I have objets have different shapes and my aim is to find ratio between object area and its Bounding Box area. There is no problem to find object area but I did not find a way to get Bounding Box area.
Is there any way or any exist function in matlab to calculate Bounding Box area?
But, you know the bounding box? If you dont, use regionprops(Imgbw,'BoundingBox')
And you will get it.
Once you have it, its kinda easy. Is just computing the area of a square. Regionprops will give you [x y] and [x_width y_width]. Im sure you are capable of calculating the area of a square with the size of its sides.
Related
I have image of robot with yellow markers as shown
The yellow points shown are the markers. There are two cameras used to view placed at an offset of 90 degrees. The robot bends in between the cameras. The crude schematic of the setup can be referred.
https://i.stack.imgur.com/aVyDq.png
Using the two cameras I am able to get its 3d co-ordinates of the yellow markers. But, I need to find the 3d-co-oridnates of the central point of the robot as shown.
I need to find the 3d position of the red marker points which is inside the cylindrical robot. Firstly, is it even feasible? If yes, what is the method I can use to achieve this?
As a bonus, is there any literature where they find the 3d location of such internal points which I can refer to (I searched, but could not find anything similar to my ask).
I am welcome to a theoretical solution as well(as long as it assures to find the central point within a reasonable error), which I can later translate to code.
If you know the actual dimensions, or at least, shape (e.g. perfect circle) of the white bands, then yes, it is feasible and possible.
You need to do the following steps, which are quite non trivial to do, and I won't do them here:
Optional but extremely suggested: calibrate your camera, and
undistort it.
find the equation of the projection of a 3D circle into a 2D camera, for any given rotation. You can simplify this by assuming the white line will be completely horizontal. You want some function that takes the parameters that make a circle and a rotation.
Find all white bands in the image, segment them, and make them horizontal (rotate them)
Fit points in the corrected white circle to the equation in (1). That should give you the parameters of the circle in 3d (radious, angle), if you wrote the equation right.
Now that you have an analytic equation of the actual circle (equation from 1 with parameters from 3), you can map any point from this circle (e.g. its center) to the image location. Remember to uncorrect for the rotations in step 2.
This requires understanding of curve fitting, some geometric analytical maths, and decent code skills. Not trivial, but this will provide a solution that is highly accurate.
For an inaccurate solution:
Find end points of white circles
Make line connecting endpoints
Chose center as mid point of this line.
This will be inaccurate because: choosing end points will have more error than fitting an equation with all points, ignores cone shape of view of the camera, ignores geometry.
But it may be good enough for what you want.
I have been able to extract the midpoint by fitting an ellipse to the arc visible to the camera. The centroid of the ellipse is the required midpoint.
There will be wrong ellipses as well, which can be ignored. The steps to extract the ellipse were:
Extract the markers
Binarise and skeletonise
Fit ellipse to the arc (found a matlab function for this)
Get the centroid of the ellipse
hsv_img=rgb2hsv(im);
bin=new_hsv_img(:,:,3)>marker_th; %was chosen 0.35
%skeletonise
skel=bwskel(bin);
%use regionprops to get the pixelID list
stats=regionprops(skel,'all');
for i=1:numel(stats)
el = fit_ellipse(stats(i).PixelList(:,1),stats(i).PixelList(:,2));
ellipse_draw(el.a, el.b, -el.phi, el.X0_in, el.Y0_in, 'g');
The link for fit_ellipse function
Link for ellipse_draw function
I calculate the rectangular bounding box coordinates for objects in my matlab code ([xmin ymin width height]). But the coordinates might not be precise. Then, I want to draw the box on the image and modify that by dragging the box and/or adjusting the borders. I tried to use imrect and imcrop, but those functions do not draw the draggable/adjustable rectangle on image, based on predefined coordinates. Is there any way to do that?
Thanks in advance for your time.
Take a look at imrect. It works much like the imcrop rectangle but you can set an initial position and get the current position by calling getPosition.
I have this image below. How can I find the original area of the manuscript? I used imfill and was able to find the area within the boundaries, but I need the maximum area of just the manuscript itself within the image
image: Damaged Manuscript
You can calculate the bounding box / bounding rectangle to approximate the original area.
Use regionprops to calculate the property BoundingBox.
ConvexHull might help as well but I guess the resulting area would tend to be to small.
http://de.mathworks.com/help/images/ref/regionprops.html
How to identify boundaries of a binary image to crop in matlab?
ie. the input binary image has no noises. only has one black object in white background.
You can use the edge command in MATLAB.
E = edge(I);
I would be an input grayscale or binary image. This will return a binary image with only the edges.
This can provide further assistance:
http://www.mathworks.com/help/images/ref/edge.html
If your image is just black-and-white and has a single object, you can likely make use of the Flood fill algorithm, for which Matlab has built-in support!
Try the imfill function (ref).
This should give you the extents of the object, which would allow you to crop at will.
You can also invert the image, then do regionprops to extract all of the properties for separate objects. You need to invert the image as regionprops assumes that the objects are white while the background is black. A good thing about this approach is that it generalizes for multiple objects and you only need about a few lines of code to do it.
As an example, let's artificially create a circle in the centre of an image that is black on a white background as you have suggested. Let's assume this is also a binary image.
im = true(200, 200);
[X,Y] = meshgrid(1:200, 1:200);
ind = (X-100).^2 + (Y-100).^2 <= 1000;
im(ind) = false;
imshow(im);
This is what your circle will look like:
Now let's go ahead and invert this so that it's a white circle on black background:
imInvert = ~im;
imshow(imInvert);
This is what your inverted circle will look like:
Now, invoke regionprops to find properties of all of the objects in our image. In this case, there should only be one.
s = regionProps(imInvert, 'BoundingBox');
As such, s contains a structure that is 1 element long, and has a single field called BoundingBox. This field is a 4 element array that is structured in the following way:
[x y w h]
x denotes the column/vertical co-ordinate while y denotes the row/horizontal co-ordinate of the top-left corner of the bounding box. w,h are the width and height of the rectangle. Our output of the above code is:
s =
BoundingBox: [68.5000 68.5000 63 63]
This means that the top-left corner of our bounding box is located at (x,y) = (68.5,68.5), and has a width and height of 63 each. Therefore, the span of our bounding box goes from rows (68.5,131.5) and columns (68.5,131.5). To make sure that we have the right bounding box, you can draw a rectangle around our shape by using the rectangle command.
imshow(im);
rectangle('Position', s.BoundingBox);
This is what your image will look like with a rectangle drawn around the object. As you can see, the bounding box given from regionprops is the minimum spanning bounding box required to fully encapsulate the object.
If you wish to crop the object, you can do the following:
imCrop = imcrop(imInvert, s.BoundingBox);
This should give you the cropped image that is defined by the bounding box that we talked about earlier.
Hope this is what you're looking for. Good luck!
I am using regionprop function in matlab to get MajorAxisLength of an image. I think logically this number should not be greater than sqrt(a^2+b^2) in wich a abd b are the width and heigth of the image. but for my image it is. My black and white image contains a black circle in the center of the image. I think this is strange. Can anybody help me?
Thanks.
If you look at the code of regionprops (subfunction ComputeEllipseParams), you see that they use the second moment to estimate the ellipsoid radius. This works very well for ellipsoid-shaped features, but not very well for features with holes. The second moment increases if you remove pixels from around the centroid (which is, btw, why they make I-beams). Thus, the bigger the 'hole' in the middle of your image, the bigger the apparent ellipsoid radius.
In your case, you may be better off using the extrema property of regionprops, and to calculate the largest radius from there.