I want to extract centreline pixels in vessel. At first I have seleted a seed point close to a vessel edge using ginput(1) command. This provides the starting point and specifies the region of interest (ROI) on a vessel segment where the analysis needs to be performed.
figure; imshow(Igreen_eq); % Main green channel Image
p = ginput(1);
Then the selected seed point is served as centre of a circle with diameter less than the expected diameter of the vessel, in order for the circle not to intersect with the opposite edge.
t = 0:pi/20:2*pi;
d = 0.8*15; %d=80% of minwidthOfVessel so that it wont intesect with opposite edge;
R0=d/2;%radius
xi = R0*cos(t)+p(1);
yi = R0*sin(t)+p(2);
line(xi,yi,'LineWidth',2,'Color',[0 1 0]);
roimask = poly2mask(double(xi), double(yi), size(Igreen_eq,1), size(Igreen_eq,2));
figure; imshow(roimask) % Binary image of region selected
Itry = Igreen_eq;
Itry(~roimask ) = 0;
imshow(Itry);
Itry = im2double(Itry);
line(xi, yi,'LineWidth', 2, 'Color', [0 1 0]);
hold on; plot(p(1), p(2),'*r')
Problem:
Hessian matrix is to be computed for the light intensity on the circumference of this circle and the eigenvectors has to be obtained.
I have calculated Dxx,Dyy,Dxy using:
[Dxx,Dxy,Dyy] = Hessian2D(Itry,2); %(sigma=2)
I need to write a code in MATLAB for following problem"
For a point inside the vessel, the eigenvectors corresponding to the largest
eigenvalues are normal to the edges and those corresponding to the smallest eigenvalues point to the direction along the vessels.
The first two consecutive vectors on the circle with maximum change in direction are considered as the pixels reflecting the vessel boundaries. The points on the tracking direction are considered as the centers for the subsequent circles. Repetition of this process gives an estimate of the vessel boundary.
How will I calculate largest eigen values and its correspoinding eigen vector of Hessian matrix to select new seed point as discussed above.
Thanks for your reply . I have used eig2image.m to find the eigen vectors at each point on the image (in my image, there is grey values on the concentric circular region and background is black ).
[Lambda1,Lambda2,Ix,Iy]=eig2image(Dxx,Dxy,Dyy)
where Ix and Iy are largest eigen vectors.
But when I try to plot eigen vectors using :
quiver(Ix, Iy)
I can also see the vectors on the black background which should be zero !!
Can you please reply how can I plot eigen vector on the top of the image.
Assuming Dxx, Dyy, Dxy are matrices of second-order partial derivatives of dimensions size(Itry) then for a given point (m,n) in Itry you can do:
H = [Dxx(m,n) Dxy(m,n); Dxy(m,n) Dyy(m,n)];
[V,D] = eig(H); % check by H*V = V*D;
eigenVal1 = D(1,1);
eigenVal2 = D(2,2);
eigenVec1 = V(1,:);
eigenVec2 = V(2,:);
This local eigen-decomposition will give you eigenvalues (and corresponding eigenvectors) which you can sort according to magnitude. You can loop across image points or for a more compact solution see eig2image.m in FileExchange.
Related
I have three sections (top, mid, bot) of grayscale images (3D). In each section, I have a point with coordinates (x,y) and intensity values [0-255]. The distance between each section is 20 pixels.
I created an illustration to show how those images were generated using a microscope:
Illustration
Illustration (side view): red line is the object of interest. Blue stars represents the dots which are visible in top, mid, bot section. The (x,y) coordinates of these dots are known. The length of the object remains the same but it can rotate in space - 'out of focus' (illustration shows a rotating line at time point 5). At time point 1, the red line is resting (in 2D image: 2 dots with a distance equal to the length of the object).
I want to estimate the x,y,z-coordinate of the end points (represents as stars) by using the changes in intensity, the knowledge about the length of the object and the information in the sections I have. Any help would be appreciated.
