沿其整个范围的轮廓宽度测量

Question

我正在研究一个测量轮廓宽度的项目。我从图像中检测到了轮廓（见图1）。下一步是沿着长度测量coutour的宽度（如图2所示）。请建议我任何想法。我真的很感谢你的帮助！

谢谢！

masked contour image

required width measurement from contour, green lines indicate width

我有一个计算轮廓的功能，然后下一步是测量所选轮廓沿其长度的宽度。以下是示例代码。

...
// image is read, thresholded and canny edges are detected. That image is input to a function that computes contours from the image. 
///Below is the code in the contour function

cv::Mat src_contour= inputImage.clone(); // input image is cloned for contour detection
cv::Mat maskContour = cv::Mat::zeros(src_contour.size(), CV_8UC3);
std::vector<std::vector<cv::Point> > contours; // stores contours points. Each contour is stored in a vector and there are number of vectors for number of contours
cv::RNG rng(12345); // random number used for random colours of contours
cv::findContours( src_contour, contours, hierarchy, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE, cv::Point(0, 0) );

int nc=contours.size();// nc: total number of countours detected  

std::vector<int>areas(nc); // stores area of each contour in a vector
std::vector<double>arclens(nc); // stores arc length (perimeter) of each contour in a vector
std::vector<double>aspect_ratio(nc); // Apect ratio of the contour , width/height 
std::vector<cv::Rect> r(nc) ; // Vector of rectangles,

int min_area=15000 ; // Minimum area for contour selection 
int min_arclen=1000; // Minimum Arc length used for filtering contours 
double min_aspRatio= 2; // Minimum Aspect ratio used for filtering , deafutl 2.0
double max_aspRatio=4; // Maximum aspect rario used for filtering , default 4.0

for (int i=0; i< nc; i++) // Loop iterates through contours , calculates properties and draws selected contours 
{  
    areas[i]=cv::contourArea(contours[i],false); // Area of each contour is stored in a vector, false: any contour, true: closed contour
    arclens[i]=cv::arcLength(contours[i],false); // Arclength of each contour is stored in a vector 
    r[i]=cv::boundingRect(contours[i]); // Stores bounding rect for each contour in a vector r
    aspect_ratio[i]=float(r[i].width)/r[i].height; // Aspect ratio of each contour is stored in a vector 

    if ((areas[i] > min_area) && (arclens[i] > min_arclen) && (aspect_ratio[i] > min_aspRatio && aspect_ratio[i] < max_aspRatio)) 
    {   
        cv::drawContours(maskContour, contours, i, cv::Scalar(255,255,255), CV_FILLED); // creates mask from contours (filterd by criteria), fills them
    }
}

// maskContour image is the image of selected contours filled , I have access to all the points on the contour. From the selected contours
// need to compute width of contours 

.....
// Now contour width measurement is required 

Answer 1

我认为（{1}}（OpenCV）和distanceTransform（也许是你自己）可以使用。

主要想法：

1. 阈值灰度图像，然后执行skeleton以获取dist-map
2. 找到dist-map的骨架，宽度是骨架值的两倍。

距离图如下。



然后，您尝试distanceTransform，find the skeleton获取宽度。

使用C ++代码更新：

double the dist value