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
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 have the following code in MATLAB:
I=imread(image);
h=fspecial('gaussian',si,sigma);
I=im2double(I);
I=imfilter(I,h,'conv');
figure,imagesc(I),impixelinfo,title('Original Image after Convolving with gaussian'),colormap('gray');
How can I define and apply a Gaussian filter to an image without imfilter, fspecial and conv2?
It's really unfortunate that you can't use the some of the built-in methods from the Image Processing Toolbox to help you do this task. However, we can still do what you're asking, though it will be a bit more difficult. I'm still going to use some functions from the IPT to help us do what you're asking. Also, I'm going to assume that your image is grayscale. I'll leave it to you if you want to do this for colour images.
Create Gaussian Mask
What you can do is create a grid of 2D spatial co-ordinates using meshgrid that is the same size as the Gaussian filter mask you are creating. I'm going to assume that N is odd to make my life easier. This will allow for the spatial co-ordinates to be symmetric all around the mask.
If you recall, the 2D Gaussian can be defined as:
The scaling factor in front of the exponential is primarily concerned with ensuring that the area underneath the Gaussian is 1. We will deal with this normalization in another way, where we generate the Gaussian coefficients without the scaling factor, then simply sum up all of the coefficients in the mask and divide every element by this sum to ensure a unit area.
Assuming that you want to create a N x N filter, and with a given standard deviation sigma, the code would look something like this, with h representing your Gaussian filter.
%// Generate horizontal and vertical co-ordinates, where
%// the origin is in the middle
ind = -floor(N/2) : floor(N/2);
[X Y] = meshgrid(ind, ind);
%// Create Gaussian Mask
h = exp(-(X.^2 + Y.^2) / (2*sigma*sigma));
%// Normalize so that total area (sum of all weights) is 1
h = h / sum(h(:));
If you check this with fspecial, for odd values of N, you'll see that the masks match.
Filter the image
The basics behind filtering an image is for each pixel in your input image, you take a pixel neighbourhood that surrounds this pixel that is the same size as your Gaussian mask. You perform an element-by-element multiplication with this pixel neighbourhood with the Gaussian mask and sum up all of the elements together. The resultant sum is what the output pixel would be at the corresponding spatial location in the output image. I'm going to use the im2col that will take pixel neighbourhoods and turn them into columns. im2col will take each of these columns and create a matrix where each column represents one pixel neighbourhood.
What we can do next is take our Gaussian mask and convert this into a column vector. Next, we would take this column vector, and replicate this for as many columns as we have from the result of im2col to create... let's call this a Gaussian matrix for a lack of a better term. With this Gaussian matrix, we will do an element-by-element multiplication with this matrix and with the output of im2col. Once we do this, we can sum over all of the rows for each column. The best way to do this element-by-element multiplication is through bsxfun, and I'll show you how to use it soon.
The result of this will be your filtered image, but it will be a single vector. You would need to reshape this vector back into matrix form with col2im to get our filtered image. However, a slight problem with this approach is that it doesn't filter pixels where the spatial mask extends beyond the dimensions of the image. As such, you'll actually need to pad the border of your image with zeroes so that we can properly do our filter. We can do this with padarray.
Therefore, our code will look something like this, going with your variables you have defined above:
N = 5; %// Define size of Gaussian mask
sigma = 2; %// Define sigma here
%// Generate Gaussian mask
ind = -floor(N/2) : floor(N/2);
[X Y] = meshgrid(ind, ind);
h = exp(-(X.^2 + Y.^2) / (2*sigma*sigma));
h = h / sum(h(:));
%// Convert filter into a column vector
h = h(:);
%// Filter our image
I = imread(image);
I = im2double(I);
I_pad = padarray(I, [floor(N/2) floor(N/2)]);
C = im2col(I_pad, [N N], 'sliding');
C_filter = sum(bsxfun(#times, C, h), 1);
out = col2im(C_filter, [N N], size(I_pad), 'sliding');
out contains the filtered image after applying a Gaussian filtering mask to your input image I. As an example, let's say N = 9, sigma = 4. Let's also use cameraman.tif that is an image that's part of the MATLAB system path. By using the above parameters, as well as the image, this is the input and output image we get:
Assume I have a 2x2 matrix filled with values which will represent a plane. Now I want to rotate the plane around itself in a 3-D way, in the "z-Direction". For a better understanding, see the following image:
I wondered if this is possible by a simple affine matrix, thus I created the following simple script:
%Create a random value matrix
A = rand*ones(200,200);
%Make a box in the image
A(50:200-50,50:200-50) = 1;
Now I can apply transformations in the 2-D room simply by a rotation matrix like this:
R = affine2d([1 0 0; .5 1 0; 0 0 1])
tform = affine3d(R);
transformed = imwarp(A,tform);
However, this will not produce the desired output above, and I am not quite sure how to create the 2-D affine matrix to create such behavior.
