How to take pixels from an input image by using Gaussian sub-sampling (shotgun pattern like)?
I want to take the locations of pixels that are to be taken like in a shotgun pattern concentrated in the middle of the image. Because I do not want to extract features of all pixels in an image. The output should be the coordinates of sampled pixels. I will be thankful if you guide me.
Is there any function or code that I can get help from that.
Your help is appreciated.
If you are looking for a method to define a Region of Interest (ROI) of an image in Matlab in order to perform some operation in a restricted are, remembering that x coordinates represent column and y is on the rows (matlab reads images as matrices):
For cut an image from x1 to x2 and from y1 to y2 try something like
ROI = image[y1:y2,x1:x2]
but how to determine these 4 values without a specific example is up to you
Related
As a result of the faster r-cnn method of object detection, I have obtained a set of boxes of intensity values(each bounding box can be thought of as a 3D matrix with depth of 3 for rgb intensity, a width and a height which can then be converted into a 2D matrix by taking gray scale) corresponding to the region containing the object. What I want to do is to obtain the corresponding co-ordinate points in the original image for each cell of intensity inside of the bounding box. Any ideas how to do so?
From what I understand, you got an R-CNN model that outputs cropped pieces of the input image and you now want to trace those output crops back to their coordinates in the original image.
What you can do is simply use a patch-similarity-measure to find the original position.
Since the output crop should look exactly like itself in the original image, just use Pixel-based distance:
Find the place in the image with the smallest distance (should be zero) and from that you can find your desired coordinates.
In python:
d_min = 10**6
crop_size = crop.shape
for x in range(org_image.shape[0]-crop_size[0]):
for y in range(org_image.shape[1]-crop_size[1]):
d = np.abs(np.sum(np.sum(org_image[x:x+crop_size[0],y:y+crop_size[0]]-crop)))
if d <= d_min:
d_min = d
coord = [x,y]
However, your model should have that info available in it (after all, it crops the output based on some coordinates). Maybe if you add some info on your implementation.
So I have been tinkering a little bit with opencv and I want to be able to use a camera image to get the position of certain objects that are lying flat on plane. These objects are simple shapes such as circles squares etc. They all have the same height of 5cm. To be able to relate real world points to pixels on the camera I painted 4 white squares on the plane with known distances between them.
So the steps I have been taking are:
Initialization:
Calibrate my camera using a checkerboard image and save the calibration data.
Get the input image. call cv::undistort with the calibration data for my camera.
Find the center points of the 4 squares in the image and pass that data and the real world coordinates of the squares to the cv::solvePnP function. Save the rvec and tvec return parameters.
Warp the perspective of the image so you can get a top down view from the image. This is essentially following this tutorial: https://docs.opencv.org/3.4.1/d9/dab/tutorial_homography.html
Use the resulting image to again find the 4 white squares and then calculate a "pixels per meter" translation constant which can relate a certain amount of difference in pixels between points to the real world distance on the plane where the 4 squares are.
Finding object, This is done after initialization:
Get the input image. call cv::undistort with the calibration data for my camera.
Warp the perspective of the image so you can get a top down view from the image. This is the same as step 4 during initialisation.
Find the centerpoint of the object to detect.
Since the centerpoint of the object is on a higher plane then where I calibrated I use the following formula to correct this(d = is the pixel offset from the center of the image. camHeight is the cameraHeight I measured by using a tape measure. h is height of the object):
d = x - (h * (x / camHeight))
So here for an illustration how I got this formule:
But still the coordinates are not matching up...
So I am wondering at all if this is the correct. Specifically I have the following questions:
Is using cv::undistort before using cv::solvenPnP correct? cv::solvePnP also takes the camera calibration data as input so I'm not sure if I have to pass an undistorted image to it or not.
Similar to 1. During Finding object I call cv::undistort -> cv::warpPerspective. Is this undistort necessary here?
Is my calculation to correct for the parallel planes in step 4 correct? I feel like I am missing something but I can't see what. One thing I am wondering is whether I can get the camera height from opencv once solvePnp is done.
I am a newbie to CV so If anything else is totally wrong please also point it out to me.
Thank you for reading this wall of text!
I read similar question in Stack Overflow. I tried, but I still can not understand how it works.
I read OpenCV document cv::HoughCircles, here are some explanation about dp parameter:
Inverse ratio of the accumulator resolution to the image resolution. For example, if dp=1 , the accumulator has the same resolution as the input image. If dp=2 , the accumulator has half as big width and height.
Here are my question. For example, if dp = 1, the size of accumulator is same as image, there is a consistent one-to-one match between pixels in image and positions in accumulator, but if dp = 2, how to match?
Thanks in advance.
There is no such thing as a one-to-one match here. You do have an image with pixels and a hough space, which is used for voting for circles. This parameter is just a convenient way to specify the size of the hough space relatively to the image size.
Please take a look at this answer for more details.
EDIT:
Your image has (x,y)-coordinates. Your circle hough space has (a,b,r)-coordinates, whereas (a,b) are the circle centers and r are the radii. Let's say you find a edge pixel. Now you vote for each circle, which could go through this edge pixel. I found this nice picture of hough space with a single vote i.e. a single edge pixel (continuous case). In practice this vote happens within the 3D accumulator matrix. You can think of it as rasterization of this continuous case.
Now, as already mentioned the dp parameter defines the size of this accumulator matrix relatively to your image size. The bigger the dp parameter the lower the resolution of your rasterization. It's like taking photos with different resolutions. If you downsize your photo multiple pixels will reduce to a single one. Same happens if you reduce your accumulator matrix respectively increase your dp parameter. Multiple votes for different circle centers (which lie next to each other) and radii (which are of similar size) are now merged, i.e. you do get less accurate circle parameters, but a more "robust" voting.
