我有一个BitmapFrames数组,需要进行直方图拉伸。我知道这与直方图均衡化不同,最终结果是什么...... sorta。问题是我得到直方图后完全不知道该怎么办。
到目前为止,我的代码为直方图创建了一个数组,所以我知道每个值有多少个像素。但在那之后我不知道该怎么做。
这是我到目前为止的代码......现在它使直方图然后直方图均衡......这不是我想要的......我只是想了解更多有关直方图的信息
[Cmdlet(VerbsData.ConvertTo, "HistoStretch")]
public class HistoStretchCmdlet : PSCmdlet
{
private BitmapFrame[] bFrame, outFrame;
private BitmapSource src;
private double pixelsize;
private byte[] pixels, outPixels;
private byte MAX_VAL;
private int[] histogram;
private int cf, start;
[Parameter(ValueFromPipeline = true,
ValueFromPipelineByPropertyName = true), ValidateNotNullOrEmpty]
public BitmapFrame[] Bitmap
{
get
{
return bFrame;
}
set
{
bFrame = value;
}
}
protected override void ProcessRecord()
{
base.ProcessRecord();
Console.Write("Applying a histogram stretch to the image...\n\n");
outFrame = new BitmapFrame[bFrame.Length];
for (int c = 0; c < bFrame.Length; c++)
{
MAX_VAL = (byte)((1 << bFrame[c].Format.BitsPerPixel) - 1);
histogram = new int[MAX_VAL + 1];
for (int i = 0; i <= MAX_VAL; i++)
{
histogram[i] = 0;
}
pixelsize = bFrame[c].PixelWidth * bFrame[c].PixelHeight;
pixels = new byte[(int)pixelsize];
outPixels = new byte[(int)pixelsize];
bFrame[c].CopyPixels(pixels,(int)bFrame[c].Width * (bFrame[c].Format.BitsPerPixel / 8),0);
for (int i = 0; i < pixelsize; i++)
{
histogram[(int)pixels[i]] = histogram[(int)pixels[i]] + 1;
}
for (int i = 0; i <= MAX_VAL; i++)
{
Console.Write("{0}: {1}\n", i, histogram[i]);
}
for (int i = 0; i <= MAX_VAL; i++)
{
if (histogram[i] >= 1)
{
start = i;
break;
}
}
for (int i = 0; i < pixelsize; i++)
{
cf = 0;
for (int g = 0; g <= MAX_VAL; g++)
{
cf += histogram[g];
if (g == pixels[i])
{
break;
}
}
outPixels[i] = (byte)(cf * (MAX_VAL / pixelsize));
}
src = BitmapSource.Create(bFrame[c].PixelWidth, bFrame[c].PixelHeight, bFrame[c].DpiX, bFrame[c].DpiY,
bFrame[c].Format, bFrame[c].Palette, outPixels, (int)(bFrame[c].Width * (bFrame[c].Format.BitsPerPixel / 8)));
outFrame[c] = BitmapFrame.Create(src);
}
WriteObject(outFrame);
}
}
根据老师的说法,这就是直方图的样子:
http://www.fileden.com/files/2009/8/18/2547657/histostretch.PNG
我运行了上面的代码......并得到了一个直的黑色图像。 这是我的代码:
outFrame = new BitmapFrame[bFrame.Length];
for (int c = 0; c < bFrame.Length; c++)
{
MAX_VAL = (byte)((1 << bFrame[c].Format.BitsPerPixel) - 1);
histogram = new int[MAX_VAL + 1];
for (int i = 0; i <= MAX_VAL; i++)
{
histogram[i] = 0;
}
pixelsize = bFrame[c].PixelWidth * bFrame[c].PixelHeight;
pixels = new byte[(int)pixelsize];
outPixels = new byte[(int)pixelsize];
bFrame[c].CopyPixels(pixels,(int)bFrame[c].Width * (bFrame[c].Format.BitsPerPixel / 8),0);
max = pixels[0];
min = pixels[0];
for (int i = 0; i < pixelsize; i++)
{
histogram[(int)pixels[i]] = histogram[(int)pixels[i]] + 1;
if((int)pixels[i] > max)
max = pixels[i];
if((int)pixels[i] < min)
min = pixels[i];
}
dynamic = max - min;
for (int i = 0; i < pixelsize; i++)
{
outPixels[i] = (byte)(((pixels[i] - min) / dynamic) * MAX_VAL);
}
答案 0 :(得分:9)
直方图拉伸是像素值的映射,以便:
换句话说,直方图拉伸意味着将图像数据的动态(84:153)拉伸到最大可能的动态(0:255)。
这不应该影响直方图峰值的高度,而只影响它们的扩散(这一点上的插图有点误导)。
histogram stretch http://cct.rncan.gc.ca/resource/tutor/fundam/images/linstre.gif
在实践中这是您应用于图像像素的映射(伪代码):
maxVal = image.maximumValue() # 153
minVal = image.minumumValue() # 84
dynamic = maxVal-minVal
for pixel in image.Pixels():
newPixel = ((pixel-minVal)/dynamic)*255
答案 1 :(得分:0)
如果您可以控制相机的光线和/或增益/偏移,您可以根据需要对其进行优化并拉伸直方图。
答案 2 :(得分:0)
别忘了考虑漂浮。所以对dabonz413的回答略有修改:
maxVal = image.maximumValue() # 153
minVal = image.minumumValue() # 84
dynamic = maxVal-minVal
for pixel in image.Pixels():
newPixel = ((float) (pixel-minVal)/dynamic)*255