在opencv

时间:2016-05-31 06:27:14

标签: c++ opencv image-processing image-segmentation opencv-contour

我有这个源图像

enter image description here

我已应用二进制阈值来获取此

enter image description here

我使用轮廓来区分具有子轮廓的轮廓和不具有轮廓的轮廓。所得到的图片是

enter image description here

但是,我如何计算每个绿色轮廓包含的儿童轮廓的数量?这是我用过的代码: -

Mat binMask = lung;// the thresholded image
Mat lung_src = imread("source.tiff");// the source image
//imshow("bin mask", binMask);
vector<std::vector<cv::Point>> contours;
vector<cv::Vec4i> hierarchy;
int count = 0, j;

double largest_area = 0;
int largest_contour_index = 0;

findContours(binMask, contours, hierarchy, CV_RETR_TREE, CV_CHAIN_APPROX_SIMPLE, cv::Point(0, 0));

for (int i = 0; i < contours.size(); i++)
{
    double a = contourArea(contours[i], false);  //  Find the area of contour
    if (a>largest_area)
    {
        largest_area = a;
        largest_contour_index = i;
    }
    for (j = 0; j <= i; j++)
    {
        if (hierarchy[j][2] != -1) // means it has child contour
        {

                drawContours(lung_src, contours, j, Scalar(0, 255, 0), 1, 8, hierarchy, 0, Point());

        }           
        else  // means it doesn't have any child contour
        {
            drawContours(lung_src, contours, j, Scalar(0, 0, 255), 1, 8, hierarchy, 0, Point());
        }
    }
}
drawContours(lung_src, contours, largest_contour_index, Scalar(255, 0, 0), 1, 8, hierarchy, 0, Point());
imshow("lung-mapped", lung_src);

EDIT-1-我在最后添加了Humam的代码来检查它:

std::vector<int> number_of_inner_contours(contours.size(), -1);
int number_of_childs = 0;
for (size_t i = 0; i < contours.size(); i++)
{

    int first_child_index = hierarchy[i][2];
    if (first_child_index >= 0)
    {
        int next_child_index = hierarchy[first_child_index][0];
        if (number_of_inner_contours[next_child_index]<0)
        {
            number_of_childs = number_of_inner_contours[next_child_index];
        }
        else
        {
            while (next_child_index >= 0)
            {
                next_child_index = hierarchy[next_child_index][0];
                ++number_of_childs;
            }
            number_of_inner_contours[i] = number_of_childs;
        }
    }
    else
    {
        number_of_inner_contours[i] = 0;
    }
    cout << "\nThe contour[" << i << "] has " << number_of_inner_contours[i] << "child contours";
}

但我得到的输出是:

      The contour[456 ] has 0 child contours
      The contour[457 ] has 0 child contours
      The contour[458 ] has 0 child contours
      The contour[459 ] has -1 child contours

1 个答案:

答案 0 :(得分:1)

来自OpenCV documentation

  

层次结构 - 可选输出向量,包含有关的信息   图像拓扑。它具有与轮廓数量一样多的元素。对于   每个第i个轮廓轮廓[i],元素层次[i] [0],   hiearchy i,hiearchy [i] [2]和hiearchy [i] [3]设置为   基于0的指数在下一轮和前一轮的轮廓中   相同的层次级别,第一个子轮廓和父级   轮廓,分别。如果轮廓i没有下一个,   previous,parent或嵌套轮廓,相应的元素   等级[i]将是否定的。

这是用于完成工作的未经测试的代码:

std::vector<size_t> number_of_inner_contours;
number_of_inner_contours.reserve(contours.size());
for (size_t i = 0; i < contours.size(); i++){
    size_t number_of_childs = 0;   
    auto first_child_index=hierarchy[i][2];
    if(first_child_index>=0){
        auto next_child_index=hierarchy[first_child_index][0];
        while (next_child_index>=0){
            next_child_index=hierarchy[next_child_index][0];
            ++number_of_childs;
        }
        number_of_inner_contours.emplace_back(number_of_childs);
    }
    else{
         number_of_inner_contours.emplace_back(0);
    }
}

通过使用动态编程的概念,可以更好地完成此代码。这是第一次尝试:

std::vector<int> number_of_inner_contours(contours.size(),-1);
for (size_t i = 0; i < contours.size(); i++){
    auto number_of_childs = 0;   
    auto first_child_index=hierarchy[i][2];
    if(first_child_index>=0){
        auto next_child_index=hierarchy[first_child_index][0];
        if(number_of_inner_contours[next_child_index]<0){
            number_of_childs=number_of_inner_contours[next_child_index];
        }
        else{
            while (next_child_index>=0){
                next_child_index=hierarchy[next_child_index][0];
                ++number_of_childs;
            }
            number_of_inner_contours[i]=number_of_childs;
        }
    }
    else{
         number_of_inner_contours[i]=0;
    }
}