OpenCV学习(18) 细化算法(6)

本章我们在学习一下基于索引表的细化算法。

假设要处理的图像为二值图,前景值为1,背景值为0。

索引表细化算法使用下面的8邻域表示法:

image

一个像素的8邻域,我们可以用8位二进制表示,比如下面的8邻域,表示为00111000=0x38=56

image

我们可以枚举出各种情况下,当前像素能否删除的表,该表大小为256。它的索引即为8邻域表示的值,表中存的值为0或1,0表示当前像素不能删除,1表示可以删除。deletemark[256]

比如下图第一个表示,索引值为0,它表示孤立点,不能删除,所以deletemark[0]=0,第二个表示索引值为17,它表示端点,也不能删除,所以deletemark[17]=0,第三个表示索引为21,删除的话会改变连通域数量,所以deletemark[21]=0,第四个表示索引值为96,此时可以删除,所以deletemark[96]=1。

image

最终我们会定义一张完整的表来表示当前像素能否删除。

索引表细化算法描述很简单。

1.找到轮廓,其值用4表示

2.查找值为4的轮廓,查找索引表判断能否删除,能删除的话把它置为0。

循环迭代1,2直到再也没有可以删除的点为止。

下面的算法的代码:

void gThin::cvidxThin1(cv::Mat& src, cv::Mat& dst)
{


if(src.type()!=CV_8UC1)
{
printf("只能处理二值或灰度图像\n");
return;
}
//非原地操作时候,copy src到dst
if(dst.data!=src.data)
{
src.copyTo(dst);
}

// P0 P1 P2
// P7 P3
// P6 P5 P4
unsigned char deletemark[256] = {
0,0,0,0,0,0,0,1, 0,0,1,1,0,0,1,1,
0,0,0,0,0,0,0,0, 0,0,1,1,1,0,1,1,
0,0,0,0,0,0,0,0, 1,0,0,0,1,0,1,1,
0,0,0,0,0,0,0,0, 1,0,1,1,1,0,1,1,
0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0, 1,0,0,0,1,0,1,1,
1,0,0,0,0,0,0,0, 1,0,1,1,1,0,1,1,
0,0,1,1,0,0,1,1, 0,0,0,1,0,0,1,1,
0,0,0,0,0,0,0,0, 0,0,0,1,0,0,1,1,
1,1,0,1,0,0,0,1, 0,0,0,0,0,0,0,0,
1,1,0,1,0,0,0,1, 1,1,0,0,1,0,0,0,
0,1,1,1,0,0,1,1, 0,0,0,1,0,0,1,1,
0,0,0,0,0,0,0,0, 0,0,0,0,0,1,1,1,
1,1,1,1,0,0,1,1, 1,1,0,0,1,1,0,0,
1,1,1,1,0,0,1,1, 1,1,0,0,1,1,0,0
};//索引
int i, j;
int width, height;
//之所以减1,是方便处理8邻域,防止越界
width = src.cols -1;
height = src.rows -1;
int step = src.step;
int p0, p1, p2,p3,p4,p5,p6,p7;
uchar* img;
bool ifEnd;
bool border = false; //交换删除的次序,防止从一边细化
while(1)
{

border = !border;
img = dst.data;
for(i = 1; i < height; i++)
{
img += step;
for(j =1; j<width; j++)
{
uchar* p = img + j;
//如果p点是背景点,继续循环
if(p[0]==0) continue;
p0 = p[-step-1]>0?1:0;
p1 = p[-step]>0?1:0;
p2 = p[-step+1]>0?1:0;
p3 = p[1]>0?1:0;
p4 = p[step+1]>0?1:0;
p5 = p[step]>0?1:0;
p6 = p[step-1]>0?1:0;
p7 = p[-1]>0?1:0;

//如果sum等于0,则不是内部点,是轮廓点,设置其像素值为2
int sum;
sum = p0 & p1 & p2 & p3 & p4 & p5 & p6 & p7;

//判断是否是邻接点或孤立点,0,1分别对于那个孤立点和端点
if(sum==0)
{
dst.at<uchar>(i,j) = 4; //满足删除条件,设置当前像素为0
}