Here is an example of images:
Bot section
Mid section
Top section
My 3D PSF data:
https://drive.google.com/file/d/1qoyhWtLDD2fUy2zThYUgkYM3vMXxNh64/view?usp=sharing
Attempt so far:
enter image description here
I guess the correct approach would be to record three images with slightly different z-coordinates for your bot and your top frame, then do a 3D-deconvolution (using Richardson-Lucy or whatever algorithm).
However, a more simple approach would be as I have outlined in my comment. If you use the data for a publication, I strongly recommend to emphasize that this is just an estimation and to include the steps how you have done it.
I'd suggest the following procedure:
Since I do not have your PSF-data, I fake some by estimating the PSF as a 3D-Gaussiamn. Of course, this is a strong simplification, but you should be able to get the idea behind it.
First, fit a Gaussian to the PSF along z:
[xg, yg, zg] = meshgrid(-32:32, -32:32, -32:32);
rg = sqrt(xg.^2+yg.^2);
psf = exp(-(rg/8).^2) .* exp(-(zg/16).^2);
% add some noise to make it a bit more realistic
psf = psf + randn(size(psf)) * 0.05;
% view psf:
%
subplot(1,3,1);
s = slice(xg,yg,zg, psf, 0,0,[]);
title('faked PSF');
for i=1:2
s(i).EdgeColor = 'none';
end
% data along z through PSF's center
z = reshape(psf(33,33,:),[65,1]);
subplot(1,3,2);
plot(-32:32, z);
title('PSF along z');
% Fit the data
% Generate a function for a gaussian distibution plus some background
gauss_d = #(x0, sigma, bg, x)exp(-1*((x-x0)/(sigma)).^2)+bg;
ft = fit ((-32:32)', z, gauss_d, ...
'Start', [0 16 0] ... % You may find proper start points by looking at your data
);
subplot(1,3,3);
plot(-32:32, z, '.');
hold on;
plot(-32:.1:32, feval(ft, -32:.1:32), 'r-');
title('fit to z-profile');
The function that relates the intensity I to the z-coordinate is
gauss_d = #(x0, sigma, bg, x)exp(-1*((x-x0)/(sigma)).^2)+bg;
You can re-arrange this formula for x. Due to the square root, there are two possibilities:
% now make a function that returns the z-coordinate from the intensity
% value:
zfromI = #(I)ft.sigma * sqrt(-1*log(I-ft.bg))+ft.x0;
zfromI2= #(I)ft.sigma * -sqrt(-1*log(I-ft.bg))+ft.x0;
Note that the PSF I have faked is normalized to have one as its maximum value. If your PSF data is not normalized, you can divide the data by its maximum.
Now, you can use zfromI or zfromI2 to get the z-coordinate for your intensity. Again, I should be normalized, that is the fraction of the intensity to the intensity of your reference spot:
zfromI(.7)
ans =
9.5469
>> zfromI2(.7)
ans =
-9.4644
Note that due to the random noise I have added, your results might look slightly different.
I am working on problem related to camera calibration. In the below image, we consider a world coordinate system with X-axis going leftward, Y-axis rightward and Z-axis upward. We select 15 points(x,y,z) distributed uniformly across the 3 planes. The distance between grid lines is 1 inch. We also obtain MATLAB coordinates for the 15 pixels(u,v). The objective is to obtain the 3x4 camera matrix (M) using homogeneous linear least squares and then project the world points (x,y,z) to the image (u',v') using M. I have written code to do this but the coordinates I'm obtaining (u',v') seem to be very small in magnitude compared to the actual coordinates (u,v). The RMS error is too large and the projected points don't even map onto the image anywhere near the actual points. Is there any scaling that I need to do to convert it to MATLAB coordinates? I am also including my code which isn't very well written since I am relatively new to MATLAB.
