I want to compare to images of the same size with some text on it.
Let's say the two words are: 'google' and 'gooogle'.
Before measuring the image difference in PS, I am blurring the images using Gauß.
The neat thing in PS is, no matter how you arrange the layers - gooogle on top or google on top - the difference of the layers stay the same.
You get a black background and the difference as (more or less) white pixels.
I am unable to reproduce this functionality in Python.
How did PS manage to get commutativity in there?
I was able to find the solution:
You need to take the absolute. Problem is, you need to convert the RGB image (uint8) to a bigger datatype.
After that you can subtract the images and take the absolute. In the end you need to convert it back to RGB, which is uint8.
def ps_like_diff:(img1, img2):
img1_ = img1.astype(int)
img2_ = img2.astype(int)
diff = img1_ - img2_
return (np.abs(diff)).astype('uint8')
Related
My aim is to stitch 1-2 thousand images together. I find the key points in all the images, then I find the matches between them. Next, I find the homography between the two images. I also take into account the current homography and all the previous homographies. Finally, I warp the images based on combined homography. (My code is written in python 2.7)
The issue I am facing is that when I overlay the warped images, they become extremely bright. The reason is that most of the area between two consecutive images is common/overalapping. So, when I overlay them, the intensities of the common areas increase by a factor of 2 and as more and more images are overalid the moew bright the values become and eventually I get a matrix where all the pixels have the value of 255.
Can I do something to adjust the brightness after every image I overlay?
I am combining/overlaying the images via open cv function named cv.addWeighted()
dst = cv.addWeighted( src1, alpha, src2, beta, gamma)
here, I am taking alpha and beta = 1
dst = cv.addWeighted( image1, 1, image2, 1, 0)
I also tried decreasing the value of alpha and beta but here a problem comes that, when around 100 images have been overlaid, the first ones start to vanish probably because the intensity of those images became zero after being multiplied by 0.5 at every iteration. The function looked as follows. Here, I also set the gamma as 5:
dst = cv.addWeighted( image1, 0.5, image2, 0.5, 5)
Can someone please help how can I solve the problem of images getting extremely bright (when aplha = beta = 1) or images vanishing after a certain point (when alpha and beta are both around 0.5).
This is the code where I am overlaying the images:
for i in range(0, len(allWarpedImages)):
for j in range(1, len(allWarpedImages[i])):
allWarpedImages[i][0] = cv2.addWeighted(allWarpedImages[i][0], 1, allWarpedImages[i][j], 1, 0)
images.append(allWarpedImages[i][0])
cv2.imwrite('/root/Desktop/thesis' + 'final.png', images[0])
When you stitch two images, the pixel values of overlapping part do not just add up. Ideally, two matching pixels should have the same value (a spot in the first image should also has the same value in the second image), so you simply keep one value.
In reality, two matching pixels may have slightly different pixel value, you may simply average them out. Better still, you adjust their exposure level to match each other before stitching.
For many images to be stitched together, you will need to adjust all of their exposure level to match. To equalize their exposure level is a rather big topic, please read about "histogram equalization" if you are not familiar with it yet.
Also, it is very possible that there is high contrast across that many images, so you may need to make your stitched image an HDR (high dynamic range) image, to prevent pixel value overflow/underflow.
I am writing a disparity matching algorithm using block matching, but I am not sure how to find the corresponding pixel values in the secondary image.
Given a square window of some size, what techniques exist to find the corresponding pixels? Do I need to use feature matching algorithms or is there a simpler method, such as summing the pixel values and determining whether they are within some threshold, or perhaps converting the pixel values to binary strings where the values are either greater than or less than the center pixel?
I'm going to assume you're talking about Stereo Disparity, in which case you will likely want to use a simple Sum of Absolute Differences (read that wiki article before you continue here). You should also read this tutorial by Chris McCormick before you read more here.
side note: SAD is not the only method, but it's really common and should solve your problem.
You already have the right idea. Make windows, move windows, sum pixels, find minimums. So I'll give you what I think might help:
To start:
If you have color images, first you will want to convert them to black and white. In python you might use a simple function like this per pixel, where x is a pixel that contains RGB.
def rgb_to_bw(x):
return int(x[0]*0.299 + x[1]*0.587 + x[2]*0.114)
You will want this to be black and white to make the SAD easier to computer. If you're wondering why you don't loose significant information from this, you might be interested in learning what a Bayer Filter is. The Bayer Filter, which is typically RGGB, also explains the multiplication ratios of the Red, Green, and Blue portions of the pixel.
Calculating the SAD:
You already mentioned that you have a window of some size, which is exactly what you want to do. Let's say this window is n x n in size. You would also have some window in your left image WL and some window in your right image WR. The idea is to find the pair that has the smallest SAD.
So, for each left window pixel pl at some location in the window (x,y) you would the absolute value of difference of the right window pixel pr also located at (x,y). you would also want some running value, which is the sum of these absolute differences. In sudo code:
SAD = 0
from x = 0 to n:
from y = 0 to n:
SAD = SAD + absolute_value|pl - pr|
After you calculate the SAD for this pair of windows, WL and WR you will want to "slide" WR to a new location and calculate another SAD. You want to find the pair of WL and WR with the smallest SAD - which you can think of as being the most similar windows. In other words, the WL and WR with the smallest SAD are "matched". When you have the minimum SAD for the current WL you will "slide" WL and repeat.
Disparity is calculated by the distance between the matched WL and WR. For visualization, you can scale this distance to be between 0-255 and output that to another image. I posted 3 images below to show you this.
Typical Results:
Left Image:
Right Image:
Calculated Disparity (from the left image):
you can get test images here: http://vision.middlebury.edu/stereo/data/scenes2003/
I am working on project what detect hematoma from skin. I am having issue with color after convertion from RGB to HSV. My algorithm detect hematoma by its color.
