LOGISTICS
opencv 2.4和Python 2.7
我正在处理的图像:
问题
我感兴趣的是隔离构成9条垂直和水平线周围的框的轮廓。我只是不确定如何去做。我看过各种教程，例如在数独游戏中完成的那些教程，并且只是假设最大的盒子就是您要寻找的盒子(因为数独游戏的盒子中没有盒子，所以要减去实际的网格)。 。我试过使用findContour函数并按大小过滤轮廓，但是没有运气。我最终得到不一致的结果，有时发现正确的轮廓，而有时发现完全错误的轮廓。谁能指出我正确的方向？谢谢。
原始图片:
最佳答案
最好放原始图像，但我尝试从您的轮廓图像中解释
我做了以下步骤
您只需要使投影平滑并针对确切的轮廓优化搜索。
如果您认为这很有用，我可以分享由opencv c++(不是python)实现的代码
编辑:
这是我用来进行水平和垂直投影的代码，您可能需要对其进行优化。
void HVprojection(Mat image)
{
// find the vertical projection
Mat smothedRes = image.clone();
vector<double> v_vl_proj; // holds the column sum values
double max_vl_proj_h = 0,max_vl_proj_v=0; // holds the maximum value
double average_v=0;
for( int i=0;i<image.cols;++i )
{
Mat col;
Scalar col_sum;
// get individual columns
col= image.col(i);
col_sum = sum( col ); // find the sum of ith column
v_vl_proj.push_back( col_sum.val[0] ); // push back to vector
if( col_sum.val[0]>max_vl_proj_v ) max_vl_proj_v = col_sum.val[0];
average_v+= col_sum.val[0];
}
average_v = average_v/image.cols;
// find the horizontal projection
vector<double> h_vl_proj; // holds the row sum values
double average_h=0;
for( int i=0;i<image.rows;++i )
{
Mat row;
Scalar row_sum;
// get individual columns
row= image.row(i);
row_sum = sum(row); // find the sum of ith row
h_vl_proj.push_back(row_sum.val[0]); // push back to vector
if( row_sum.val[0]>max_vl_proj_h ) max_vl_proj_h = row_sum.val[0];
average_h+= row_sum.val[0];
}
average_h = average_h/image.rows;
//******************Plotting vertical projection*******************
for(int j =1;j<image.cols;j++)
{
int y1 = int(image.rows*v_vl_proj[j-1]/max_vl_proj_v);
int y2 = int(image.rows*v_vl_proj[j]/max_vl_proj_v);
line(image,Point(j-1,y1),Point(j,y2),Scalar(255,255,255),1,8);
}
int average_y = int(image.rows*average_v/max_vl_proj_v); // zero level
line(image,Point(0,average_y),Point(image.cols,average_y),Scalar(255,255,255),1,8);
//***************Plotting horizontal projection**************
for(int j =1;j<image.rows;j++)
{
int x1 = int(0.25*image.cols*h_vl_proj[j-1]/max_vl_proj_h);
int x2 = int(0.25*image.cols*h_vl_proj[j]/max_vl_proj_h);
line(image,Point(x1,j-1),Point(x2,j),Scalar(255,0,0),1,8);
}
int average_x = int(0.25*image.cols*average_h/max_vl_proj_h);
line(image,Point(average_x,0),Point(average_x,image.rows),Scalar(255,0,0),1,8);
imshow("horizontal_projection",image);
imwrite("h_p.jpg",image);
// if you want to smooth the signal of projection in case of noisu signal
v_vl_proj = smoothing(v_vl_proj);
for(int j =1;j<image.cols;j++)
{
int y1 = int(image.rows*v_vl_proj[j-1]/max_vl_proj_v);
int y2 = int(image.rows*v_vl_proj[j]/max_vl_proj_v);
line(smothedRes,Point(j-1,y1),Point(j,y2),Scalar(0,255,0),1,8);
}
int average_y1 = int(smothedRes.rows*average_v/max_vl_proj_v); // zero level
line(smothedRes,Point(0,average_y1),Point(smothedRes.cols,average_y1),Scalar(0,255,0),1,8);
imshow("SMoothed",smothedRes);
imwrite("Vertical_projection.jpg",smothedRes);
waitKey(0);
}
要平滑投影信号:
vector<double> smoothing(vector<double> a)
{
//How many neighbors to smooth
int NO_OF_NEIGHBOURS=5;
vector<double> tmp=a;
vector<double> res=a;
for(int i=0;i<a.size();i++)
{
if(i+NO_OF_NEIGHBOURS+1<a.size())
{
for(int j=1;j<NO_OF_NEIGHBOURS;j++)
{
res.at(i)+=res.at(i+j+1);
}
res.at(i)/=NO_OF_NEIGHBOURS;
}
else
{
for(int j=1;j<NO_OF_NEIGHBOURS;j++)
{
res.at(i)+=tmp.at(i-j);
}
res.at(i)/=NO_OF_NEIGHBOURS;
}
}
return res;
}