How to know the width of the line

edit

Using 0.8 instead of 3 :

#include <opencv2/opencv.hpp> 
#include <opencv2/ximgproc.hpp>
#include <iostream>

using namespace cv;
using namespace std;


int main(int argc, char **argv)
{
    Mat select = imread("line.png", IMREAD_GRAYSCALE);
    Mat thinning_select;
    ximgproc::thinning(select, thinning_select, ximgproc::THINNING_ZHANGSUEN);
    Mat select_float;
    distanceTransform(select, select_float, DIST_C, 5);
    Mat img;
    // Quantize the hue to 30 levels
    // and the saturation to 32 levels
    int bins = 256;
    int histSize[] = { bins};
    float range[] = { 0, 256 };
    const float* ranges[] = { range};
    // we compute the histogram from the 0-th and 1-st channels
    int channels[] = { 0 };
    Mat hist;
    calcHist(&select_float, 1, channels, thinning_select, hist, 1, histSize, ranges,true,false);
    double tailleSquelette = countNonZero(thinning_select);
    double ratio = 0;
    int i = 0;
    while (ratio < tailleSquelette*0.8 && i<hist.rows)
    {
        ratio += hist.at<float>(i, 0);
        i++;
    }


    Mat mask = select_float <= i;

    imshow("No line", mask);
    waitKey(0);
return 0;
}

I've a simple idea, may this can help you:)

1. Do some morphology to eliminate the noise.
2. Straighten the line, correct its angle to make it horizontal: use radon_transformation or houghline to get the rotating angle then warpaffine it.
3. Finally get the number of white pixels on one column which is just the width (or you can get the averge of thoses on the whole line or part of it to make the result more stable).