for my project i have to segment the abnormalities in a CT brain image.
I want to do that by comparing the right side of the brain with the left side. This could be done
by using the intensity difference of the image. For example, blood is brighter than the brain tissue in
CT images. Due to the fact that the right and left side of the brain are nearly symmetric, it is possible
to find a abnormality in one side by comparing that with the other side. Using Matlab, I want to work
with the Dicom files of the CT images. I want to segment the abnormal area by comparing both sides of the brain.
After segmenting the abnormalities in 2D, I want to register the 2D images and create a 3D reconstruction.
Does anyone perhaps know, what the best coding method is (in Matlab) for comparing the left and right side of a Dicom image?
Maybe take a look at this paper: http://www.sciencedirect.com/science/article/pii/S0167865503000497
It explains how to find the symmetry plane in 3D MRI images, but the method should also work on CT. First you search for the center of mass in your image. Next you calculate the axes of the ellipsoid of inertia and evaluate the symmetry. Finally, you can improve the symmetry plane using the downhill simplex method.
Hope this helps!
EDIT: here is how I would tackle this problem:
search for the center of mass R
ind = find(ones(size(image)));
ind = reshape(ind, size(image,1), size(image,2), size(image,3)); %for a 3D volume
[x,y,z] = ind2sub(size(image), ind); %for a 3D volume
Rx = image.*x;
Ry = image.*y;
Rz = image.*z;
Rx = round(1/sum(image(:)) * sum(Rx(:)));
Ry = round(1/sum(image(:)) * sum(Ry(:)));
Rz = round(1/sum(image(:)) * sum(Rz(:)));
Rx, Ryand Rznow contain the position of the center of mass in your image. The code is easily adaptable to 2D.
Now, look for the axes of the ellipsoid of inertia:
for p=0:2
for q=0:2
for r=0:2
if p+q+r==2
integr = image.*(x-Rx).^p.*(y-Ry).^q.*(z-Rz).^r;
m = sum(integr(:));
if p==2, xx=1; yy=1;
elseif p>0 && q>0, xx=1; yy=2;
elseif p>0 && r>0, xx=1; yy=3;
elseif q==2, xx=2; yy=2;
elseif q>0 && r>0, xx=2; yy=3;
elseif r==2, xx=3; yy=3;
end
M(xx,yy) = m;
M(yy,xx) = m;
end
end
end
end
[V,~] = eig(M);
The matrix V contains the directions of the axes of the ellipsoid of inertia. These are first guesses for the symmetry plane.
Evaluate the symmetry. This is the hard part, because you have to rotate the image around all three (or two, in 2D) possible symmetry planes. I have used the affine3D and imwarp commands, but it's quite cumbersome. Make sure to define the different axes through the center of mass found before. A possible measure of symmetry is mu = 1 - ||image - mirrored_ image||^2 / (2*||image||^2). The axis with the highest mu value is the best symmetry plane.
If you are not happy about the symmetry axis, you can improve it using the downhill simplex method, see e.g. https://en.wikipedia.org/wiki/Nelder%E2%80%93Mead_method
Now you have you original image and the mirrored image around the midsagittal plane. Subtracting both should give you an idea of abnormalities.
I hope this is clear. For more information, please check the excellent paper by Tuzikov et al. mentioned above.
Related
I want to detect edges (with sub-pixel accuracy) in images like the one displayed:
The resolution would be around 600 X 1000.
I came across a comment by Mark Ransom here, which mentions about edge detection algorithms for vertical edges. I haven't come across any yet. Will it be useful in my case (since the edge isn't strictly a straight line)? It will always be a vertical edge though. I want it to be accurate till 1/100th of a pixel at least. I also want to have access to these sub-pixel co-ordinate values.
I have tried "Accurate subpixel edge location" by Agustin Trujillo-Pino. But this does not give me a continuous edge.
Are there any other algorithms available? I will be using MATLAB for this.
I have attached another similar image which the algorithm has to work on:
Any inputs will be appreciated.
Thank you.
Edit:
I was wondering if I could do this:
Apply Canny / Sobel in MATLAB and get the edges of this image (note that it won't be a continuous line). Then, somehow interpolate this Sobel edges and get the co-ordinates in subpixel. Is it possible?
A simple approach would be to project your image vertically and fit the projected profile with an appropriate function.
Here is a try, with an atan shape:
% Load image
Img = double(imread('bQsu5.png'));
% Project
x = 1:size(Img,2);
y = mean(Img,1);
% Fit
f = fit(x', y', 'a+b*atan((x0-x)/w)', 'Startpoint', [150 50 10 150])
% Display
figure
hold on
plot(x, y);
plot(f);
legend('Projected profile', 'atan fit');
And the result:
I get x_0 = 149.6 pix for your first image.
However, I doubt you will be able to achieve a subpixel accuracy of 1/100th of pixel with those images, for several reasons:
As you can see on the profile, your whites are saturated (grey levels at 255). As you cut the real atan profile, the fit is biased. If you have control over the experiments, I suggest you do it again again with a smaller exposure time for instance.
There are not so many points on the transition, so there is not so many information on where the transition is. Typically, your resolution will be the square root of the width of the atan (or whatever shape you prefer). In you case this limits the subpixel resolution at 1/5th of a pixel, at best.
Finally, your edges are not stricly vertical, they are slightly titled. If you choose to use this projection method, to increase the accuracy you should look for a way to correct this tilt before projecting. This won't increase your accuracy by several orders of magnitude, though.
Best,
There is a problem with your image. At pixel level, it seems like there are four interlaced subimages (odd and even rows and columns). Look at this zoomed area close to the edge.
