how to find the outline of union rectangles

Question

I have coordinates of each rectangles and want to find the outlines of union rectangles.

So what I did is that just drew and colored all rectangles with the coordinates. And found the contour of all rectangles.

But I want to know that if there are some kind of algorithms doing the same thing only with coordinates information. So that I don't need to draw and color the whole rectangles. (there are more than 4000 rectangles so it is computationally expensive)

Answer 1

Using contour detection you can achieve what you want. For that you should have a good knowledge about contour hierarchy style used in OpenCV. There are a few contour retrieval modes. Such as cv2.RETR_LIST , cv2.RETR_TREE , cv2.RETR_CCOMP , cv2.RETR_EXTERNAL . You can check the documentation for more information.

In your case you should use cv2.RETR_EXTERNAL in order to retrieve extreme outer contour of the image.

This is the solution code:

import cv2
im = cv2.imread('test.png')
imgray = cv2.cvtColor(im, cv2.COLOR_BGR2GRAY)
cv2.bitwise_not(imgray,imgray)
ret, thresh = cv2.threshold(imgray, 127, 255, 0)
im2, contours, hierarchy = cv2.findContours(thresh, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_NONE)
cv2.drawContours(im, contours, -1, (0,255,0), 3)
cv2.imshow("image", im)
cv2.waitKey(0)

OUTPUT:

Answer 2

findContours is the function you are looking for. The code bellow explains how it works.

import cv2
import numpy as np
import matplotlib.pyplot as plt


img = cv2.imread("./outline_rect.png")
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)

ret, thresh = cv2.threshold(gray, 0, 255, cv2.THRESH_BINARY | cv2.THRESH_OTSU)
thresh = 255-thresh
_, cnts, _ = cv2.findContours(thresh.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)


maxArea = 0
best = None
for contour in cnts:
  area = cv2.contourArea(contour)
  if area > maxArea :
    maxArea = area
    best = contour

cv2.drawContours(img, [best], 0, (0, 0, 255), 3)

while True:
  cv2.imshow("result", img)
  k = cv2.waitKey(30) & 0xff
  if k == 27:
      break

The result: