多级(4) Otsu阈值
多级(4) Otsu阈值
提问于 2016-01-28 08:52:18
我试图实现多级Otsu的阈值，更具体地说，我需要3个阈值/4个类。
我知道有两个类似的问题：#34856019和#22706742。问题是我没有得到好的结果:我读过几篇文章，其中有一些样本图像和阈值，这些代码与这些文章中的不同。
假设我有一张在黑色背景上有三个圆圈的图片，圆圈的亮度从很亮到很暗不等：
样本图像
我认为这样做是对的吗?根据圆圈的强度，我可以得到4个等级:黑色背景和3个以上的等级？
我的程序给出了以下阈值：226,178，68
因此，第三个圆圈是完全看不见的--它和背景在同一个类中。
有人能检查一下这些值和/或源代码吗？也许可以用Matlab或者其他的方法来检查这幅图像..。顺便问一下，用0来处理除法最好的方法是什么，这种除法在直方图中通常是零值的。源代码：
void MultilevelThresholding(cv::Mat& src)
{
int histogram[256] = { 0 };
int pixelsCount = src.cols * src.rows;
for (int y = 0; y < src.rows; y++)
{
for (int x = 0; x < src.cols; x++)
{
uchar value = src.at<uchar>(y, x);
histogram[value]++;
}
}
double c = 0;
double Mt = 0;
double p[256] = { 0 };
for (int i = 0; i < 256; i++)
{
p[i] = (double) histogram[i] / (double) pixelsCount;
Mt += i * p[i];
}
int optimalTreshold1 = 0;
int optimalTreshold2 = 0;
int optimalTreshold3 = 0;
double maxBetweenVar = 0;
double w0 = 0;
double m0 = 0;
double c0 = 0;
double p0 = 0;
double w1 = 0;
double m1 = 0;
double c1 = 0;
double p1 = 0;
double w2 = 0;
double m2 = 0;
double c2 = 0;
double p2 = 0;
for (int tr1 = 0; tr1 < 256; tr1++)
{
p0 += p[tr1];
w0 += (tr1 * p[tr1]);
if (p0 != 0)
{
m0 = w0 / p0;
}
c0 = p0 * (m0 - Mt) * (m0 - Mt);
c1 = 0;
w1 = 0;
m1 = 0;
p1 = 0;
for (int tr2 = tr1 + 1; tr2 < 256; tr2++)
{
p1 += p[tr2];
w1 += (tr2 * p[tr2]);
if (p1 != 0)
{
m1 = w1 / p1;
}
c1 = p1 * (m1 - Mt) * (m1 - Mt);
c2 = 0;
w2 = 0;
m2 = 0;
p2 = 0;
for (int tr3 = tr2 + 1; tr3 < 256; tr3++)
{
p2 += p[tr3];
w2 += (tr3 * p[tr3]);
if (p2 != 0)
{
m2 = w2 / p2;
}
c2 = p2 * (m2 - Mt) * (m2 - Mt);
c = c0 + c1 + c2;
if (maxBetweenVar < c)
{
maxBetweenVar = c;
optimalTreshold1 = tr1;
optimalTreshold2 = tr2;
optimalTreshold3 = tr3;
}
}
}
}回答 1
Stack Overflow用户
回答已采纳
发布于 2016-01-28 11:56:15
所以，我想通了。4类(3个阈值) Otsu阈值的最终源代码：
// cv::Mat& src - source image's matrix
int histogram[256] = { 0 };
int pixelsCount = src.cols * src.rows;
for (int y = 0; y < src.rows; y++)
{
for (int x = 0; x < src.cols; x++)
{
uchar value = src.at<uchar>(y, x);
histogram[value]++;
}
}
double c = 0;
double Mt = 0;
double p[256] = { 0 };
for (int i = 0; i < 256; i++)
{
p[i] = (double) histogram[i] / (double) pixelsCount;
Mt += i * p[i];
}
int optimalTreshold1 = 0;
int optimalTreshold2 = 0;
int optimalTreshold3 = 0;
double maxBetweenVar = 0;
double w0 = 0;
double m0 = 0;
double c0 = 0;
double p0 = 0;
double w1 = 0;
double m1 = 0;
double c1 = 0;
double p1 = 0;
double w2 = 0;
double m2 = 0;
double c2 = 0;
double p2 = 0;
for (int tr1 = 0; tr1 < 256; tr1++)
{
p0 += p[tr1];
w0 += (tr1 * p[tr1]);
if (p0 != 0)
{
m0 = w0 / p0;
}
c0 = p0 * (m0 - Mt) * (m0 - Mt);
c1 = 0;
w1 = 0;
m1 = 0;
p1 = 0;
for (int tr2 = tr1 + 1; tr2 < 256; tr2++)
{
p1 += p[tr2];
w1 += (tr2 * p[tr2]);
if (p1 != 0)
{
m1 = w1 / p1;
}
c1 = p1 * (m1 - Mt) * (m1 - Mt);
c2 = 0;
w2 = 0;
m2 = 0;
p2 = 0;
for (int tr3 = tr2 + 1; tr3 < 256; tr3++)
{
p2 += p[tr3];
w2 += (tr3 * p[tr3]);
if (p2 != 0)
{
m2 = w2 / p2;
}
c2 = p2 * (m2 - Mt) * (m2 - Mt);
double p3 = 1 - (p0 + p1 + p2);
double w3 = Mt - (w0 + w1 + w2);
double m3 = w3 / p3;
double c3 = p3 * (m3 - Mt) * (m3 - Mt);
double c = c0 + c1 + c2 + c3;
if (maxBetweenVar < c)
{
maxBetweenVar = c;
optimalTreshold1 = tr1;
optimalTreshold2 = tr2;
optimalTreshold3 = tr3;
}
}
}
}
阈值：179，92，25
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/35056760
复制相关文章
相似问题
腾讯云开发者