In order to avoid this artifact, I just have taken the even rows and columns of your image, and compute subpixel edges. And finally, I look for the best fitting straight line, using the function clsq whose code is in this page:
%load image
url='http://i.stack.imgur.com/bQsu5.png';
image = imread(url);
imageEvenEven = image(1:2:end,1:2:end);
imshow(imageEvenEven, 'InitialMagnification', 'fit');
% subpixel detection
threshold = 25;
edges = subpixelEdges(imageEvenEven, threshold);
visEdges(edges);
% compute fit line
A = [ones(size(edges.x)) edges.x edges.y];
[c n] = clsq(A,2);
y = [1,200];
x = -(n(2)*y+c) / n(1);
hold on;
plot(x,y,'g');
When executing this code, you can see the green line that best aproximate all the edge points. The line is given by the equation c + n(1)*x + n(2)*y = 0
Take into account that this image has been scaled by 1/2 when taking only even rows and columns, so the right coordinates must be scaled.
Besides, you can try with the other tree subimages (imageEvenOdd, imageOddEven and imageOddOdd) and combine the four straigh lines to obtain the best solution.
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 want to get a metric of straightness of contour in my binary image (relatively faster). The image looks as follows:
Now, the contours in the red box are the ones which I would like to be removed preferably. Since they are not straight. These are the things I have tried. I am as of now implementing in MATLAB.
1.Collect row and column coordinates of each contour and then take derivative. For straight objects (such as rectangle), derivative will be mostly low with a few spikes (along the corners of the rectangle).
Problem: The coordinates collected are not in order i.e. the order in which the contour will be traversed if we imaging it as a path. Therefore, derivative gives absurdly high values sometimes. Also, the contour is not absolutely straight, its an output of edge detection algorithm, so you can imagine that there might be some discontinuity (see the rectangle at the bottom, human eye can understand that it is a rectangle though it is not absolutely straight).
2.Tried to think about polyfit, but again this contour issue comes up. Since its a rectangle I don't know how to apply polyfit to that point set.
Also, I would like to remove contours which are distributed vertically/horizontally. Basically this is a lane detection algorithm. So lanes cannot be absolutely vertical/horizontal.
Any ideas?
You should look into the features of regionprops more. To be fair I stole the script from this answer, but here it is:
BW = imread('lanes.png');
BW = im2bw(BW);
figure(1),
subplot(1,2,1);
imshow(BW);
cc = bwconncomp(BW);
l = labelmatrix(cc);
a_rp = regionprops(CC,'Area','MajorAxisLength','MinorAxislength','Orientation','PixelList','Eccentricity');
idx = ([a_rp.Eccentricity] > 0.99 & [a_rp.Area] > 100 & [a_rp.Orientation] < 70 & [a_rp.Orientation] > -90);
BW2 = ismember(l,find(idx));
subplot(1,2,2);
imshow(BW2);
You can mess around with the properties. 'Orientation', 'Eccentricity', and 'Area' are probably the parameters you want to mess with. I also messed with the ratios of the major/minor axis lengths but eccentricity basically does this (eccentricity is a measure of how "circular" an ellipse is). Here's the output:
I actually saw a good video specifically from matlab for lane detection using regionprops. I'll try to see if I can find it and link it.
You can segment your image using bwlabel, then work separately on each bwlabel connected object, using find. This should help solve your order problem.
About a metric, the only thing that come to mind at the moment is to fit to an ellipse, and set the a/b (major axis/minor axis) ratio (basically eccentricity) a parameter. For example a straight line (even if not perfect) will be fitted to an ellipse with a very big major axis and a very small minor axis. So say you set a ratio threshold of >10 etc... Fitting to an ellipse can be done using this FEX submission for example.
In a project I'm currently working on I have to calculate the 5 moments of the contour of an image. Which I can then for example use to get the centroid. To do this I used matlab :
f = imread(Is);
%Edge detection with prewitt
contourImage = edge(f,'prewitt');
% Morphological operation to close the open spaces
se = strel('disk',2);
closecontourImage = imclose(contourImage,se);
imshow(closecontourImage);
%Find the x y positions of all the nonzero elements of the edge
[row,col] = find(closecontourImage);
% 3 moments
m10= 0;
m00= 0;
m01= 0;
mu00 =0;
% Calculate the 3 moments based on the given paper
for r=1:length(row)
for c=1:length(col)
m10 = m10 + ((row(r)^1)*(col(c)^0));
m00 = m00 + ((row(r)^0)*(col(c)^0));
m01 = m01 + ((row(r)^0)*(col(c)^1));
end
end
% Calculate centroid (zwaartepunt) based on the given formulas
x = m10/m00;
y= m01/m00;
Original image (in png, i use pgm in matlab):
The edge (which I assume is the contour):
The plot with the Image and the centroid
When I compare this to matlabs built in centroid calculation it is pretty close.
Though my problem is concerning the area calculation. I read that the 0th moment = area. though my m00 is not the same as the area. Which is logical because the 0th moment is a summation of all the white pixels ... which only represent the edge of the image, therefore this couldn't result in the area. My question is now , is there a difference in contour moments and moments on the entire image ? And is it possible to get the area based on the contour in this representation ?
In my assignment they explicitly say that the moments of the contour should be calculated and that the 1ste moment is equal to the centroid (which is also not the case in my algorithm). But what I read here is that the 1st order central moment = the centroid. So does this mean that the contour moments are the same as the central moments ? And a more general question, can i use this edge as a contour ?
I find these moments very confusing
There is a difference between moments of filled area or its contour. Think about the following case:
The contour is the exactly the same for both of these objects. Nevertheless, it is obvious that the center of weight in the right square is biased to the right, since it is "more full to the right".
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.