使用 Opencv 检测图像中矩形的中心和角度

2022-01-09 00:00:00 python opencv computer-vision image-processing

问题描述

我有一张如下图:

我需要找出矩形的数量，每个矩形的中心，并测量与穿过中心的矩形长边平行的轴之间的角度，并测量从水平方向逆时针方向的角度.我发现了图像中矩形的数量.我很惊讶地发现了反射的中心和角度.通过瞬间找到中心并没有给我正确的答案.

I need to find out the number of rectangles,centre of each rectangle and the measure the angle between the axis parallel to the longer edge of the rectangle passing through centre and measure the angle in anti­clockwise direction from the horizontal.I found out the number of rectangles in the image.I'm struck in finding out the centre and angle of reflection.Finding the centre through moments is not giving me the correct answer.

我的代码:

import cv2
import numpy as np 
import sys

img = cv2.imread(str(sys.argv[1]),0)
ret,thresh = cv2.threshold(img,127,255,0)
contours,hierarchy = cv2.findContours(thresh,1,2)



for contour in contours:
    area = cv2.contourArea(contour)
    if area>100000:
        contours.remove(contour)




cnt = contours[0]

epsilon = 0.02*cv2.arcLength(cnt,True)
approx = cv2.approxPolyDP(cnt,epsilon,True)

print 'No of rectangles',len(approx)


#finding the centre of the contour
M = cv2.moments(cnt)

cx = int(M['m10']/M['m00'])
cy = int(M['m01']/M['m00'])

print cx,cy

解决方案

使用openCV的minAreaRect函数可以做到这一点.它是用 C++ 编写的，但您可能很容易适应它，因为几乎只使用了 OpenCV 函数.

This is how you can do it with minAreaRect function of openCV. It's written in C++ but probably you can adapt that easily, since nearly only OpenCV functions were used.

    cv::Mat input = cv::imread("../inputData/rectangles.png");

    cv::Mat gray;
    cv::cvtColor(input,gray,CV_BGR2GRAY);

    // since your image has compression artifacts, we have to threshold the image
    int threshold = 200;
    cv::Mat mask = gray > threshold;

    cv::imshow("mask", mask);

    // extract contours
    std::vector<std::vector<cv::Point> > contours;
    cv::findContours(mask, contours, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_NONE);

    for(int i=0; i<contours.size(); ++i)
    {
        // fit bounding rectangle around contour
        cv::RotatedRect rotatedRect = cv::minAreaRect(contours[i]);

        // read points and angle
        cv::Point2f rect_points[4]; 
        rotatedRect.points( rect_points );

        float  angle = rotatedRect.angle; // angle

        // read center of rotated rect
        cv::Point2f center = rotatedRect.center; // center

        // draw rotated rect
        for(unsigned int j=0; j<4; ++j)
            cv::line(input, rect_points[j], rect_points[(j+1)%4], cv::Scalar(0,255,0));

        // draw center and print text
        std::stringstream ss;   ss << angle; // convert float to string
        cv::circle(input, center, 5, cv::Scalar(0,255,0)); // draw center
        cv::putText(input, ss.str(), center + cv::Point2f(-25,25), cv::FONT_HERSHEY_COMPLEX_SMALL, 1, cv::Scalar(255,0,255)); // print angle
    }

生成此图像:

如您所见，角度可能不是您想要的(因为它们随机使用较长或较小的线作为参考).您可以改为提取矩形的长边并手动计算角度.

as you can see, the angles are probably not what you want (because they randomly use the longer or the smaller line as reference). You can instead extract the longer sides of the rectangles and compute the angle manually.

如果您选择旋转矩形的较长边缘并从中计算角度，它看起来像这样:

If you choose the longer edge of the rotated rects and compute the angle from it it looks like this:

// choose the longer edge of the rotated rect to compute the angle
        cv::Point2f edge1 = cv::Vec2f(rect_points[1].x, rect_points[1].y) - cv::Vec2f(rect_points[0].x, rect_points[0].y);
        cv::Point2f edge2 = cv::Vec2f(rect_points[2].x, rect_points[2].y) - cv::Vec2f(rect_points[1].x, rect_points[1].y);

        cv::Point2f usedEdge = edge1;
        if(cv::norm(edge2) > cv::norm(edge1))
            usedEdge = edge2;

        cv::Point2f reference = cv::Vec2f(1,0); // horizontal edge


        angle = 180.0f/CV_PI * acos((reference.x*usedEdge.x + reference.y*usedEdge.y) / (cv::norm(reference) *cv::norm(usedEdge)));

给出这个结果，这应该是你要找的！

giving this result, which should be what you are looking for!

看起来操作没有使用他发布的输入图像，因为参考矩形中心将位于图像之外.

It looks like the op doesn't use the input image he posted, because reference rectangle centres would lie outside of the image.

使用此输入(手动重新调整但可能仍不是最佳):

Using this input (manually rescaled but probably still not optimal):

我得到了这些结果(蓝点是操作提供的参考矩形中心):

I get those results (blue dots are reference rectangle centers provided by the op):

比较参考与检测:

reference (x,y,angle)    detection (x,y,angle)
(320,240,0)              (320, 240, 180) // angle 180 is equal to angle 0 for lines
(75,175,90)              (73.5, 174.5, 90)
(279,401,170)            (279.002, 401.824, 169.992)
(507,379,61)             (507.842, 379.75, 61.1443)
(545,95,135)             (545.75, 94.25, 135)
(307,79,37)              (306.756, 77.8384, 37.1042)

我很想看到真实的输入图像，也许结果会更好.

I would love to see the REAL input image though, maybe the result will be even better.

相关文章