P=[];% 2nx12 matrix - 30x12 matrix
for i=1:15 %compute P
world_row = world_coords(i,:); % 3d homogeneous coordinates (x,y,z,1)
zeroelem = repelem(0,4);
image_coord = image_coords(i,:);
img_u = image_coord(1);
prod = -img_u*world_row;
row1 = [world_row,zeroelem,prod];
zeroelem = repelem(0,3);
img_v = image_coord(2);
prod = -img_v*world_row;
row2 = [0,world_row,zeroelem,prod];
P=[P;row1;row2];
end
var1 = P'*P;
[V,D] = eig(var1');//compute eigen vector corresponding to least eigen value
m = V(:,1); //unit vector of norm 1
M = reshape(m,3,4); //camera matrix of 3x4 size
%get projected points
proj = M*world_coords';
U = proj (1,:);
V = proj (2,:);
W = proj (3,:);
for i=1:15
U(i) = U(i)/W(i);
V(i) = V(i)/W(i);
end
final = [U;V];//(u',v')
I am also including the image with the 15 points I have selected. Take P1(u,v) = (286,260) and P1(x,y,z) = (4,0,3). The (u',v') I obtained for this has low values. Can anyone point me what I'm doing wrong?
It was a silly error from my me that was giving me the wrong camera matrix. I noted down the world coordinates of the point P wrongly ((7,0,1) instead of (1,0,1)). This led to wrongly formed 30x12 matrix which we use to form an equation to be solved by homogeneous linear least squares. I have obtained the calibration matrix which projects the 3D points with a low RMS error after correcting this mistake.
I have a list of 2D points (x1,y1),(x2,y2)......(Xn,Yn) representing a curved segment, is there any formula to determine whether the direction of drawing that segment is clockwise or anti clockwise ?
any help is appreciated
Alternately, you can use a bit of linear algebra. If you have three points a, b, and c, in that order, then do the following:
1) create the vectors u = (b-a) = (b.x-a.x,b.y-a.y) and v = (c-b) ...
2) calculate the cross product uxv = u.x*v.y-u.y*v.x
3) if uxv is -ve then a-b-c is curving in clockwise direction (and vice-versa).
by following a longer curve along in the same manner, you can even detect when as 's'-shaped curve changes from clockwise to anticlockwise, if that is useful.
One possible approach. It should work reasonably well if the sampling of the line represented by your list of points is uniform and smooth enough, and if the line is sufficiently simple.
Subtract the mean to "center" the line.
Convert to polar coordinates to get the angle.
Unwrap the angle, to make sure its increments are meaningful.
Check if total increment is possitive or negative.
I'm assuming you have the data in x and y vectors.
theta = cart2pol(x-mean(x), y-mean(y)); %// steps 1 and 2
theta = unwrap(theta); %// step 3
clockwise = theta(end)<theta(1); %// step 4. Gives 1 if CW, 0 if ACW
This only considers the integrated effect of all points. It doesn't tell you if there are "kinks" or sections with different directions of turn along the way.
A possible improvement would be to replace the average of x and y by some kind of integral. The reason is: if sampling is denser in a region the average will be biased towards that, whereas the integral wouldn't.
Now this is my approach, as mentioned in a comment to the question -
Another approach: draw a line from starting point to ending point. This line is indeed a vector. A CW curve has most of its part on RHS of this line. For CCW, left.
