CODEWITHSUNDEEP
pythonopencvimage-processingcomputer-vision
Kaushik Ramachandran
Kaushik Ramachandran

Reputation: 501

Detect centre and angle of rectangles in an image using Opencv

I have an image as below :

Sample image containing many rectangle contours

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.

My code :

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

Upvotes: 15

Views: 28441

Answers (3)

Micka
Micka

Reputation: 20130

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
}

resulting in this image:

enter image description here

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.

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)));

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

enter image description here

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

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

enter image description here

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

enter image description here

Comparing the reference with the detections:

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)

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

Upvotes: 22

FiReTiTi
FiReTiTi

Reputation: 5888

Here is how you can do it:

  1. Connected component labeling in order to detect each pattern (in your case the rectangles)
  2. Separate the patterns in different images
  3. (optional) if the pattern are not all rectangles, then use shape indexes to discriminate them
  4. Compute the main axis using Principal Component Analysis (PCA), it will give you the angle you are looking for.

Upvotes: 2

Ajay
Ajay

Reputation: 370

approx = cv2.approxPolyDP(cnt,epsilon,True) creates an approximated polygon of a given closed contour. The line segments in the polygon are of variable length, resulting in incorrect moment computation as it expects the points to be sampled from a regular grid to give you the correct center.

There are three solutions to your problem:

  1. Use moments of original contours before calling the method for polygon approximation.
  2. Use drawContours to generate the mask of the regions inside each closed contour and then use moments of the generated mask to compute the center.
  3. Sample points at unit distance along each line segment of your closed polygon and use the resulting collections of points to compute the moments yourself. This should give you the same center.

Upvotes: 1

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