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。
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。