This is supposed to calculate the histogram of an 8-bit grayscale image. With a 1024x770 test bitmap, CreateTime ends up at around 890ms. How can I make this go (way, way) faster?
这应该计算8位灰度图像的直方图。使用1024x770测试位图,CreateTime最终在890ms左右。我怎样才能更快地完成这个(方式,方式)?
EDIT: I should mention that this doesn't actually compute the histogram yet, it only gets the values out of the bitmap. So I really should have asked, what is the fastest way to retrieve all pixel values from an 8-bit grayscale image?
编辑:我应该提到,这实际上并没有计算直方图,它只从位图中获取值。所以我真的应该问,从8位灰度图像中检索所有像素值的最快方法是什么?
public class Histogram {
private static int[,] values;
public Histogram(Bitmap b) {
var sw = Stopwatch.StartNew();
values = new int[b.Width, b.Height];
for (int w = 0; w < b.Width; ++w) {
for (int h = 0; h < b.Height; ++h) {
values[w, h] = b.GetPixel(w, h).R;
}
}
sw.Stop();
CreateTime = (sw.ElapsedTicks /
(double)Stopwatch.Frequency) * 1000;
}
public double CreateTime { get; set; }
}
3 个解决方案
#1
The basic histogram algorithm is something like:
基本直方图算法类似于:
int[] hist = new hist[256];
//at this point dont forget to initialize your vector with 0s.
for(int i = 0; i < height; ++i)
{
for(int j = 0 ; j < widthl ++j)
{
hist[ image[i,j] ]++;
}
}
The algorithm sums how many pixels with value 0 you have, how many with value=1 and so on. The basic idea is to use the pixel value as the index to the position of the histogram where you will count.
该算法总结了您拥有的值为0的像素数,具有值= 1的多少像素,依此类推。基本思想是使用像素值作为您将计算的直方图位置的索引。
I have one version of this algorithm written for C# using unmanaged code (which is fast) I dont know if is faster than your but feel free to take it and test, here is the code:
我有一个版本的这个算法使用非托管代码为C#编写(这很快)我不知道是否比你更快但随意采取它并测试,这里是代码:
public void Histogram(double[] histogram, Rectangle roi)
{
BitmapData data = Util.SetImageToProcess(image, roi);
if (image.PixelFormat != PixelFormat.Format8bppIndexed)
return;
if (histogram.Length < Util.GrayLevels)
return;
histogram.Initialize();
int width = data.Width;
int height = data.Height;
int offset = data.Stride - width;
unsafe
{
byte* ptr = (byte*)data.Scan0;
for (int y = 0; y < height; ++y)
{
for (int x = 0; x < width; ++x, ++ptr)
histogram[ptr[0]]++;
ptr += offset;
}
}
image.UnlockBits(data);
}
static public BitmapData SetImageToProcess(Bitmap image, Rectangle roi)
{
if (image != null)
return image.LockBits(
roi,
ImageLockMode.ReadWrite,
image.PixelFormat);
return null;
}
I hope I could help you.
我希望我能帮助你。
#2
You'll want to use the Bitmap.LockBits method to access the pixel data. This is a good reference on the process. Essentially, you're going to need to use unsafe
code to iterate over the bitmap data.
您将需要使用Bitmap.LockBits方法来访问像素数据。这是该过程的一个很好的参考。实际上,您将需要使用不安全的代码来迭代位图数据。
#3
Here's a copy/pastable version of the function I've come up w/ based on on this thread.
这是我基于这个帖子提出的函数的复制/可管理版本。
The unsafe code expects the bitmap to be Format24bppRgb, and if it's not, it'll convert the bitmap to that format and operate on the cloned version.
不安全的代码期望位图是Format24bppRgb,如果不是,它会将位图转换为该格式并对克隆版本进行操作。
Note that the call to image.Clone() will throw if you pass in a bitmap using an indexed pixel format, such as Format4bppIndexed.
请注意,如果使用索引像素格式(例如Format4bppIndexed)传入位图,则会调用image.Clone()。
Takes ~200ms to get a histogram from an image 9100x2048 on my dev machine.
需要〜200ms从我的开发机器上的图像9100x2048获取直方图。
private long[] GetHistogram(Bitmap image)
{
var histogram = new long[256];
bool imageWasCloned = false;
if (image.PixelFormat != PixelFormat.Format24bppRgb)
{
//the unsafe code expects Format24bppRgb, so convert the image...
image = image.Clone(new Rectangle(0, 0, image.Width, image.Height), PixelFormat.Format24bppRgb);
imageWasCloned = true;
}
BitmapData bmd = null;
try
{
bmd = image.LockBits(new Rectangle(0, 0, image.Width, image.Height), ImageLockMode.ReadOnly,
PixelFormat.Format24bppRgb);
const int pixelSize = 3; //pixels are 3 bytes each w/ Format24bppRgb
//For info on locking the bitmap bits and finding the
//pixels using unsafe code, see http://www.bobpowell.net/lockingbits.htm
int height = bmd.Height;
int width = bmd.Width;
int rowPadding = bmd.Stride - (width * pixelSize);
unsafe
{
byte* pixelPtr = (byte*)bmd.Scan0;//starts on the first row
for (int y = 0; y < height; ++y)
{
for (int x = 0; x < width; ++x)
{
histogram[(pixelPtr[0] + pixelPtr[1] + pixelPtr[2]) / 3]++;
pixelPtr += pixelSize;//advance to next pixel in the row
}
pixelPtr += rowPadding;//advance ptr to the next pixel row by skipping the padding @ the end of each row.