I wrote a sample code to elaborate this idea. Most of the explanation can be found in comments in the code.
clear;clc;close all
%% draw a spiral curve
N = 30;
theta = linspace(0,pi/2,N); % a CCW curve
rho = linspace(1,.5,N);
[x,y] = pol2cart(theta,rho);
clearvars theta rho N
plot(x,y);
hold on
%% find "the vector"
vec(:,:,1) = [x(1), y(1); x(end), y(end)]; % "the vector"
scatter(x(1),y(1), 200,'s','r','fill') % square is the starting point
scatter(x(end),y(end), 200,'^','r','fill') % triangle is the ending point
line(vec(:,1,1), vec(:,2,1), 'LineStyle', '-', 'Color', 'r')
%% find center of mass
com = [mean(x), mean(y)]; % center of mass
vec(:,:,2) = [x(1), y(1); com]; % secondary vector (start -> com)
scatter(com(1), com(2), 200,'d','k','fill') % diamond is the com
line(vec(:,1,2), vec(:,2,2), 'LineStyle', '-', 'Color', 'k')
%% find rotation angle
dif = diff(vec,1,1);
[ang, ~] = cart2pol(reshape(dif(1,1,:),1,[]), reshape(dif(1,2,:),1,[]));
clearvars dif
% now you can tell the answer by the rotation angle
if ( diff(ang)>0 )
disp('CW!')
else
disp('CCW!')
end
One can always tell on which side of the directed line (the vector) a point is, by comparing two vectors, namely, rotating vector [starting point -> center of mass] to the vector [starting point -> ending point], and then comparing the rotation angle to 0. A few seconds of mind-animating can help understand.
this is my situation: I have a 30x30 image and I want to calculate the radial and tangent component of the gradient of each point (pixel) along the straight line passing through the centre of the image (15,15) and the same (i,j) point.
[dx, dy] = gradient(img);
for i=1:30
for j=1:30
pt = [dx(i, j), dy(i,j)];
line = [i-15, j-15];
costh = dot(line, pt)/(norm(line)*norm(pt));
par(i,j) = norm(costh*line);
tang(i,j) = norm(sin(acos(costh))*line);
end
end
is this code correct?
I think there is a conceptual error in your code, I tried to get your results with a different approach, see how it compares to yours.
[dy, dx] = gradient(img);
I inverted x and y because the usual convention in matlab is to have the first dimension along the rows of a matrix while gradient does the opposite.
I created an array of the same size as img but with each pixel containing the angle of the vector from the center of the image to this point:
[I,J] = ind2sub(size(img), 1:numel(img));
theta=reshape(atan2d(I-ceil(size(img,1)/2), J-ceil(size(img,2)/2)), size(img))+180;
The function atan2d ensures that the 4 quadrants give distinct angle values.
Now the projection of the x and y components can be obtained with trigonometry:
par=dx.*sind(theta)+dy.*cosd(theta);
tang=dx.*cosd(theta)+dy.*sind(theta);
Note the use of the .* to achieve point-by-point multiplication, this is a big advantage of Matlab's matrix computations which saves you a loop.
Here's an example with a well-defined input image (no gradient along the rows and a constant gradient along the columns):
img=repmat(1:30, [30 1]);
The results:
subplot(1,2,1)
imagesc(par)
subplot(1,2,2)
imagesc(tang)
colorbar
I have a .txt file with about 100,000 points in the 2-D plane. When I plot the points, there is a clearly defined 2-D region (think of a 2-D disc that has been morphed a bit).
What is the easiest way to compute the area of this region? Any way of doing easily in Matlab?
I made a polygonal approximation by finding a bunch (like 40) points on the boundary of the region and computing the area of the polygonal region in Matlab, but I was wondering if there is another, less tedious method than finding 40 points on the boundary.
Consider this example:
%# random points
x = randn(300,1);
y = randn(300,1);
%# convex hull
dt = DelaunayTri(x,y);
k = convexHull(dt);
%# area of convex hull
ar = polyarea(dt.X(k,1),dt.X(k,2))
%# plot
plot(dt.X(:,1), dt.X(:,2), '.'), hold on
fill(dt.X(k,1),dt.X(k,2), 'r', 'facealpha', 0.2);
hold off
title( sprintf('area = %g',ar) )
There is a short screencast By Doug Hull which solves this exact problem.
EDIT:
I am posting a second answer inspired by the solution proposed by #Jean-FrançoisCorbett.