}
}
//printf("\n");
//PrintMat(dst);
//执行删除操作
ifEnd = false;

img = dst.data;
for(i = 1; i < height; i++)
{
img += step;
for(j =1; j<width; j++)
{
uchar* p = img + j;
//如果p点是背景点,继续循环
if(p[0]!=4) continue;
p0 = p[-step-1]>0?1:0;
p1 = p[-step]>0?1:0;
p2 = p[-step+1]>0?1:0;
p3 = p[1]>0?1:0;
p4 = p[step+1]>0?1:0;
p5 = p[step]>0?1:0;
p6 = p[step-1]>0?1:0;
p7 = p[-1]>0?1:0;

p1 = p1<<1;
p2 = p2<<2;
p3 = p3 <<3;
p4 = p4<<4;
p5 = p5<<5;
p6 = p6 <<6;
p7 = p7 << 7;

//求的8邻域在索引表中的索引
int sum;
sum = p0 | p1 | p2 | p3 | p4 | p5 | p6 | p7;

//判断是否是邻接点或孤立点,0,1分别对于那个孤立点和端点
if(deletemark[sum] == 1)
{
dst.at<uchar>(i,j) = 0; //满足删除条件,设置当前像素为0
ifEnd = true;
}

}
}

//printf("\n");
//PrintMat(dst);
//printf("\n");

//已经没有可以细化的像素了,则退出迭代
if(!ifEnd) break;
}

image

上面的算法可以看到细化后的轮廓偏右了,我们可以更改删除的循环条件,把循环拆分成三个,修改后的代码如下:

void gThin::cvidxThin(cv::Mat& src, cv::Mat& dst)
{


if(src.type()!=CV_8UC1)
{
printf("只能处理二值或灰度图像\n");
return;
}
//非原地操作时候,copy src到dst
if(dst.data!=src.data)
{
src.copyTo(dst);
}

// P0 P1 P2
// P7 P3
// P6 P5 P4
unsigned char deletemark[256] = {
0,0,0,0,0,0,0,1, 0,0,1,1,0,0,1,1,
0,0,0,0,0,0,0,0, 0,0,1,1,1,0,1,1,
0,0,0,0,0,0,0,0, 1,0,0,0,1,0,1,1,
0,0,0,0,0,0,0,0, 1,0,1,1,1,0,1,1,
0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0, 1,0,0,0,1,0,1,1,
1,0,0,0,0,0,0,0, 1,0,1,1,1,0,1,1,
0,0,1,1,0,0,1,1, 0,0,0,1,0,0,1,1,
0,0,0,0,0,0,0,0, 0,0,0,1,0,0,1,1,
1,1,0,1,0,0,0,1, 0,0,0,0,0,0,0,0,
1,1,0,1,0,0,0,1, 1,1,0,0,1,0,0,0,
0,1,1,1,0,0,1,1, 0,0,0,1,0,0,1,1,
0,0,0,0,0,0,0,0, 0,0,0,0,0,1,1,1,
1,1,1,1,0,0,1,1, 1,1,0,0,1,1,0,0,
1,1,1,1,0,0,1,1, 1,1,0,0,1,1,0,0
};//索引
int i, j;
int width, height;
//之所以减1,是方便处理8邻域,防止越界
width = src.cols -1;
height = src.rows -1;
int step = src.step;
int p0, p1, p2,p3,p4,p5,p6,p7;
uchar* img;
bool ifEnd;
bool border = false; //交换删除的次序,防止从一边细化
while(1)
{

border = !border;
img = dst.data;
for(i = 1; i < height; i++)
{
img += step;
for(j =1; j<width; j++)
{
uchar* p = img + j;
//如果p点是背景点,继续循环
if(p[0]==0) continue;
p0 = p[-step-1]>0?1:0;
p1 = p[-step]>0?1:0;
p2 = p[-step+1]>0?1:0;
p3 = p[1]>0?1:0;
p4 = p[step+1]>0?1:0;
p5 = p[step]>0?1:0;
p6 = p[step-1]>0?1:0;
p7 = p[-1]>0?1:0;

//如果sum等于0,则不是内部点,是轮廓点,设置其像素值为2
int sum;
sum = p0 & p1 & p2 & p3 & p4 & p5 & p6 & p7;