With some images I have good results like here:
Original img: http://imgur.com/WHiOWdj
Result img: http://imgur.com/PujbnHa
But with some images i have bad result like this:
Original img: http://imgur.com/OshB99r
Result img: http://imgur.com/CuNzAId
The same original image after convertion to HSV: http://imgur.com/lkVwtCs
Do you have any ideas how to fix it?
Thanks
Looking at your result image I think that you are only using the H channel of the original image in your algorithm. The false positive detection can inherit from that the some part of the healty skin has quite the same H value than the hematoma has. You can see on the qrey-scale image of H channel that both parts have similar values:
The difference between the two parts is the saturation value. On the following image you can see the S channel of the original image and it shows perfectly that at the hematoma the saturation is much higher than at other the part of the arm:
This was expected because the hematoma has much stronger color than the healty skin has.
So, I suggest you to use both H and S channel in your algorithm that is you have to take into account only that parts of H image where the S image contains high saturation values. A possible and simple solution to do that is that you binarize both H and S images and with an AND operation you can execute this filtering:
H image after binarisation:
S image after binarisation:
Image after H&S operation:
You can see that on the result image only the hematoma part is white (except some noise but you can eliminate easily, for example by size or by morphological filtering).
EDIT
Important to note that binarization is one of most important (and sometimes also very complicated) step in the object detection algorithms namely binarization is the first highlight of the objects to detect.
If the the external conditions (lighting, color of objects etc.) do not change significantly from image to image you can use fix binaraziation thresholds. If this constant environment can not be issured you have to use more complicated methods. There are a lot of possibilies you can use, here you can read some examples:
Wikipedia - Thresholding
Wikipedia - Balanced histogram thresholding
Several solutions are based on the histogram analysis: on the histograms with objects there are always more local maximums which positions can vary depend on the environment and if you find them you can adapt the binarization threshold easily.
For example the histogram of the H channel of the original image is the following:
The first maximum belongs to the background, the second to the skin and the last to the hematome. It can be supposed that these 3 thresholds can be found in each image only their positions vary depend on the lighting or on other conditions. To put a threshold between the 2nd and the 3rd local maximum it can be a good choice to highlight the hematome.
Finally I offer you the read the following articel about thresholding in OpenCV:
OpenCV - Thresholding
I'm trying to merge/stitch 2 images together but found that the default stitcher class in OpenCV could not handle my images.
So I started to write my own..
Unfortunately the images are too large to attach to this message (they are both 12600x9000 pixels in size).. so I'll try to explain as good as possible.
The 2 images are not pictures takes by a camera but are tiff files extracted from a PDF file.
The images themselves were actually CAD drawings, so not much gradients in there and therefore I think the default stitcher class could not handle them.
So far, I managed to extract the features and match them.
Also I used the following well known example to stitch them together:
Mat WarpedImage;
cv::warpPerspective(img_2,WarpedImage,homography,cv::Size(2*img_2.cols,2*img_2.rows));
Mat half(WarpedImage,Rect(0,0,img_1.cols,img_1.rows));
img_1.copyTo(half);
I sort of made it fit.. because my problem is that in my case the 2 images could be aligned vertically or horizontally.
By default, all stitch examples on the internet assume the first image is the left image and the 2nd image is the right image.
So my first question would be:
How can I detect if the image is to the left, right, above or below the first image and create a proper sized new image?
Secondly..
Currently I'm getting the proper image.. however, because I'm not having some decent code to check the ideal width and height of the new image, I have a lot of black/empty space in the new image.
What would be the best C++ code to remove those black area's?
(I'm seeing a lot of Python scripts on the net.. but no C++ examples of this.. and I have 0 Python skills....)
Thank you very much in advance for your help.
Greetings,
Floris.
You can reproject the corners of the second image with perspectiveTransform. With the transformed points you can find the relative position of your image and calculate the new image size that will fit both images. This will also let you deal with the black areas, since you have the boundaries of the two images.
I want to merge 2 images. How can i remove the same area between 2 images?
Can you tell me an algorithm to solve this problem. Thanks.
Two image are screenshoot image. They have the same width and image 1 always above image 2.
When two images have the same width and there is no X-offset at the left side this shouldn't be too difficult.
You should create two vectors of integer and store the CRC of each pixel row in the corresponding vector element. After doing this for both pictures you find the CRC of the first line of the lower image in the first vector. This is the offset in the upper picture. Then you check that all following CRCs from both pictures are identical. If not, you have to look up the next occurrence of the initial CRC in the upper image again.
After checking that the CRCs between both pictures are identical when you apply the offset you can use the bitblit function of your graphics format and build the composite picture.
I haven't come across something similar before but I think the following might work:
Convert both to grey-scale.
Enhance the contrast, the grey box might become white for example and the text would become more black. (This is just to increase the confidence in the next step)
Apply some threshold, converting the pictures to black and white.
afterwards, you could find the similar areas (and thus the offset of overlap) with a good degree of confidence. To find the similar parts, you could harper's method (which is good but I don't know how reliable it would be without the said filtering), or you could apply some DSP operation(s) like convolution.
Hope that helps.
If your images are same width and image 1 is always on top. I don't see how that hard could it be..
Just store the bytes of the last line of image 1.
from the first line to the last of the image 2, make this test :
If the current line of image 2 is not equal to the last line of image 1 -> continue
else -> break the loop
you have to define a new byte container for your new image :
Just store all the lines of image 1 + all the lines of image 2 that start at (the found line + 1).
What would make you sweat here is finding the libraries to manipulate all these data structures. But after a few linkage and documentation digging, you should be able to easily implement that.