沿其整个长度测量轮廓宽度

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，然后下一步是沿着它们的长度测量所选轮廓的宽度。下面是示例代码。

...
// 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distanceTransform（OpenCV）和skeleton（也许是你自己）会奏效。

大意：

1. 阈值你灰色图像，然后做distanceTransform得到dist-map
2. 找到dist-map的骨架，宽度是骨架值的两倍。

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距离图如下。

然后你尝试find the skeleton，double the dist value来获得宽度。

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使用C ++代码更新：

int main() {
    // read as gray and threshold 
    Mat gray, threshed, dist;
    gray = imread("img01.png", 0);
    threshold(gray, threshed, 100, 255, THRESH_BINARY);
    imshow("threshed", threshed);

    //distanceTransform
    distanceTransform(threshed, dist, DIST_L2, 3);

    // normalize for display
    Mat dst;
    normalize(dist, dst, 255, 0, NORM_MINMAX,CV_8UC1);
    imshow("dst", dst);
    waitKey();

    return 0;
}