6

我已经实现了一个简单函数的普通和并行版本，该函数从 32bppArgb 位图计算直方图。普通版本在 1920x1080 图像上大约需要 0.03 秒，而并行版本需要 0.07 秒。

线程开销真的那么重吗？除了 Parallel.For 之外还有其他构造可以加快这个过程吗？我需要加快速度，因为我正在处理 30fps 视频。

这是简化的代码：

public sealed class Histogram
{
    public int MaxA = 0;
    public int MaxR = 0;
    public int MaxG = 0;
    public int MaxB = 0;
    public int MaxT = 0;

    public int [] A = null;
    public int [] R = null;
    public int [] G = null;
    public int [] B = null;

    public Histogram ()
    {
        this.A = new int [256];
        this.R = new int [256];
        this.G = new int [256];
        this.B = new int [256];

        this.Initialize();
    }

    public void Initialize ()
    {
        this.MaxA = 0;
        this.MaxR = 0;
        this.MaxG = 0;
        this.MaxB = 0;
        this.MaxT = 0;

        for (int i = 0; i < this.A.Length; i++)
            this.A [i] = 0;
        for (int i = 0; i < this.R.Length; i++)
            this.R [i] = 0;
        for (int i = 0; i < this.G.Length; i++)
            this.G [i] = 0;
        for (int i = 0; i < this.B.Length; i++)
            this.B [i] = 0;
    }

    public void ComputeHistogram (System.Drawing.Bitmap bitmap, bool parallel = false)
    {
        System.Drawing.Imaging.BitmapData data = null;

        data = bitmap.LockBits
        (
            new System.Drawing.Rectangle(0, 0, bitmap.Width, bitmap.Height),
            System.Drawing.Imaging.ImageLockMode.ReadOnly,
            System.Drawing.Imaging.PixelFormat.Format32bppArgb
        );

        try
        {
            ComputeHistogram(data, parallel);
        }
        catch
        {
            bitmap.UnlockBits(data);

            throw;
        }

        bitmap.UnlockBits(data);
    }

    public void ComputeHistogram (System.Drawing.Imaging.BitmapData data, bool parallel = false)
    {
        int stride = System.Math.Abs(data.Stride);

        this.Initialize();

        if (parallel)
        {
            unsafe
            {
                System.Threading.Tasks.Parallel.For
                (
                    0,
                    data.Height,
                    new System.Threading.Tasks.ParallelOptions() { MaxDegreeOfParallelism = System.Environment.ProcessorCount },
                    y =>
                    {
                        byte* pointer = ((byte*) data.Scan0) + (stride * y);

                        for (int x = 0; x < stride; x += 4)
                        {
                            this.B [pointer [x + 0]]++;
                            this.G [pointer [x + 1]]++;
                            this.R [pointer [x + 2]]++;
                            this.A [pointer [x + 3]]++;
                        }
                    }
                );
            }
        }
        else
        {
            unsafe
            {
                for (int y = 0; y < data.Height; y++)
                {
                    byte* pointer = ((byte*) data.Scan0) + (stride * y);

                    for (int x = 0; x < stride; x += 4)
                    {
                        this.B [pointer [x + 0]]++;
                        this.G [pointer [x + 1]]++;
                        this.R [pointer [x + 2]]++;
                        this.A [pointer [x + 3]]++;
                    }
                }
            }
        }

        for (int i = 0; i < this.A.Length; i++)
            if (this.MaxA < this.A [i]) this.MaxA = this.A [i];
        for (int i = 0; i < this.R.Length; i++)
            if (this.MaxR < this.R [i]) this.MaxR = this.R [i];
        for (int i = 0; i < this.G.Length; i++)
            if (this.MaxG < this.G [i]) this.MaxG = this.G [i];
        for (int i = 0; i < this.B.Length; i++)
            if (this.MaxB < this.B [i]) this.MaxB = this.B [i];

        if (this.MaxT < this.MaxA) this.MaxT = this.MaxA;
        if (this.MaxT < this.MaxR) this.MaxT = this.MaxR;
        if (this.MaxT < this.MaxG) this.MaxT = this.MaxG;
        if (this.MaxT < this.MaxB) this.MaxT = this.MaxB;
    }
}
4

