获取数组最后 N 个元素的最有效方法

Question

对于一个项目，我将不得不经常使用包含大量数据的数组的最后 N 个元素。

我试着做

myArray.Skip(myArray.Length - toTake).Take(toTake)

但我发现它很慢。

我将其与此进行了比较：

public static int[] TakeLast(this int[] inputArray, int count)
{
    int[] returnArray = new int[count];
    int startIndex = Math.Max(inputArray.Count() - count, 0);
    unsafe
    {
        fixed (int* itemArrayPtr = &(inputArray[startIndex]))
        {
            fixed (int* arrayPtr = &(returnArray[0]))
            {
                int* itemValuePtr = itemArrayPtr;
                int* valuePtr = arrayPtr;

                for (int i = 0; i < count; i++)
                {
                    *valuePtr++ = *itemValuePtr++;
                }
            }
        }
    }
    return returnArray;
}

这很有效，因为它不能是通用的（我希望这适用于任何原始类型（int、float、double 等）。

有没有办法实现具有通用/linq/... 方法的可比性能？ 我不需要让它在 IEnumerable 上工作，Array 对我来说就足够了。

编辑我目前正在测试你给我的所有方法，现在它是 Array.Copy 似乎更快：

Generating array for 100000000 elements.
SkipTake: 00:00:00.3009047
Unsafe: 00:00:00.0006289
Array.Copy: 00:00:00.0000012
Buffer.BlockCopy: 00:00:00.0001860
Reverse Linq: 00:00:00.2201143
Finished

Answer 1

来自评论：

public static T[] TakeLast<T>(this T[] inputArray, int count)
{
    var result = new T[count];
    Array.Copy(inputArray, inputArray.Length - count, result, 0, count);
    return result;
}

似乎表现不错。 值得指出的是，根据具体需求，可能可以完全避免新数组，并迭代原始inputArray 。 你不能复制比根本不复制更快。 :)

Answer 2

这个怎么样？ 应该相当快：

public static class ArrayExt
{
    public static T[] TakeLast<T>(this T[] inputArray, int count) where T: struct
    {
        count = Math.Min(count, inputArray.Length);
        int size = Marshal.SizeOf(typeof(T));

        T[] result = new T[count];
        Buffer.BlockCopy(inputArray, (inputArray.Length-count)*size, result, 0, count*size);

        return result;
    }
}

（我认为这比原始类型的 Array.Copy() 快一点。但我不会认为这是理所当然的 - 5 分钟后返回一些时间。;）

---

[编辑] 计时显示 Array.Copy() 具有相似的速度，但结果因运行而异并取决于数组大小。

这是一些示例代码：

using System;
using System.Diagnostics;
using System.Linq;
using System.Runtime.InteropServices;

namespace Demo
{
    internal class Program
    {
        private void run()
        {
            const int ARRAY_SIZE = 10000;
            var array = Enumerable.Range(0, ARRAY_SIZE).Select(x => x).ToArray();
            Stopwatch sw = new Stopwatch();
            const int COUNT = 100000;

            for (int i = 0; i < 8; ++i)
            {
                sw.Restart();

                for (int j = 0; j < COUNT; ++j)
                    array.TakeLastViaArrayCopy(ARRAY_SIZE/2);

                Console.WriteLine("TakeLastViaArrayCopy took " + sw.Elapsed);

                sw.Restart();

                for (int j = 0; j < COUNT; ++j)
                    array.TakeLastViaBlockCopy(ARRAY_SIZE/2);

                Console.WriteLine("TakeLastViaBlockCopy took " + sw.Elapsed);
                Console.WriteLine();
            }
        }

        private static void Main()
        {
            new Program().run();
        }
    }

    public static class ArrayExt
    {
        public static T[] TakeLastViaBlockCopy<T>(this T[] inputArray, int count) where T: struct
        {
            count = Math.Min(count, inputArray.Length);
            int size = Marshal.SizeOf(typeof(T));

            T[] result = new T[count];
            Buffer.BlockCopy(inputArray, (inputArray.Length-count)*size, result, 0, count*size);

            return result;
        }

        public static T[] TakeLastViaArrayCopy<T>(this T[] inputArray, int count) where T: struct
        {
            count = Math.Min(count, inputArray.Length);

            T[] result = new T[count];
            Array.Copy(inputArray, inputArray.Length-count, result, 0, count);

            return result;
        }
    }
}

结果（像往常一样发布构建）：

TakeLastViaArrayCopy took 00:00:00.3028503
TakeLastViaBlockCopy took 00:00:00.3052196

TakeLastViaArrayCopy took 00:00:00.2969425
TakeLastViaBlockCopy took 00:00:00.3000117

TakeLastViaArrayCopy took 00:00:00.2906120
TakeLastViaBlockCopy took 00:00:00.2987753

TakeLastViaArrayCopy took 00:00:00.2954674
TakeLastViaBlockCopy took 00:00:00.3005010

TakeLastViaArrayCopy took 00:00:00.2944490
TakeLastViaBlockCopy took 00:00:00.3006893

TakeLastViaArrayCopy took 00:00:00.3041998
TakeLastViaBlockCopy took 00:00:00.2920206

TakeLastViaArrayCopy took 00:00:00.3115137
TakeLastViaBlockCopy took 00:00:00.2996884

TakeLastViaArrayCopy took 00:00:00.2906820
TakeLastViaBlockCopy took 00:00:00.2985933

Array.Copy()更简单，因此可以使用它。

Answer 3

从C#8 开始，您可以像这样使用Range 。

var lastN = array[^n..];

Answer 4

myArray.Reverse().Take(toTake).Reverse();