[C#] 使用Linq获取集合的最后N个元素?


Answers

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,但是如果你真的想在没有额外方法的情况下做,那么你只需要认识到Enumerable.Reverse()可以用作扩展方法,您不需要这样使用它:

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

鉴于一个集合,有没有办法获得该集合的最后N个元素? 如果框架中没有方法,那么编写扩展方法来做到这一点最好的方法是什么?




这里有一种方法适用于任何枚举,但只使用O(N)临时存储:

public static class TakeLastExtension
{
    public static IEnumerable<T> TakeLast<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];
        }
    }
}

用法:

List<int> l = new List<int> {4, 6, 3, 6, 2, 5, 7};
List<int> lastElements = l.TakeLast(3).ToList();

它通过使用大小为N的环形缓冲区来存储元素,因为它看到它们,用新元素覆盖旧元素。 当到达枚举的末尾时,环形缓冲区包含最后N个元素。




这是我的解决方案:

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();
            }
        }
    }
}

代码有点笨重,但作为一个可重用的组件,它应该在大多数场景中表现得很好,并且它会保持使用它的代码的简洁明了。 :-)

TakeLast for non IList`1基于与@Mark Byers和@MackieChan的答案相同的环缓冲算法。 有趣的是他们有多相似 - 我完全独立写了我的。 猜猜真的只有一种方法可以正确地完成环形缓冲区。 :-)

查看@ kbrimington的答案,可以为IQuerable<T>添加一个额外的检查,以回溯到适用于Entity Framework的方法 - 假设我在这一点上没有。




在RX的System.Interactive程序集中使用EnumerableEx.TakeLast。 这是一个O(N)的实现,比如@ Mark's,但是它使用一个队列而不是一个环形缓冲区结构(并在达到缓冲区容量时将项目出队)。

(注意:这是IEnumerable版本 - 不是IObservable版本,尽管两者的实现几乎完全相同)




使用这种方法获得所有范围没有错误

 public List<T> GetTsRate( List<T> AllT,int Index,int Count)
        {
            List<T> Ts = null;
            try
            {
                Ts = AllT.ToList().GetRange(Index, Count);
            }
            catch (Exception ex)
            {
                Ts = AllT.Skip(Index).ToList();
            }
            return Ts ;
        }



使用LINQ取得集合的最后N个是有点低效的,因为所有上述解决方案都需要迭代整个集合。 System.Interactive TakeLast(int n)也有这个问题。

如果你有一个列表,更有效率的事情是使用下面的方法进行分片

/// 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);
}

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;
}

和一些测试用例

[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});
}






如果使用第三方库是一个选项, MoreLinq定义TakeLast() ,它完全是这样做的。