First I create random data, and using the interactive brush tool, I remove some points to make it look like the desired "kidney" shape...
To have a baseline to compare against, we can manually trace the enclosing region using the IMFREEHAND function (I'm doing this using my laptop's touchpad, so not the most accurate drawing!). Then we find the area of this polygon using POLYAREA. Just like my previous answer, I compute the convex hull as well:
Now, and based on a previous SO question I had answered (2D histogram), the idea is to lay a grid over the data. The choice of the grid resolution is very important, mine was numBins = [20 30]; for the data used.
Next we count the number of squares containing enough points (I used at least 1 point as threshold, but you could try a higher value). Finally we multiply this count by the area of one grid square to obtain the approximated total area.
%### DATA ###
%# some random data
X = randn(100000,1)*1;
Y = randn(100000,1)*2;
%# HACK: remove some point to make data look like a kidney
idx = (X<-1 & -4<Y & Y<4 ); X(idx) = []; Y(idx) = [];
%# or use the brush tool
%#brush on
%### imfreehand ###
figure
line('XData',X, 'YData',Y, 'LineStyle','none', ...
'Color','b', 'Marker','.', 'MarkerSize',1);
daspect([1 1 1])
hROI = imfreehand('Closed',true);
pos = getPosition(hROI); %# pos = wait(hROI);
delete(hROI)
%# total area
ar1 = polyarea(pos(:,1), pos(:,2));
%# plot
hold on, plot(pos(:,1), pos(:,2), 'Color','m', 'LineWidth',2)
title('Freehand')
%### 2D histogram ###
%# center of bins
numBins = [20 30];
xbins = linspace(min(X), max(X), numBins(1));
ybins = linspace(min(Y), max(Y), numBins(2));
%# map X/Y values to bin-indices
Xi = round( interp1(xbins, 1:numBins(1), X, 'linear', 'extrap') );
Yi = round( interp1(ybins, 1:numBins(2), Y, 'linear', 'extrap') );
%# limit indices to the range [1,numBins]
Xi = max( min(Xi,numBins(1)), 1);
Yi = max( min(Yi,numBins(2)), 1);
%# count number of elements in each bin
H = accumarray([Yi(:), Xi(:)], 1, [numBins(2) numBins(1)]);
%# total area
THRESH = 0;
sqNum = sum(H(:)>THRESH);
sqArea = (xbins(2)-xbins(1)) * (ybins(2)-ybins(1));
ar2 = sqNum*sqArea;
%# plot 2D histogram/thresholded_histogram
figure, imagesc(xbins, ybins, H)
axis on, axis image, colormap hot; colorbar; %#caxis([0 500])
title( sprintf('2D Histogram, bins=[%d %d]',numBins) )
figure, imagesc(xbins, ybins, H>THRESH)
axis on, axis image, colormap gray
title( sprintf('H > %d',THRESH) )
%### convex hull ###
dt = DelaunayTri(X,Y);
k = convexHull(dt);
%# total area
ar3 = polyarea(dt.X(k,1), dt.X(k,2));
%# plot
figure, plot(X, Y, 'b.', 'MarkerSize',1), daspect([1 1 1])
hold on, fill(dt.X(k,1),dt.X(k,2), 'r', 'facealpha',0.2); hold off
title('Convex Hull')
%### plot ###
figure, hold on
%# plot histogram
imagesc(xbins, ybins, H>=1)
axis on, axis image, colormap gray
%# plot grid lines
xoff = diff(xbins(1:2))/2; yoff = diff(ybins(1:2))/2;
xv1 = repmat(xbins+xoff,[2 1]); xv1(end+1,:) = NaN;
yv1 = repmat([ybins(1)-yoff;ybins(end)+yoff;NaN],[1 size(xv1,2)]);
yv2 = repmat(ybins+yoff,[2 1]); yv2(end+1,:) = NaN;
xv2 = repmat([xbins(1)-xoff;xbins(end)+xoff;NaN],[1 size(yv2,2)]);
xgrid = [xv1(:);NaN;xv2(:)]; ygrid = [yv1(:);NaN;yv2(:)];
line(xgrid, ygrid, 'Color',[0.8 0.8 0.8], 'HandleVisibility','off')
%# plot points
h(1) = line('XData',X, 'YData',Y, 'LineStyle','none', ...
