How to show the whole image when using OpenCV warpPerspective

First, follow the earlier solution to compute the homography matrix. After you have the homography matrix, you need to warp the image with respect to the Homography matrix. Lastly, merge the warped image.

Here, I will share another idea that could be used to merge the warped images. (The earlier answer, uses a range of indices to be overlay, here I'm using the masking of ROI)

Mask the Region of Interest (ROI) and the Image with Black. Then add the image with the ROI. (See OpenCV Bitmask Tutorial)

def copyOver(source, destination):
    result_grey = cv2.cvtColor(source, cv2.COLOR_BGR2GRAY)
    ret, mask = cv2.threshold(result_grey, 10, 255, cv2.THRESH_BINARY)
    mask_inv = cv2.bitwise_not(mask)
    roi = cv2.bitwise_and(source, source, mask=mask)
    im2 = cv2.bitwise_and(destination, destination, mask=mask_inv)
    result = cv2.add(im2, roi)
    return result


warpedImageB = cv2.warpPerspective(imageB, H, (imageA.shape[1], imageA.shape[0]))
result = copyOver(imageA, warpedImageB)

First Image:

Second Image:

Stitched Image:

Yes, but you should realise that the output image might be very large. I quickly wrote the following Python code, but even a 3000 x 3000 image could not fit the output, it is just way too big due to the transformation. Although, here is my code, I hope it will be of use to you.

import cv2
import numpy as np
import cv           #the old cv interface

img1_square_corners = np.float32([[253,211], [563,211], [563,519],[253,519]])
img2_quad_corners = np.float32([[234,197], [520,169], [715,483], [81,472]])

h, mask = cv2.findHomography(img1_square_corners, img2_quad_corners)
im = cv2.imread("image1.png")

Create an output image here, I used (3000, 3000) as an example.

out_2 = cv.fromarray(np.zeros((3000,3000,3),np.uint8))

By using the old cv interface, I wrote directly to the output, and so it does not get cropped. I tried this using the cv2 interface, but for some reason it did not work... Maybe someone can shed some light on that?

cv.WarpPerspective(cv.fromarray(im), out_2, cv.fromarray(h))
cv.ShowImage("test", out_2)
cv.SaveImage("result.png", out_2)
cv2.waitKey()

Anyway, this gives a very large image, that contains your original image 1, warped. The entire image will be visible if you specify the output image to be large enough. (Which might be very large indeed!)

I hope that this code may help you.

My solution is to calculate the result image size, and then do a translation.

def warpTwoImages(img1, img2, H):
    '''warp img2 to img1 with homograph H'''
    h1,w1 = img1.shape[:2]
    h2,w2 = img2.shape[:2]
    pts1 = float32([[0,0],[0,h1],[w1,h1],[w1,0]]).reshape(-1,1,2)
    pts2 = float32([[0,0],[0,h2],[w2,h2],[w2,0]]).reshape(-1,1,2)
    pts2_ = cv2.perspectiveTransform(pts2, H)
    pts = concatenate((pts1, pts2_), axis=0)
    [xmin, ymin] = int32(pts.min(axis=0).ravel() - 0.5)
    [xmax, ymax] = int32(pts.max(axis=0).ravel() + 0.5)
    t = [-xmin,-ymin]
    Ht = array([[1,0,t[0]],[0,1,t[1]],[0,0,1]]) # translate

    result = cv2.warpPerspective(img2, Ht.dot(H), (xmax-xmin, ymax-ymin))
    result[t[1]:h1+t[1],t[0]:w1+t[0]] = img1
    return result

dst_pts = float32([kp1[m.queryIdx].pt for m in good]).reshape(-1,1,2)
src_pts = float32([kp2[m.trainIdx].pt for m in good]).reshape(-1,1,2)
M, mask = cv2.findHomography(src_pts, dst_pts, cv2.RANSAC, 5.0)

result = warpTwoImages(img1_color, img2_color, M)