使用LINQ获取集合的最后N有点低效，因为所有上述解决方案都需要遍历集合。TakeLast(int n) in System。Interactive也存在这个问题。

如果你有一个列表，更有效的方法是使用下面的方法进行切片

/// Select from start to end exclusive of end using the same semantics
/// as python slice.
/// <param name="list"> the list to slice</param>
/// <param name="start">The starting index</param>
/// <param name="end">The ending index. The result does not include this index</param>
public static List<T> Slice<T>
(this IReadOnlyList<T> list, int start, int? end = null)
{
    if (end == null)
    {
        end = list.Count();
    }
     if (start < 0)
    {
        start = list.Count + start;
    }
     if (start >= 0 && end.Value > 0 && end.Value > start)
    {
        return list.GetRange(start, end.Value - start);
    }
     if (end < 0)
    {
        return list.GetRange(start, (list.Count() + end.Value) - start);
    }
     if (end == start)
    {
        return new List<T>();
    }
     throw new IndexOutOfRangeException(
        "count = " + list.Count() + 
        " start = " + start +
        " end = " + end);
}

with

public static List<T> GetRange<T>( this IReadOnlyList<T> list, int index, int count )
{
    List<T> r = new List<T>(count);
    for ( int i = 0; i < count; i++ )
    {
        int j=i + index;
        if ( j >= list.Count )
        {
            break;
        }
        r.Add(list[j]);
    }
    return r;
}

以及一些测试用例

[Fact]
public void GetRange()
{
    IReadOnlyList<int> l = new List<int>() { 0, 10, 20, 30, 40, 50, 60 };
     l
        .GetRange(2, 3)
        .ShouldAllBeEquivalentTo(new[] { 20, 30, 40 });
     l
        .GetRange(5, 10)
        .ShouldAllBeEquivalentTo(new[] { 50, 60 });

}
 [Fact]
void SliceMethodShouldWork()
{
    var list = new List<int>() { 1, 3, 5, 7, 9, 11 };
    list.Slice(1, 4).ShouldBeEquivalentTo(new[] { 3, 5, 7 });
    list.Slice(1, -2).ShouldBeEquivalentTo(new[] { 3, 5, 7 });
    list.Slice(1, null).ShouldBeEquivalentTo(new[] { 3, 5, 7, 9, 11 });
    list.Slice(-2)
        .Should()
        .BeEquivalentTo(new[] {9, 11});
     list.Slice(-2,-1 )
        .Should()
        .BeEquivalentTo(new[] {9});
}

下面是如何从一个集合(数组)中获取最后3个元素的实际示例:

// split address by spaces into array
string[] adrParts = adr.Split(new string[] { " " },StringSplitOptions.RemoveEmptyEntries);
// take only 3 last items in array
adrParts = adrParts.SkipWhile((value, index) => { return adrParts.Length - index > 3; }).ToArray();

老实说，我对这个答案并不是特别自豪，但对于小的集合，你可以使用以下方法:

var lastN = collection.Reverse().Take(n).Reverse();

有点俗气，但它的工作;)

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。

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();

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。