Perspective correction in OpenCV using python

Question

I am trying to do a perspective correction of a tilted rectangle ( a credit card), which is tilted in all the 4 directions. I could find its four corners and the respective

Answer 1

Here is the way you need to follow...

For easiness I have resized your image to smaller size,

• Compute quadrangle vertices for source image, here I find out manually, you can choose edge detection, hough line etc..

  Q1=manual calculation;
  Q2=manual calculation;
  Q3=manual calculation;
  Q4=manual calculation;

• Compute quadrangle vertices in the destination image by keeping aspect ratio, here you can to take width of card from above quadrangle vertices of source, and compute height by multiplying with aspect ratio.

   // compute the size of the card by keeping aspect ratio.
    double ratio=1.6;
    double cardH=sqrt((Q3.x-Q2.x)*(Q3.x-Q2.x)+(Q3.y-Q2.y)*(Q3.y-Q2.y)); //Or you can give your own height
    double cardW=ratio*cardH;
    Rect R(Q1.x,Q1.y,cardW,cardH);

• Now you got quadrangle vertices for source and destination, then apply warpPerspective.

You can refer below C++ code,

   //Compute quad point for edge
    Point Q1=Point2f(90,11);
    Point Q2=Point2f(596,135);
    Point Q3=Point2f(632,452);
    Point Q4=Point2f(90,513);

    // compute the size of the card by keeping aspect ratio.
    double ratio=1.6;
    double cardH=sqrt((Q3.x-Q2.x)*(Q3.x-Q2.x)+(Q3.y-Q2.y)*(Q3.y-Q2.y));//Or you can give your own height
    double cardW=ratio*cardH;
    Rect R(Q1.x,Q1.y,cardW,cardH);

    Point R1=Point2f(R.x,R.y);
    Point R2=Point2f(R.x+R.width,R.y);
    Point R3=Point2f(Point2f(R.x+R.width,R.y+R.height));
    Point R4=Point2f(Point2f(R.x,R.y+R.height));

    std::vector<Point2f> quad_pts;
    std::vector<Point2f> squre_pts;

    quad_pts.push_back(Q1);
    quad_pts.push_back(Q2);
    quad_pts.push_back(Q3);
    quad_pts.push_back(Q4);

    squre_pts.push_back(R1);
    squre_pts.push_back(R2);
    squre_pts.push_back(R3);
    squre_pts.push_back(R4);


    Mat transmtx = getPerspectiveTransform(quad_pts,squre_pts);
    int offsetSize=150;
    Mat transformed = Mat::zeros(R.height+offsetSize, R.width+offsetSize, CV_8UC3);
    warpPerspective(src, transformed, transmtx, transformed.size());

    //rectangle(src, R, Scalar(0,255,0),1,8,0);

    line(src,Q1,Q2, Scalar(0,0,255),1,CV_AA,0);
    line(src,Q2,Q3, Scalar(0,0,255),1,CV_AA,0);
    line(src,Q3,Q4, Scalar(0,0,255),1,CV_AA,0);
    line(src,Q4,Q1, Scalar(0,0,255),1,CV_AA,0);

    imshow("quadrilateral", transformed);
    imshow("src",src);
    waitKey();

Answer 2

I'm writing the answer provided by @Haris in python.

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

img = cv2.imread('test.jpg')
rows,cols,ch = img.shape

pts1 = np.float32([[360,50],[2122,470],[2264, 1616],[328,1820]])

ratio=1.6
cardH=math.sqrt((pts1[2][0]-pts1[1][0])*(pts1[2][0]-pts1[1][0])+(pts1[2][1]-pts1[1][1])*(pts1[2][1]-pts1[1][1]))
cardW=ratio*cardH;
pts2 = np.float32([[pts1[0][0],pts1[0][1]], [pts1[0][0]+cardW, pts1[0][1]], [pts1[0][0]+cardW, pts1[0][1]+cardH], [pts1[0][0], pts1[0][1]+cardH]])

M = cv2.getPerspectiveTransform(pts1,pts2)

offsetSize=500
transformed = np.zeros((int(cardW+offsetSize), int(cardH+offsetSize)), dtype=np.uint8);
dst = cv2.warpPerspective(img, M, transformed.shape)

plt.subplot(121),plt.imshow(img),plt.title('Input')
plt.subplot(122),plt.imshow(dst),plt.title('Output')
plt.show()

Answer 3

I have a better solution which is much easy:

• The red rectangle on original image and the corners points of the rectangle are source points

• We use cv2.getPerspectiveTransform(src, dst) that takes source points and destination points as arguments and returns the transformation matrix which transforms any image to destination image as show in the diagram

• We use this transformation matrix in cv2.warpPerspective()
  - As you can see results are better. You get a very nice bird view of the image

  import cv2
  import matplotlib.pyplot as plt
  import numpy as np
  
  
  def unwarp(img, src, dst, testing):
      h, w = img.shape[:2]
      # use cv2.getPerspectiveTransform() to get M, the transform matrix, and Minv, the inverse
      M = cv2.getPerspectiveTransform(src, dst)
      # use cv2.warpPerspective() to warp your image to a top-down view
      warped = cv2.warpPerspective(img, M, (w, h), flags=cv2.INTER_LINEAR)
  
      if testing:
          f, (ax1, ax2) = plt.subplots(1, 2, figsize=(20, 10))
          f.subplots_adjust(hspace=.2, wspace=.05)
          ax1.imshow(img)
          x = [src[0][0], src[2][0], src[3][0], src[1][0], src[0][0]]
          y = [src[0][1], src[2][1], src[3][1], src[1][1], src[0][1]]
          ax1.plot(x, y, color='red', alpha=0.4, linewidth=3, solid_capstyle='round', zorder=2)
          ax1.set_ylim([h, 0])
          ax1.set_xlim([0, w])
          ax1.set_title('Original Image', fontsize=30)
          ax2.imshow(cv2.flip(warped, 1))
          ax2.set_title('Unwarped Image', fontsize=30)
          plt.show()
      else:
          return warped, M
  im = cv2.imread("so.JPG")
  w, h = im.shape[0], im.shape[1]
  # We will first manually select the source points 
  # we will select the destination point which will map the source points in
  # original image to destination points in unwarped image
  src = np.float32([(20,     1),
                    (540,  130),
                    (20,    520),
                    (570,  450)])
  
  dst = np.float32([(600, 0),
                    (0, 0),
                    (600, 531),
                    (0, 531)])
  
  unwarp(im, src, dst, True)
  
  cv2.imshow("so", im)
  cv2.waitKey(0)[![enter image description here][1]][1]
  cv2.destroyAllWindows()