'Color','b', 'Marker','.', 'MarkerSize',1);
%# plot convex hull
h(2) = patch('XData',dt.X(k,1), 'YData',dt.X(k,2), ...
'LineWidth',2, 'LineStyle','-', ...
'EdgeColor','r', 'FaceColor','r', 'FaceAlpha',0.5);
%# plot freehand polygon
h(3) = plot(pos(:,1), pos(:,2), 'g-', 'LineWidth',2);
%# compare results
title(sprintf('area_{freehand} = %g, area_{grid} = %g, area_{convex} = %g', ...
ar1,ar2,ar3))
legend(h, {'Points' 'Convex Jull','FreeHand'})
hold off
Here is the final result of all three methods overlayed, with the area approximations displayed:
My answer is the simplest and perhaps the least elegant and precise. But first, a comment on previous answers:
Since your shape is usually kidney-shaped (not convex), calculating the area of its convex hull won't do, and an alternative is to determine its concave hull (see e.g. http://www.concavehull.com/home.php?main_menu=1) and calculate the area of that. But determining a concave hull is far more difficult than a convex hull. Plus, straggler points will cause trouble in both he convex and concave hull.
Delaunay triangulation followed by pruning, as suggested in #Ed Staub's answer, may a bit be more straightforward.
My own suggestion is this: How precise does your surface area calculation have to be? My guess is, not very. With either concave hull or pruned Delaunay triangulation, you'll have to make an arbitrary choice anyway as to where the "boundary" of your shape is (the edge isn't knife-sharp, and I see there are some straggler points sprinkled around it).
Therefore a simpler algorithm may be just as good for your application.
Divide your image in an orthogonal grid. Loop through all grid "pixels" or squares; if a given square contains at least one point (or perhaps two points?), mark the square as full, else empty. Finally, add the area of all full squares. Bingo.
The only parameter is the resolution length (size of the squares). Its value should be set to something similar to the pruning length in the case of Delaunay triangulation, i.e. "points within my shape are closer to each other than this length, and points further apart than this length should be ignored".
Perhaps an additional parameter is the number of points threshold for a square to be considered full. Maybe 2 would be good to ignore straggler points, but that may define the main shape a bit too tightly for your taste... Try both 1 and 2, and perhaps take an average of both. Or, use 1 and prune away the squares that have no neighbours (game-of-life-style). Simlarly, empty squares whose 8 neighbours are full should be considered full, to avoid holes in the middle of the shape.
There is no end to how much this algorithm can be refined, but due to the arbitrariness intrinsic to the problem definition in your particular application, any refinement is probably the algorithm equivalent of "polishing a turd".
I know next to nothing, so don't put much stock in this... consider doing a Delaunay triangulation. Then remove any hull (outer) edges longer than some maximum. Repeat until nothing to remove. Fill the remaining triangles.
This will orphan some outlier points.
I suggest using a space-filling-curve, for example a z-curve or better a moore curve. A sfc fills the full space and is good to index each points. For example for all f(x)=y you can sort the points of the curve in ascendending order and from that result you take as many points until you get a full roundtrip. These points you can then use to compute the area. Because you have many points maybe you want to use less points and use a cluster which make the result less accurate.
I think you can get the border points using convex hull algorithm with restriction to the edge length (you should sort points by vertical axis). Thus it will follow nonconvexity of your region. I propose length round 0.02. In any case you can experiment a bit with different lengths drawing the result and examining it visually.