I have an image and I want to crop the circular region of interest around a point. I did following in MATLAB:
vessel=imread('vessel.jpg');
imshow( vessel)
t = 0:pi/20:2*pi;
xc=230; % point around which I want to extract/crop image
yc=79;
r=20; %Radium of circular region of interest
xcc = r*cos(t)+xc;
ycc = r*sin(t)+yc;
roimaskcc = poly2mask(double(xcc),double(ycc), size(vessel,1),size(vessel,2));
pr_gccc = find(roimaskcc);
roimean_cc= mean(vessel(pr_gccc));
figure, imshow(roimaskcc)
roimaskcc is correct but when I do the following it gives in nX1 matrix but not the region of interest under mask:
vessel_undermask=vessel(roimaskcc==1);
Can anybody pls. help to extract the circular region of interest around point of interest (xc, yc).
Thanks
What you're doing is extract everything from vessel that's outside the mask. Since there's no such thing as circular matrices, Matlab's solution is to output a vector of all values inside the mask.
While that is technically correct, that vector can be pretty hard to work with. An alternative solution is to leave the data matrix square, and set everything outside the mask to NaN:
% make a copy
vessel_undermask = vessel;
% NaN everything outside the mask (in R, G and B layers)
vessel_undermask(repmat(~roimaskcc,[1,1,3])) = NaN;
imshow(vessel_undermask)
This should give you a matrix that's somewhat simpler to work with.
Note that
vessel_undermask(~isnan(vessel_undermask)) == vessel(roimaskcc)
Yes I got it. I did something like that:
vesseltry=vessel;
vesseltry(~roimaskcc)=0;
vesseltry is now my new image with circular region of interest...
Related
I am trying to detect elevators on floor plans in MATLAB. The code I have now is not detecting elevators, it is instead just pointing at the edges of the image. I am expecting to detect all the elevators on a floor plan. Elevators are represented by a square or rectangle with an x inside, similar to the template image. I have attached the template, image and a result screenshot.
Template image:
Image:
Results:
Code:
template= rgb2gray(imread('ele7.png'));
image = rgb2gray(imread('floorplan.jpg'));
%imshowpair(image,template,'montage')
c = normxcorr2(template,image);% perform cross-correlation
figure, surf(c), shading flat
[ypeak, xpeak] = find(c==max(c(:)));%peak of correlation
%Compute translation from max location in correlation matrix, =padding
yoffSet = ypeak-size(template,1);
xoffSet = xpeak-size(template,2);
%Display matched area
figure
hAx = axes;
imshow(image,'Parent', hAx);
imrect(hAx, [xoffSet+1, yoffSet+1, size(template,2), size(template,1)]);
To check if everything runs smoothly, you should plot the correlation:
figure, surf(c)
As mention by #cris-luengo , it's easy to fail with the sizes of the image and so on. However, I've seen that you followed the tutorial on https://es.mathworks.com/help/images/ref/normxcorr2.html . Since both images, already black and white images (or 2-colour images), normxcorr2 works well with rgb images (with textures and objects, etc...). Thus, I think that is not a correct approach to use normxcorr2.
An approach I would consider is look for branches. Using Matlab's help and bwmorph:
BW = imread('circles.png');
imshow(BW);
BW1 = bwmorph(BW,'skel',Inf);
You first skeletonize the image, then you can use any of the functions that displayed on bwmorph's help (https://es.mathworks.com/help/images/ref/bwmorph.html). In this case, I'd search for branch points, i.e. crosslinks. It is as simple as:
BW2 = bwmorph(BW1,'branchpoints');
branchPointsPixels = find(BW2 == 1);
The indices of the branch points pixels, will be where it finds an X. However, it can be any rotated X (or +, ...). So you'll find more points that you desire, you would need to filter the points in order to get what you want.
I have this BW image:
And using the function RegionProps, it shows that some objetcs are connected:
So I used morphological operations like imerode to separte the objects to get their centroids:
Now I have all the centroids of each object separated, but to that I lost a lot of information when eroding the region, like you can see in picture 3 in comparison with picture 1.
So I was thinking if is there anyway to "dilate" the picture 3 till get closer to picture 1 but without connecting again the objects.
You might want to take a look at bwmorph(). With the 'thicken', inf name-value pair it will thicken the labels until they would overlap. It's a neat tool for segmentation. We can use it to create segmentation borders for the original image.
bw is the original image.
labels is the image of the eroded labels.
lines = bwmorph(labels, 'thicken', inf);
segmented_bw = bw & lines
You could also skip a few phases and achieve similiar results with a marker based watershed. Or even better, as the morphological seesaw has destroyed some information as seen on the poorly segmented cluster on the lower right.
You can assign each white pixel in the mask to the closest centroid and work with the resulting label map:
[y x]= find(bw); % get coordinates of mask pixels
D = pdist2([x(:), y(:)], [cx(:), cy(:)]); % assuming cx, cy are centers' coordinates
[~, lb] = min(D, [], 2); % find index of closest center
lb_map = 0*bw;
lb_map(bw) = lb; % should give you the map.