Please be aware that the OpenCV implementation is a little bit more complicated (they use the Hough gradient method instead of the standard Hough transform) but the considerations still apply.
I am not able to under stand the formula ,
What is W (window) and intensity in the formula mean,
I found this formula in opencv doc
http://docs.opencv.org/trunk/doc/py_tutorials/py_feature2d/py_features_harris/py_features_harris.html
For a grayscale image, intensity levels (0-255) tells you how bright is the pixel..hope that you already know about it.
So, now the explanation of your formula is below:
Aim: We want to find those points which have maximum variation in terms of intensity level in all direction i.e. the points which are very unique in a given image.
I(x,y): This is the intensity value of the current pixel which you are processing at the moment.
I(x+u,y+v): This is the intensity of another pixel which lies at a distance of (u,v) from the current pixel (mentioned above) which is located at (x,y) with intensity I(x,y).
I(x+u,y+v) - I(x,y): This equation gives you the difference between the intensity levels of two pixels.
W(u,v): You don't compare the current pixel with any other pixel located at any random position. You prefer to compare the current pixel with its neighbors so you chose some value for "u" and "v" as you do in case of applying Gaussian mask/mean filter etc. So, basically w(u,v) represents the window in which you would like to compare the intensity of current pixel with its neighbors.
This link explains all your doubts.
For visualizing the algorithm, consider the window function as a BoxFilter, Ix as a Sobel derivative along x-axis and Iy as a Sobel derivative along y-axis.
http://docs.opencv.org/doc/tutorials/imgproc/imgtrans/sobel_derivatives/sobel_derivatives.html will be useful to understand the final equations in the above pdf.
Assuming that I have a grayscale (8-bit) image and assume that I have an integral image created from that same image.
Image resolution is 720x576. According to SURF algorithm, each octave is composed of 4 box filters, which are defined by the number of pixels on their side. The
first octave uses filters with 9x9, 15x15, 21x21 and 27x27 pixels. The
second octave uses filters with 15x15, 27x27, 39x39 and 51x51 pixels.The third octave uses filters with 27x27, 51x51, 75x75 and 99x99 pixels. If the image is sufficiently large and I guess 720x576 is big enough (right??!!), a fourth octave is added, 51x51, 99x99, 147x147 and 195x195. These
octaves partially overlap one another to improve the quality of the interpolated results.
// so, we have:
//
// 9x9 15x15 21x21 27x27
// 15x15 27x27 39x39 51x51
// 27x27 51x51 75x75 99x99
// 51x51 99x99 147x147 195x195
The questions are:What are the values in each of these filters? Should I hardcode these values, or should I calculate them? How exactly (numerically) to apply filters to the integral image?
Also, for calculating the Hessian determinant I found two approximations:
det(HessianApprox) = DxxDyy − (0.9Dxy)^2 anddet(HessianApprox) = DxxDyy − (0.81Dxy)^2Which one is correct?
(Dxx, Dyy, and Dxy are Gaussian second order derivatives).
I had to go back to the original paper to find the precise answers to your questions.
Some background first
SURF leverages a common Image Analysis approach for regions-of-interest detection that is called blob detection.
The typical approach for blob detection is a difference of Gaussians.
There are several reasons for this, the first one being to mimic what happens in the visual cortex of the human brains.
The drawback to difference of Gaussians (DoG) is the computation time that is too expensive to be applied to large image areas.
In order to bypass this issue, SURF takes a simple approach. A DoG is simply the computation of two Gaussian averages (or equivalently, apply a Gaussian blur) followed by taking their difference.
A quick-and-dirty approximation (not so dirty for small regions) is to approximate the Gaussian blur by a box blur.
A box blur is the average value of all the images values in a given rectangle. It can be computed efficiently via integral images.
Using integral images
Inside an integral image, each pixel value is the sum of all the pixels that were above it and on its left in the original image.
The top-left pixel value in the integral image is thus 0, and the bottom-rightmost pixel of the integral image has thus the sum of all the original pixels for value.
Then, you just need to remark that the box blur is equal to the sum of all the pixels inside a given rectangle (not originating in the top-lefmost pixel of the image) and apply the following simple geometric reasoning.
If you have a rectangle with corners ABCD (top left, top right, bottom left, bottom right), then the value of the box filter is given by:
boxFilter(ABCD) = A + D - B - C,
where A, B, C, D is a shortcut for IntegralImagePixelAt(A) (B, C, D respectively).
Integral images in SURF
SURF is not using box blurs of sizes 9x9, etc. directly.
What it uses instead is several orders of Gaussian derivatives, or Haar-like features.
Let's take an example. Suppose you are to compute the 9x9 filters output. This corresponds to a given sigma, hence a fixed scale/octave.
The sigma being fixed, you center your 9x9 window on the pixel of interest. Then, you compute the output of the 2nd order Gaussian derivative in each direction (horizontal, vertical, diagonal). The Fig. 1 in the paper gives you an illustration of the vertical and diagonal filters.
The Hessian determinant
There is a factor to take into account the scale differences. Let's believe the paper that the determinant is equal to:
Det = DxxDyy - (0.9 * Dxy)^2.
Finally, the determinant is given by: Det = DxxDyy - 0.81*Dxy^2.
Look at page 17 of this document
http://www.sci.utah.edu/~fletcher/CS7960/slides/Scott.pdf
If you made a code for normal Gaussian 2D convolution, just use the box filter as a Gaussian kernel and the input image will be the same original image not integral image. The results from this method will be same with the one you asked.