Stitch multiple images using OpenCV (Python)
Hi have seen a lot of tutorials how to do simple image stitching using two photos and that is no problem.
But what to do when I want to make a panorama from 4-6 images or more?
I have code that takes in list of image files(the images are in order from the first image in the sequence to the last). Then for each image I compute the SIFT feature descriptors . But then I am stuck, for two images I would set up a matcher using FLANN kd-tree and find matches between the images and calculate the Homography. Similar to this tutorial http://docs.opencv.org/trunk/doc/py_tutorials/py_feature2d/py_feature_homography/py_feature_homography.html#py-feature-homography
But in stead of showing the lines between feature points at the end I have used this https://stackoverflow.com/a/20355545/622194 function to make a panorama from 2 images. But I am not sure what to do when I want to add the third and the fourth image to the panorama.
EDIT:
From the answers I have tried to implement my image stitching script to calculate a homography matrix between images that are next to each other in the image sequence. So if I have I1 I2 I3 and I4 I now have H_12, H_23 and H_34. Then I start by stitching I1 and I2 using H_12. Then I want to find cumulative homography to stitch I3 to the current panorama. I fing H_13 = H_12*H_23 and stitch the image 3 to the current panorama but here I get very apparent gap in my panorama image and when next image is stitched it is even bigger gap and the images is very stretched. Here is my code http://pastebin.com/dQjhE5VD
Can anyone tell me if I am using right approach for this or can someone spot the error or see what I am doing wrong.
Step by step, assuming you want to stitch four images I0,I1,I2,I3, your goal is to compute homographies H_0,H_1,H_2,H_3;
See section 4 of this seminal paper https://www.cs.bath.ac.uk/brown/papers/ijcv2007.pdf for an in depth explanation.
Hacky approach
The easiest way (though not super efficient) given the functions you've written, is to just grow the panorama image by stitching it with each successive image. Something like this pseudocode:
panorama = images[0]
for i in 1:len(images)-1
panorama = stitch(panorama,images[i])
This method basically attempts to match the next image to any part of the current panorama. It should work decently well, assuming each new image is somewhere on the border of the current panorama, and there isn't too much perspective distortion.
Mathematical approach
The other option, if you know the order that you want to stitch, is to find the Homography from one image to the next, and then multiply them. The result is the Homography from that image to image 0.
For example: the H that transforms image 3 to line up with image 0 is H_03 = H_01 * H_12 * H_23. Where H_01 is the H that transforms image 1 to line up with image 0. (Depending on the way their code defines H, you might need to reverse the above multiplication order.) So you would multiply to obtain H_0i and then use it to transform image i to line up with image 0.
For background on why you multiply the transformations, see: https://www.math.lsu.edu/~verrill/teaching/linearalgebra/linalg/linalg5.html specifically the "Composition of tranformations" part.
I had the similar problem with gaps between images. The first thing you should do is to init your accumulated homography matrix to identity at first frame. Then, with every new frame you should multiply it by homography matrix between current and next frame. Be aware to use numpy matrices and not numpy arrays. IDK why but they have different multiplication routines.
Here is my code:
def addFramePair(self, images, ratio=0.75, reprojThresh=4.0, showMatches=False):
(imageA, imageB) = images
(kpsA, featuresA) = self.detectAndDescribe(imageA)
(kpsB, featuresB) = self.detectAndDescribe(imageB)
H = self.matchKeypoints(kpsA, kpsB, featuresA, featuresB, ratio, reprojThresh)
self.accHomography *= np.asmatrix(H)
result = cv2.warpPerspective(imageA, np.linalg.inv(self.accHomography), (1600, 900))
return result
imageA is current, imageB is the next one.
Hope this helps.
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