See pdist2 for more information.
I am doing a road sign recognition program in Matlab and I want to recognize circular roadsigns. Therefore I use the matlab function imfindcircles. I would like to crop only the circular roadsigns and to put them in an isolate figure. However, we have other roadsigns on each figure (triangles or squares) but I don't want them. I have no idea how to do this. Here my code :
[im_bw,map] = imread('roadsign.JPG'); %image black and white
S = regionprops(im_bw,'Extrema','Centroid','BoundingBox');
[centers, radii] = imfindcircles(im_bw,[12 40]);
for k = 1:length(S)
im_cercle = imcrop(im_bw, S(k).BoundingBox);
im_cercle = padarray(im_cercle, [20 20]); % put each roadsigns in a small figure
if radii(k) ~= 0 % Error
figure,imshow(im_cercle); title 'Circle spotted'; % Show every circular roadsigns in a figure
else
figure('visible','off'),imshow(im_cercle); title 'wrong raodsign';
end
end
I tried some other conditions with centres and radii but when I execute the code, i get dimension errors or sometimes it shows me a shape which is not a circle. I also tried to do a variable that only sets when he finds a circle, but without results. Can you help me please ?
Thanks in advance.
What you do you expect the output of regionprops to be? It isn't picking out the circles - it's picking out all the "areas" (anything that is a connected area in im_bw). In addition while imfindcircles can find circles that overlap with other things, regionprops will detect overlapping areas as a single object.
On the other hand, you call imfindcircles and then do nothing with the output.
Rather than doing anything with regionprops, just use the values of centers, radii to define a bounding box around each detected circle (optionally with some additional padding), crop that area out of the image and save/display it.
What we want is to draw several solid circles at random locations, with random gray scale colors, on a dark gray background. How can we do this? Also, if the circles overlap, we need them to change color in the overlapping part.
Since this is an assignment for school, we are not looking for ready-made answers, but for a guide which tools to use in MATLAB!
Here's a checklist of things I would investigate if you want to do this properly:
Figure out how to draw circles in MATLAB. Because you don't have the Image Processing Toolbox (see comments), you will probably have to make a function yourself. I'll give you some starter code:
function [xout, yout] = circle(x,y,r,rows,cols)
[X,Y] = meshgrid(x-r:x+r, y-r:y+r);
ind = find(X.^2 + Y.^2 <= r^2 & X >= 1 & X <= cols & Y >= 1 & Y <= rows);
xout = X(ind);
yout = Y(ind);
end
What the above function does is that it takes in an (x,y) co-ordinate as well as the radius of
the circle. You also will need to specify how many rows and how many columns you want in your image. The reason why is because this function will prevent giving you co-ordinates that are out of bounds in the image that you can't draw. The final output of this will give you co-ordinates of all values inside and along the boundary of the circle. These co-ordinates will already be in integer so there's no need for any rounding and such things. In addition, these will perfectly fit when you're assigning these co-ordinates to locations in your image. One caveat to note is that the co-ordinates assume an inverted Cartesian. This means that the top left corner is the origin (0,0). x values increase from left to right, and y values increase from top to bottom. You'll need to keep this convention in mind when drawing circles in your image.
Take a look at the rand class of functions. rand will generate random values for you and so you can use these to generate a random set of co-ordinates - each of these co-ordinates can thus serve as your centre. In addition, you can use this class of functions to help you figure out how big you want your circles and also what shade of gray you want your circles to be.
Take a look at set operations (logical AND, logical OR) etc. You can use a logical AND to find any circles that are intersecting with each other. When you find these areas, you can fill each of these areas with a different shade of gray. Again, the rand functions will also be of use here.
As such, here is a (possible) algorithm to help you do this:
Take a matrix of whatever size you want, and initialize all of the elements to dark gray. Perhaps an intensity of 32 may work.
Generate a random set of (x,y) co-ordinates, a random set of radii and a random set of intensity values for each circle.
For each pair of circles, check to see if there are any co-ordinates that intersect with each other. If there are such co-ordinates, generate a random shade of gray and fill in these co-ordinates with this new shade of gray. A possible way to do this would be to take each set of co-ordinates of the two circles and draw them on separate temporary images. You would then use the logical AND operator to find where the circles intersect.
Now that you have your circles, you can plot them all. Take a look at how plot works with plotting matrices. That way you don't have to loop through all of the circles as it'll be inefficient.
Good luck!
Let's get you home, shall we? Now this stays away from the Image Processing Toolbox functions, so hopefully these must work for you too.