}
}
}
finally
{
if (bmd != null)
image.UnlockBits(bmd);
if (imageWasCloned)
image.Dispose();
}
return histogram;
}
#1
The basic histogram algorithm is something like:
基本直方图算法类似于:
int[] hist = new hist[256];
//at this point dont forget to initialize your vector with 0s.
for(int i = 0; i < height; ++i)
{
for(int j = 0 ; j < widthl ++j)
{
hist[ image[i,j] ]++;
}
}
The algorithm sums how many pixels with value 0 you have, how many with value=1 and so on. The basic idea is to use the pixel value as the index to the position of the histogram where you will count.
该算法总结了您拥有的值为0的像素数,具有值= 1的多少像素,依此类推。基本思想是使用像素值作为您将计算的直方图位置的索引。
I have one version of this algorithm written for C# using unmanaged code (which is fast) I dont know if is faster than your but feel free to take it and test, here is the code:
我有一个版本的这个算法使用非托管代码为C#编写(这很快)我不知道是否比你更快但随意采取它并测试,这里是代码:
public void Histogram(double[] histogram, Rectangle roi)
{
BitmapData data = Util.SetImageToProcess(image, roi);
if (image.PixelFormat != PixelFormat.Format8bppIndexed)
return;
if (histogram.Length < Util.GrayLevels)
return;
histogram.Initialize();
int width = data.Width;
int height = data.Height;
int offset = data.Stride - width;
unsafe
{
byte* ptr = (byte*)data.Scan0;
for (int y = 0; y < height; ++y)
{
for (int x = 0; x < width; ++x, ++ptr)
histogram[ptr[0]]++;
ptr += offset;
}
}
image.UnlockBits(data);
}
static public BitmapData SetImageToProcess(Bitmap image, Rectangle roi)
{
if (image != null)
return image.LockBits(
roi,
ImageLockMode.ReadWrite,
image.PixelFormat);
return null;
}
I hope I could help you.
我希望我能帮助你。
#2
You'll want to use the Bitmap.LockBits method to access the pixel data. This is a good reference on the process. Essentially, you're going to need to use unsafe
code to iterate over the bitmap data.
您将需要使用Bitmap.LockBits方法来访问像素数据。这是该过程的一个很好的参考。实际上,您将需要使用不安全的代码来迭代位图数据。
#3
Here's a copy/pastable version of the function I've come up w/ based on on this thread.
这是我基于这个帖子提出的函数的复制/可管理版本。
The unsafe code expects the bitmap to be Format24bppRgb, and if it's not, it'll convert the bitmap to that format and operate on the cloned version.
不安全的代码期望位图是Format24bppRgb,如果不是,它会将位图转换为该格式并对克隆版本进行操作。
Note that the call to image.Clone() will throw if you pass in a bitmap using an indexed pixel format, such as Format4bppIndexed.
请注意,如果使用索引像素格式(例如Format4bppIndexed)传入位图,则会调用image.Clone()。
Takes ~200ms to get a histogram from an image 9100x2048 on my dev machine.
需要〜200ms从我的开发机器上的图像9100x2048获取直方图。
private long[] GetHistogram(Bitmap image)
{
var histogram = new long[256];
bool imageWasCloned = false;
if (image.PixelFormat != PixelFormat.Format24bppRgb)
{
//the unsafe code expects Format24bppRgb, so convert the image...
image = image.Clone(new Rectangle(0, 0, image.Width, image.Height), PixelFormat.Format24bppRgb);
imageWasCloned = true;
}
BitmapData bmd = null;
try
{
bmd = image.LockBits(new Rectangle(0, 0, image.Width, image.Height), ImageLockMode.ReadOnly,
PixelFormat.Format24bppRgb);
const int pixelSize = 3; //pixels are 3 bytes each w/ Format24bppRgb
//For info on locking the bitmap bits and finding the
//pixels using unsafe code, see http://www.bobpowell.net/lockingbits.htm
int height = bmd.Height;
int width = bmd.Width;
int rowPadding = bmd.Stride - (width * pixelSize);
unsafe
{
byte* pixelPtr = (byte*)bmd.Scan0;//starts on the first row
for (int y = 0; y < height; ++y)
{
for (int x = 0; x < width; ++x)
{
histogram[(pixelPtr[0] + pixelPtr[1] + pixelPtr[2]) / 3]++;
pixelPtr += pixelSize;//advance to next pixel in the row
}
pixelPtr += rowPadding;//advance ptr to the next pixel row by skipping the padding @ the end of each row.
}
}
}
finally
{
if (bmd != null)
image.UnlockBits(bmd);
if (imageWasCloned)
image.Dispose();
}
return histogram;
}