在二值图像上分割重叠的粗线

Segmenting overlapping thick lines on a binary image

在对原始图像应用各种预处理和检测管道后,我确实得到了如下所示的二值图像。

如图所示,实际上有 2 条飞机跑道(停机坪)在交叉区域相互交叉。我需要的是分割跑道和 return 它们的轮廓。我已经检查了关于轮廓特征的 opencv 函数,但没有运气。 cv2.fitLine 似乎没问题,但只有在轮廓中只有一条线时才有效。应用蒙版后的结果图像应如下所示:

我试图通过引用 my old answer.

的 C++ 来解决您的问题

一些步骤:

--after finding contours find defect points by convexityDefects

approxPolyDP(contours[i], contours[i], 9, true);
convexHull(contours[i], contoursHull, true);
convexityDefects(contours[i], contoursHull, defects);

创建两个二进制图像副本并使用缺陷点画线

Vec4i defpoint0 = defects[0];
Vec4i defpoint1 = defects[1];
Vec4i defpoint2 = defects[2];
Vec4i defpoint3 = defects[3];
line(bw0, contours[i][defpoint0[2]], contours[i][defpoint1[2]], Scalar(0), 2);
line(bw0, contours[i][defpoint2[2]], contours[i][defpoint3[2]], Scalar(0), 2);

line(bw1, contours[i][defpoint0[2]], contours[i][defpoint3[2]], Scalar(0), 2);
line(bw1, contours[i][defpoint1[2]], contours[i][defpoint2[2]], Scalar(0), 2);

从图像中找到轮廓并绘制它们(我硬编码找到轮廓索引,需要改进)

findContours(bw0, contours, RETR_EXTERNAL, CHAIN_APPROX_NONE);
drawContours(src, contours, 1, Scalar((rand() & 255), (rand() & 255), (rand() & 255)), 2);

findContours(bw1, contours, RETR_EXTERNAL, CHAIN_APPROX_NONE);
drawContours(src, contours, 2, Scalar((rand() & 255), (rand() & 255), (rand() & 255)), 2);

#include <opencv2/highgui.hpp>
#include <opencv2/imgproc.hpp>
#include <iostream>

using namespace cv;
using namespace std;

int main(int argc, char** argv)
{
    Mat src = imread("e:/test/crossing_lines.png");
    if (src.empty())
        return -1;

    Mat bw,bw0,bw1;
    cvtColor(src, bw, COLOR_BGR2GRAY);
    bw = bw > 127;
    bw0 = bw.clone();
    bw1 = bw.clone();
    // Find contours
    vector<vector<Point> > contours;
    vector<int> contoursHull;
    vector<Vec4i> defects;
    findContours(bw, contours, RETR_EXTERNAL, CHAIN_APPROX_NONE);

    for (size_t i = 0; i < contours.size(); i++)
    {
        if (contourArea(contours[i]) > 500)
        {
            approxPolyDP(contours[i], contours[i], 9, true);
            convexHull(contours[i], contoursHull, true);
            convexityDefects(contours[i], contoursHull, defects);
 
            Vec4i defpoint0 = defects[0];
            Vec4i defpoint1 = defects[1];
            Vec4i defpoint2 = defects[2];
            Vec4i defpoint3 = defects[3];
            line(bw0, contours[i][defpoint0[2]], contours[i][defpoint1[2]], Scalar(0), 2);
            line(bw0, contours[i][defpoint2[2]], contours[i][defpoint3[2]], Scalar(0), 2);

            line(bw1, contours[i][defpoint0[2]], contours[i][defpoint3[2]], Scalar(0), 2);
            line(bw1, contours[i][defpoint1[2]], contours[i][defpoint2[2]], Scalar(0), 2);
        }
    }
    findContours(bw0, contours, RETR_EXTERNAL, CHAIN_APPROX_NONE);
    drawContours(src, contours, 1, Scalar((rand() & 255), (rand() & 255), (rand() & 255)), 2);

    findContours(bw1, contours, RETR_EXTERNAL, CHAIN_APPROX_NONE);
    drawContours(src, contours, 2, Scalar((rand() & 255), (rand() & 255), (rand() & 255)), 2);
    imshow("src", src);
    imshow("bw0", bw0);
    imshow("bw1", bw1);
    waitKey();
    return 0;
}

这是一种可能的方法,刚刚在终端中使用 ImageMagick 完成,但您应该能够在 Python 中使用 Wand or with scikit-image and the medial_axis 完成几乎相同的操作.

首先,对图像进行骨架化:

magick runways.png -threshold 50% -morphology Thinning:-1 Skeleton skeleton.png

然后运行一个“Hough Line Detection”寻找长于130像素的线并以表格形式询问结果:

magick skeleton.png -hough-lines 9x9+130 mvg:-

输出

# Hough line transform: 9x9+130
viewbox 0 0 464 589
# x1,y1 x2,y2 # count angle distance
line 297.15,0 286.869,589  # 255 1 476
line 0,591.173 464,333.973  # 189 61 563

这意味着它检测到 2 行:

  • 一条从坐标 297,0 到坐标 286,589 的线,长度为 255 像素,与垂直方向成 1 度角
  • 一条从坐标 0,591 到坐标 464,333 的线,长度为 189 像素,与垂直方向成 61 度角

为了说明,我将第一个画成红色,第二个画成绿色:

magick runways.png                       \
   -fill red  -draw "line 297,0 286,589" \
   -fill lime -draw "line 0,591 464,333" result.png

关键字:Python,图像处理,骨架,骨架化,细化,运行方式,运行方式,相交,霍夫线检测。