Python - 从旋转角度对OpenCV进行透视变换

Question

I'm working on depth map with OpenCV . I can obtain it but it is reconstructed from the left camera origin and there is a little tilt of this latter and as you can see on the figure, the depth is "shifted" (the depth should be close and no horizontal gradient):

I would like to express it as with a zero angle, i try with the warp perspective function as you can see below but i obtain a null field...

P = np.dot(cam,np.dot(Transl,np.dot(Rot,A1)))

dst = cv2.warpPerspective(depth, P, (2048, 2048))

with :

#Projection 2D -> 3D matrix
A1 = np.zeros((4,3))
A1[0,0] = 1 
A1[0,2] = -1024
A1[1,1] = 1
A1[1,2] = -1024
A1[3,2] = 1

#Rotation matrice around the Y axis
theta = np.deg2rad(5) 
Rot = np.zeros((4,4))
Rot[0,0] = np.cos(theta)
Rot[0,2] = -np.sin(theta)
Rot[1,1] = 1
Rot[2,0] = np.sin(theta)
Rot[2,2] = np.cos(theta)
Rot[3,3] = 1

#Translation matrix on the X axis 
dist = 0
Transl = np.zeros((4,4))
Transl[0,0] = 1 
Transl[0,2] = dist
Transl[1,1] = 1
Transl[2,2] = 1
Transl[3,3] = 1

#Camera Intrisecs matrix 3D -> 2D
cam = np.concatenate((C1,np.zeros((3,1))),axis=1)
cam[2,2] = 1
P = np.dot(cam,np.dot(Transl,np.dot(Rot,A1)))

dst = cv2.warpPerspective(Z0_0, P, (2048*3, 2048*3))

EDIT LATER :

You can download the 32MB field dataset here: https://filex.ec-lille.fr/get?k=cCBoyoV4tbmkzSV5bi6 . Then, load and view the image with:

from matplotlib import pyplot as plt
import numpy as np

img = np.load('testZ0.npy')
plt.imshow(img)
plt.show()

Answer 1

I have got a rough solution in place. You can modify it later.

I used the mouse handling operations available in OpenCV to crop the region of interest in the given heatmap.

(Did I just say I used a mouse to crop the region?) Yes, I did. To learn more about mouse functions in OpenCV SEE THIS . Besides, there are many other SO questions that can help you in this regard.:)

Using those functions I was able to obtain the following:

Now to your question of removing the tilt. I used the homography principal by taking the corner points of the image above and using it on a 'white' image of a definite size. I used the cv2.findHomography() function for this.

Now using the cv2.warpPerspective() function in OpenCV, I was able to obtain the following:

Now you can the required scale to this image as you wanted.

CODE:

I have also attached some snippets of code for your perusal:

#First I created an image of white color of a definite size
back = np.ones((435, 379, 3)) # size
back[:] = (255, 255, 255)     # white color

Next I obtained the corner points pts_src on the tilted image below :

pts_src = np.array([[25.0, 2.0],[403.0,22.0],[375.0,436.0],[6.0,433.0]])

I wanted the points above to be mapped to the points 'pts_dst' given below :

pts_dst = np.array([[2.0, 2.0], [379.0, 2.0], [379.0, 435.0],[2.0, 435.0]])

Now I used the principal of homography:

h, status = cv2.findHomography(pts_src, pts_dst)

Finally I mapped the original image to the white image using perspective transform.

fin = cv2.warpPerspective(img, h, (back.shape[1],back.shape[0]))
# img -> original tilted image.
# back -> image of white color.

Hope this helps! I also got to learn a great deal from this question.

Note: The points fed to the 'cv2.findHomography()' must be in float . For more info on Homography , visit THIS PAGE