2 回答 2

8

好吧，首先，您的 Parallel 循环中有一个巨大的错误：

您将有多个线程访问、递增和更新共享数组 - 由于固有的竞争条件，只需在同一个图像上多次运行示例代码会导致截然不同的结果。

但这不是你问的。

至于为什么你看到使用并行实现的性能下降，简单的答案是你可能没有在每个并行任务的主体中做足够的工作来抵消创建新任务的“启动成本”，调度它，等等

可能更关键的是，我相信您正在通过内存中的所有跳跃从 L1/L2 缓存中彻底崩溃 - 每个任务线程都将尝试将它认为需要的内容加载到缓存内存中，但是就像您一样到处索引，您不再创建一致的访问模式，因此每次尝试访问位图缓冲区或内部数组时，您都可能会遇到缓存未命中。

还有一种同样高效的方法可以在不使用不安全代码的情况下获取位图的只读数据……实际上，让我们先这样做：

因此，通过调用LockBits，您有一个指向非托管内存的指针。让我们复制它：

System.Drawing.Imaging.BitmapData data = null;
data = bitmap.LockBits
(
    new System.Drawing.Rectangle(0, 0, bitmap.Width, bitmap.Height),
    System.Drawing.Imaging.ImageLockMode.ReadOnly,
    System.Drawing.Imaging.PixelFormat.Format32bppArgb
);

// For later usage
var imageStride = data.Stride;
var imageHeight = data.Height;

// allocate space to hold the data
byte[] buffer = new byte[data.Stride * data.Height];

// Source will be the bitmap scan data
IntPtr pointer = data.Scan0;

// the CLR marshalling system knows how to move blocks of bytes around, FAST.
Marshal.Copy(pointer, buffer, 0, buffer.Length);

// and now we can unlock this since we don't need it anymore
bitmap.UnlockBits(data);

ComputeHistogram(buffer, imageStride, imageHeight, parallel);

现在，至于竞争条件 - 您可以通过使用Interlocked调用增加计数以合理的性能方式克服这个问题（注意！！！多线程编程很困难，我的解决方案完全有可能不是完美的！）

public void ComputeHistogram (byte[] data, int stride, int height, bool parallel = false)
{
    this.Initialize();

    if (parallel)
    {
        System.Threading.Tasks.Parallel.For
        (
            0,
            height,
            new ParallelOptions() { MaxDegreeOfParallelism = Environment.ProcessorCount },
            y =>
            {
                int startIndex = (stride * y);
                int endIndex = stride * (y+1);
                for (int x = startIndex; x < endIndex; x += 4)
                {
                    // Interlocked actions are more-or-less atomic 
                    // (caveats abound, but this should work for us)
                    Interlocked.Increment(ref this.B[data[x]]);
                    Interlocked.Increment(ref this.G[data[x+1]]);
                    Interlocked.Increment(ref this.R[data[x+2]]);
                    Interlocked.Increment(ref this.A[data[x+3]]);
                }
            }
        );
    }
    else
    {
        // the original way is ok for non-parallel, since only one
        // thread is mucking around with the data
    }

    // Sorry, couldn't help myself, this just looked "cleaner" to me
    this.MaxA = this.A.Max();
    this.MaxR = this.R.Max();
    this.MaxG = this.G.Max();
    this.MaxB = this.B.Max();
    this.MaxT = new[] { this.MaxA, this.MaxB, this.MaxG, this.MaxR }.Max();
}

那么，这对运行时行为有什么影响呢？

不是很多，但至少并行分叉现在计算出正确的结果。:)

使用非常便宜的测试台：

void Main()
{    
    foreach(var useParallel in new[]{false, true})
    {
        var totalRunTime = TimeSpan.Zero;
        var sw = new Stopwatch();
        var runCount = 10;
        for(int run=0; run < runCount; run++)
        {
            GC.Collect();
            GC.WaitForPendingFinalizers();
            GC.Collect();
            sw.Reset();
            sw.Start();
            var bmp = Bitmap.FromFile(@"c:\temp\banner.bmp") as Bitmap;
            var hist = new Histogram();
            hist.ComputeHistogram(bmp, useParallel);
            sw.Stop();
            totalRunTime = totalRunTime.Add(sw.Elapsed);
        }
        Console.WriteLine("Parallel={0}, Avg={1} ms", useParallel, totalRunTime.TotalMilliseconds / runCount);
    }
}