//判断是否是邻接点或孤立点,0,1分别对于那个孤立点和端点
if(sum==0)
{
dst.at<uchar>(i,j) = 4; //满足删除条件,设置当前像素为0
}

}
}
//printf("\n");
//PrintMat(dst);
//执行删除操作
ifEnd = false;

img = dst.data;
for(i = 1; i < height; i++)
{
img += step;
for(j =1; j<width; j+=3)
{
uchar* p = img + j;
//如果p点是背景点,继续循环
if(p[0]!=4) continue;
p0 = p[-step-1]>0?1:0;
p1 = p[-step]>0?1:0;
p2 = p[-step+1]>0?1:0;
p3 = p[1]>0?1:0;
p4 = p[step+1]>0?1:0;
p5 = p[step]>0?1:0;
p6 = p[step-1]>0?1:0;
p7 = p[-1]>0?1:0;

p1 = p1<<1;
p2 = p2<<2;
p3 = p3 <<3;
p4 = p4<<4;
p5 = p5<<5;
p6 = p6 <<6;
p7 = p7 << 7;

//求的8邻域在索引表中的索引
int sum;
sum = p0 | p1 | p2 | p3 | p4 | p5 | p6 | p7;

//判断是否是邻接点或孤立点,0,1分别对于那个孤立点和端点
if(deletemark[sum] == 1)
{
dst.at<uchar>(i,j) = 0; //满足删除条件,设置当前像素为0
ifEnd = true;
}

}
}

img = dst.data;
for(i = 1; i < height; i++)
{
img += step;
for(j =2; j<width; j+=3)
{
uchar* p = img + j;
//如果p点是背景点,继续循环
if(p[0]!=4) continue;
p0 = p[-step-1]>0?1:0;
p1 = p[-step]>0?1:0;
p2 = p[-step+1]>0?1:0;
p3 = p[1]>0?1:0;
p4 = p[step+1]>0?1:0;
p5 = p[step]>0?1:0;
p6 = p[step-1]>0?1:0;
p7 = p[-1]>0?1:0;

p1 = p1<<1;
p2 = p2<<2;
p3 = p3 <<3;
p4 = p4<<4;
p5 = p5<<5;
p6 = p6 <<6;
p7 = p7 << 7;

//求的8邻域在索引表中的索引
int sum;
sum = p0 | p1 | p2 | p3 | p4 | p5 | p6 | p7;

//判断是否是邻接点或孤立点,0,1分别对于那个孤立点和端点
if(deletemark[sum] == 1)
{
dst.at<uchar>(i,j) = 0; //满足删除条件,设置当前像素为0
ifEnd = true;
}

}
}

img = dst.data;
for(i = 1; i < height; i++)
{
img += step;
for(j =3; j<width; j+=3)
{
uchar* p = img + j;
//如果p点是背景点,继续循环
if(p[0]!=4) continue;
p0 = p[-step-1]>0?1:0;
p1 = p[-step]>0?1:0;
p2 = p[-step+1]>0?1:0;
p3 = p[1]>0?1:0;
p4 = p[step+1]>0?1:0;
p5 = p[step]>0?1:0;
p6 = p[step-1]>0?1:0;
p7 = p[-1]>0?1:0;

p1 = p1<<1;
p2 = p2<<2;
p3 = p3 <<3;
p4 = p4<<4;
p5 = p5<<5;
p6 = p6 <<6;
p7 = p7 << 7;

//求的8邻域在索引表中的索引
int sum;
sum = p0 | p1 | p2 | p3 | p4 | p5 | p6 | p7;

//判断是否是邻接点或孤立点,0,1分别对于那个孤立点和端点
if(deletemark[sum] == 1)
{
dst.at<uchar>(i,j) = 0; //满足删除条件,设置当前像素为0
ifEnd = true;
}

}
}

//printf("\n");
//PrintMat(dst);
//printf("\n");

//已经没有可以细化的像素了,则退出迭代
if(!ifEnd) break;
}

}

修改后的结果:

image

imageimage

程序源代码:工程FirstOpenCV11

posted on 2013-09-19 19:02  迈克老狼2012  阅读(2808)  评论(0编辑  收藏  举报

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