Code
%%// Paramters
numc = 5;
graph_size = [300 300];
max_r = 100;
r_arr = randperm(max_r/2,numc)+max_r/2
cpts = [randperm(graph_size(1)-max_r,numc)' randperm(graph_size(2)-max_r,numc)']
color1 = randperm(155,numc)+100
prev = zeros(graph_size(1),graph_size(2));
for k = 1:numc
r = r_arr(k);
curr = zeros(graph_size(1),graph_size(2));
curr(cpts(k,1):cpts(k,1)+r-1,cpts(k,2):cpts(k,2)+r-1)= color1(k)*imcircle(r);
common_blob = prev & curr;
curr = prev + curr;
curr(common_blob) = min(color1(1),color1(2))-50;
prev = curr;
end
figure,imagesc(curr), colormap gray
%// Please note that the code uses a MATLAB file-exchange tool called
%// imcircle, which is available at -
%// http://www.mathworks.com/matlabcentral/fileexchange/128-imcircle
Screenshot of a sample run
As you said that your problem is an assignment for school I will therefore not tell you exactly how to do it but what you should look at.
you should be familiar how 2d arrays (matrices) work and how to plot them using image/imagesc/imshow ;
you should look at the strel function ;
you should look at the rand/randn function;
such concepts should be enough for the assignment.
I have a set of points which I want to propagate on to the edge of shape boundary defined by a binary image. The shape boundary is defined by a 1px wide white edge.
I have the coordinates of these points stored in a 2 row by n column matrix. The shape forms a concave boundary with no holes within itself made of around 2500 points. I have approximately 80 to 150 points that I wish to propagate on the shape boundary.
I want to cast a ray from each point from the set of points in an orthogonal direction and detect at which point it intersects the shape boundary at. The orthogonal direction has already been determined. For the required purposes it is calculated taking the normal of the contour calculated for point, using point-1 and point+1.
What would be the best method to do this?
Are there some sort of ray tracing algorithms that could be used?
Thank you very much in advance for any help!
EDIT: I have tried to make the question much clearer and added a image describing the problem. In the image the grey line represents the shape contour, the red dots the points
I want to propagate and the green line an imaginary orthongally cast ray.
alt text http://img504.imageshack.us/img504/3107/orth.png
ANOTHER EDIT: For clarification I have posted the code used to calculate the normals for each point. Where the xt and yt are vectors storing the coordinates for each point. After calculating the normal value it can be propagated by using the linspace function and the requested length of the orthogonal line.
%#derivaties of contour
dx=[xt(2)-xt(1) (xt(3:end)-xt(1:end-2))/2 xt(end)-xt(end-1)];
dy=[yt(2)-yt(1) (yt(3:end)-yt(1:end-2))/2 yt(end)-yt(end-1)];
%#normals of contourpoints
l=sqrt(dx.^2+dy.^2);
nx = -dy./l;
ny = dx./l;
normals = [nx,ny];
It depends on how many unit vectors you want to test against one shape. If you have one shape and many tests, the easiest thing to do is probably to convert your shape coordinates to polar coordinates which implicitly represent your solution already. This may not be a very effective solution however if you have different shapes and only a few tests for every shape.
Update based on the edited question:
If the rays can start from arbitrary points, not only from the origin, you have to test against all the points. This can be done easily by transforming your shape boundary such that your ray to test starts in the origin in either coordinate direction (positive x in my example code)
% vector of shape boundary points (assumed to be image coordinates, i.e. integers)
shapeBoundary = [xs, ys];
% define the start point and direction you want to test
startPoint = [xsp, ysp];
testVector = unit([xv, yv]);
% now transform the shape boundary
shapeBoundaryTrans(:,1) = shapeBoundary(:,1)-startPoint(1);
shapeBoundaryTrans(:,2) = shapeBoundary(:,2)-startPoint(2);
rotMatrix = [testVector(2), testVector(1); ...
testVector(-1), testVector(2)];
% somewhat strange transformation to keep it vectorized
shapeBoundaryTrans = shapeBoundaryTrans * rotMatrix';
% now the test is easy: find the points close to the positive x-axis
selector = (abs(shapeBoundaryTrans(:,2)) < 0.5) & (shapeBoundaryTrans(:,1) > 0);
shapeBoundaryTrans(:,2) = 1:size(shapeBoundaryTrans, 1)';
shapeBoundaryReduced = shapeBoundaryTrans(selector, :);
if (isempty(shapeBoundaryReduced))
[dummy, idx] = min(shapeBoundaryReduced(:, 1));
collIdx = shapeBoundaryReduced(idx, 2);
% you have a collision with point collIdx of your shapeBoundary
else
% no collision
end
This could be done in a nicer way probably, but you get the idea...
If I understand your problem correctly (project each point onto the closest point of the shape boundary), you can
use sub2ind to convert the "2 row by n column matrix" description to a BW image with white pixels, something like
myimage=zeros(imagesize);
myimage(imagesize, x_coords, y_coords) = 1
use imfill to fill the outside of the boundary
run [D,L] = bwdist(BW) on the resulting image, and just read the answers from L.
Should be fairly straightforward.