title: How can I set OpenCV to use origin as bottom-left corner of image?
body: <p>Is it possible to set OpenCV to use the bottom-left corner of the image as the origin?</p>
  
  <p>I'm using cv2.imshow() to display an image in a window that allows me to click on the image and get coordinates on mouseclick events. I can use part of the answer from <a href="https://stackoverflow.com/a/43589954/2694260">this question</a> to get the image displayed how I would like it (inverted on the x-axis) however when I click on the image to retrieve coordinates the origin is still in the top-left corner of the image.</p>
  
  <pre><code>import cv2 # Version 3.4.5
  import imutils as im
  import numpy as np
  import matplotlib.pyplot as plt
  
  list_of_coords = []
  def store_mouse_coords(event, x, y, flags, params):
      global list_of_coords
      if event == cv2.EVENT_LBUTTONDOWN:
          list_of_coords.append((x, y))
          print(list_of_coords[-3:])
      return list_of_coords
  
  def select_points_on_im(image, window_name='Select points'):
      image = image.copy()
      image = image[::-1, :, :]
      while True:
          cv2.namedWindow(window_name)
          cv2.setMouseCallback(window_name, store_mouse_coords)
          cv2.imshow(window_name, image)
          #plt.imshow(window_name, image, plt.gca().invert_yaxis,plt.show())
          key = cv2.waitKey(0) &amp; 0xFF
          if key == ord("q"):
              break
      cv2.destroyAllWindows()
      return
  
  select_points_on_im(image)
  </code></pre>
  
  <p>The line commented out line <code>plt.imshow(window_name, image, plt.gca().invert_yaxis,plt.show())</code> is what I'd like to acheive but using <code>imshow</code> from OpenCV so that I can use is with my callback function. I'm using Python3 and OpenCV version 3.4.5.</p>