MarkByers()
        {
            var lastElements = l.TakeLastMarkByers(3).ToList();
        }

        [Benchmark]
        public void Benchmark2_Kbrimington()
        {
            var lastElements = l.TakeLastKbrimington(3).ToList();
        }

        [Benchmark]
        public void Benchmark3_JamesCurran()
        {
            var lastElements = l.TakeLastJamesCurran(3).ToList();
        }

        [Benchmark]
        public void Benchmark4_Alex()
        {
            var lastElements = l.TakeLastAlex(3).ToList();
        }
    }

Program.cs:

var summary = BenchmarkRunner.Run(typeof(TakeLastBenchmark).Assembly);

命令dotnet运行——project .\TestsConsole2。csproj -c Release——logBuildOutput

结果如下:

// *摘要* BenchmarkDotNet=v0.13.2, OS=Windows 10 (10.0.19044.1889/21H2/ novber2021update) AMD Ryzen 5 5600X, 1个CPU, 12个逻辑核和6个物理核 . net SDK = 6.0.401 [主机]:.NET 6.0.9 (6.0.922.41905)， X64 RyuJIT AVX2 DefaultJob: .NET 6.0.9 (6.0.922.41905)， X64 RyuJIT AVX2

事实证明，@Kbrimington提出的解决方案在内存分配和原始性能方面是最有效的。

与循环缓冲区的使用略有不同的实现。基准测试表明，该方法比使用Queue(在System.Linq中实现TakeLast)的方法快大约两倍，但是也有代价——它需要一个随着所请求的元素数量而增长的缓冲区，即使你只有一个小的集合，你也可以得到巨大的内存分配。

public IEnumerable<T> TakeLast<T>(IEnumerable<T> source, int count)
{
    int i = 0;

    if (count < 1)
        yield break;

    if (source is IList<T> listSource)
    {
        if (listSource.Count < 1)
            yield break;

        for (i = listSource.Count < count ? 0 : listSource.Count - count; i < listSource.Count; i++)
            yield return listSource[i];

    }
    else
    {
        bool move = true;
        bool filled = false;
        T[] result = new T[count];

        using (var enumerator = source.GetEnumerator())
            while (move)
            {
                for (i = 0; (move = enumerator.MoveNext()) && i < count; i++)
                    result[i] = enumerator.Current;

                filled |= move;
            }

        if (filled)
            for (int j = i; j < count; j++)
                yield return result[j];

        for (int j = 0; j < i; j++)
            yield return result[j];

    }
}

以下是我的解决方案:

public static class EnumerationExtensions
{
    public static IEnumerable<T> TakeLast<T>(this IEnumerable<T> input, int count)
    {
        if (count <= 0)
            yield break;

        var inputList = input as IList<T>;

        if (inputList != null)
        {
            int last = inputList.Count;
            int first = last - count;

            if (first < 0)
                first = 0;

            for (int i = first; i < last; i++)
                yield return inputList[i];
        }
        else
        {
            // Use a ring buffer. We have to enumerate the input, and we don't know in advance how many elements it will contain.
            T[] buffer = new T[count];

            int index = 0;

            count = 0;

            foreach (T item in input)
            {
                buffer[index] = item;

                index = (index + 1) % buffer.Length;
                count++;
            }

            // The index variable now points at the next buffer entry that would be filled. If the buffer isn't completely
            // full, then there are 'count' elements preceding index. If the buffer *is* full, then index is pointing at
            // the oldest entry, which is the first one to return.
            //
            // If the buffer isn't full, which means that the enumeration has fewer than 'count' elements, we'll fix up
            // 'index' to point at the first entry to return. That's easy to do; if the buffer isn't full, then the oldest
            // entry is the first one. :-)
            //
            // We'll also set 'count' to the number of elements to be returned. It only needs adjustment if we've wrapped
            // past the end of the buffer and have enumerated more than the original count value.

            if (count < buffer.Length)
                index = 0;
            else
                count = buffer.Length;

            // Return the values in the correct order.
            while (count > 0)
            {
                yield return buffer[index];

                index = (index + 1) % buffer.Length;
                count--;
            }
        }
    }

    public static IEnumerable<T> SkipLast<T>(this IEnumerable<T> input, int count)
    {
        if (count <= 0)
            return input;
        else
            return input.SkipLastIter(count);
    }

    private static IEnumerable<T> SkipLastIter<T>(this IEnumerable<T> input, int count)
    {
        var inputList = input as IList<T>;

        if (inputList != null)
        {
            int first = 0;
            int last = inputList.Count - count;

            if (last < 0)
                last = 0;

            for (int i = first; i < last; i++)
                yield return inputList[i];
        }
        else
        {
            // Aim to leave 'count' items in the queue. If the input has fewer than 'count'
            // items, then the queue won't ever fill and we return nothing.

            Queue<T> elements = new Queue<T>();

            foreach (T item in input)
            {
                elements.Enqueue(item);

                if (elements.Count > count)
                    yield return elements.Dequeue();
            }
        }
    }
}

代码有点粗，但作为一个可重用的插入组件，它应该在大多数场景中表现得很好，并且它将使使用它的代码保持良好和简洁。: -)

我的TakeLast for non-IList ' 1是基于与@Mark Byers和@MackieChan回答中相同的环形缓冲算法。有趣的是，它们是如此相似——我是完全独立写的。我猜只有一种方法可以正确地使用环形缓冲区。: -)

看看@kbrimington的答案，可以为IQuerable<T>添加一个额外的检查，以回到与实体框架一起工作的方法——假设我在这一点上没有。

coll.Reverse().Take(N).Reverse().ToList();


public static IEnumerable<T> TakeLast<T>(this IEnumerable<T> coll, int N)
{
    return coll.Reverse().Take(N).Reverse();
}

更新:为了解决clintp的问题:a)使用我上面定义的TakeLast()方法解决了这个问题，但如果你真的想在没有额外方法的情况下做到这一点，那么你只需要认识到Enumerable.Reverse()可以用作扩展方法，你不需要这样使用它:

List<string> mystring = new List<string>() { "one", "two", "three" }; 
mystring = Enumerable.Reverse(mystring).Take(2).Reverse().ToList();

如果你正在处理一个有键的集合(例如，来自数据库的条目)，一个快速(即比选择的答案更快)的解决方案将是

collection.OrderByDescending(c => c.Key).Take(3).OrderBy(c => c.Key);

我试图将效率和简单性结合起来，最后得到了这样的结果:

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

    Queue<T> lastElements = new Queue<T>();
    foreach (T element in source)
    {
        lastElements.Enqueue(element);
        if (lastElements.Count > count)
        {
            lastElements.Dequeue();
        }
    }

    return lastElements;
}

关于 在c#中，Queue<T>是使用循环缓冲区实现的，因此每次循环都没有对象实例化(只有当队列增长时)。我没有设置队列容量(使用专用构造函数)，因为有人可能使用count = int调用此扩展。MaxValue。为了获得额外的性能，您可以检查源实现IList是否<T>，如果是，则直接使用数组索引提取最后的值。