我得到这样的结果：

Parallel=False, Avg=1.69777 ms
Parallel=True, Avg=5.33584 ms

如您所见，我们仍未解决您最初的问题。:)

因此，让我们努力使并行工作“更好”：

让我们看看“给任务更多的工作”做了什么：

if (parallel)
{
    var batchSize = 2;
    System.Threading.Tasks.Parallel.For
    (
        0,
        height / batchSize,
        new ParallelOptions() { MaxDegreeOfParallelism = Environment.ProcessorCount },
        y =>
        {
            int startIndex = (stride * y * batchSize);
            int endIndex = startIndex + (stride * batchSize);
            for (int x = startIndex; x < endIndex; x += 4)
            {
                // Interlocked actions are more-or-less atomic 
                // (caveats abound, but this should work for us)
                Interlocked.Increment(ref this.B[data[x]]);
                Interlocked.Increment(ref this.G[data[x+1]]);
                Interlocked.Increment(ref this.R[data[x+2]]);
                Interlocked.Increment(ref this.A[data[x+3]]);
            }
        }
    );
}

结果：

Parallel=False, Avg=1.70273 ms
Parallel=True, Avg=4.82591 ms

哦，这看起来很有希望......我想知道随着我们的变化会发生什么batchSize？

让我们改变我们的测试台：

void Main()
{    
    foreach(var useParallel in new[]{false, true})
    {
        for(int batchSize = 1; batchSize < 1024; batchSize <<= 1)
        {
            var totalRunTime = TimeSpan.Zero;
            var sw = new Stopwatch();
            var runCount = 10;
            for(int run=0; run < runCount; run++)
            {
                GC.Collect();
                GC.WaitForPendingFinalizers();
                GC.Collect();
                sw.Reset();
                sw.Start();
                var bmp = Bitmap.FromFile(@"c:\temp\banner.bmp") as Bitmap;
                var hist = new Histogram();
                hist.ComputeHistogram(bmp, useParallel, batchSize);
                sw.Stop();
                totalRunTime = totalRunTime.Add(sw.Elapsed);
            }
            Console.WriteLine("Parallel={0}, BatchSize={1} Avg={2} ms", useParallel, batchSize, totalRunTime.TotalMilliseconds / runCount);
        }        
    }
}

结果：（只显示parallel=true，因为non-parallel不会改变）

Parallel=True, BatchSize=1 Avg=5.57644 ms
Parallel=True, BatchSize=2 Avg=5.49982 ms
Parallel=True, BatchSize=4 Avg=5.20434 ms
Parallel=True, BatchSize=8 Avg=5.1721 ms
Parallel=True, BatchSize=16 Avg=5.00405 ms
Parallel=True, BatchSize=32 Avg=4.44973 ms
Parallel=True, BatchSize=64 Avg=2.28332 ms
Parallel=True, BatchSize=128 Avg=1.39957 ms
Parallel=True, BatchSize=256 Avg=1.29156 ms
Parallel=True, BatchSize=512 Avg=1.28656 ms

一旦我们在批量大小中达到 64-128 范围，我们似乎正在接近某种渐近线，尽管当然您的里程可能会根据您的位图大小等而有所不同。

我希望这有帮助！这是我等待生产构建完成的一天的有趣分心！:)

于 2013-02-15T20:34:10.223 回答
1

创建线程有相当大的开销。执行可能比单线程版本运行得快得多，但完成速度太快而无法弥补这个初始开销。

如果你每帧都这样做，它只会减慢你的速度。

但是，如果您手动创建线程池、手动分配工作并为每个帧重用线程，您可能会发现到第二或第三帧时，您的代码会飞速超过单线程版本。

于 2013-02-15T17:10:38.757 回答