I guess that a 3-D affine matrix can do the trick. However, if I define a 3-D affine matrix I cannot work with the 2-D representation of the matrix anymore, since MATLAB will throw the error:
The number of dimensions of the input image A must be 3 when the
specified geometric transformation is 3-D.
So how can I code the desired output with an affine matrix?
The answer from m3tho correctly addresses how you would apply the transformation you want: using fitgeotrans with a 'projective' transform, thus requiring that you specify 4 control points (i.e. 4 pairs of corresponding points in the input and output image). You can then apply this transform using imwarp.
The issue, then, is how you select these pairs of points to create your desired transformation, which in this case is to create a perspective projection. As shown below, a perspective projection takes into account that a viewing position (i.e. "camera") will have a given view angle defining a conic field of view. The scene is rendered by taking all 3-D points within this cone and projecting them onto the viewing plane, which is the plane located at the camera target which is perpendicular to the line joining the camera and its target.
Let's first assume that your image is lying in the viewing plane and that the corners are described by a normalized reference frame such that they span [-1 1] in each direction. We need to first select the degree of perspective we want by choosing a view angle and then computing the distance between the camera and the viewing plane. A view angle of around 45 degrees can mimic the sense of perspective of normal human sight, so using the corners of the viewing plane to define the edge of the conic field of view, we can compute the camera distance as follows:
camDist = sqrt(2)./tand(viewAngle./2);
Now we can use this to generate a set of control points for the transformation. We first apply a 3-D rotation to the corner points of the viewing plane, rotating around the y axis by an amount theta. This rotates them out of plane, so we now project the corner points back onto the viewing plane by defining a line from the camera through each rotated corner point and finding the point where it intersects the plane. I'm going to spare you the mathematical derivations (you can implement them yourself from the formulas in the above links), but in this case everything simplifies down to the following set of calculations:
term1 = camDist.*cosd(theta);
term2 = camDist-sind(theta);
term3 = camDist+sind(theta);
outP = [-term1./term2 camDist./term2; ...
term1./term3 camDist./term3; ...
term1./term3 -camDist./term3; ...
-term1./term2 -camDist./term2];
And outP now contains your normalized set of control points in the output image. Given an image of size s, we can create a set of input and output control points as follows:
scaledInP = [1 s(1); s(2) s(1); s(2) 1; 1 1];
scaledOutP = bsxfun(#times, outP+1, s([2 1])-1)./2+1;
And you can apply the transformation like so:
tform = fitgeotrans(scaledInP, scaledOutP, 'projective');
outputView = imref2d(s);
newImage = imwarp(oldImage, tform, 'OutputView', outputView);
The only issue you may come across is that a rotation of 90 degrees (i.e. looking end-on at the image plane) would create a set of collinear points that would cause fitgeotrans to error out. In such a case, you would technically just want a blank image, because you can't see a 2-D object when looking at it edge-on.
Here's some code illustrating the above transformations by animating a spinning image:
img = imread('peppers.png');
s = size(img);
outputView = imref2d(s);
scaledInP = [1 s(1); s(2) s(1); s(2) 1; 1 1];
viewAngle = 45;
camDist = sqrt(2)./tand(viewAngle./2);
for theta = linspace(0, 360, 360)
term1 = camDist.*cosd(theta);
term2 = camDist-sind(theta);
term3 = camDist+sind(theta);
outP = [-term1./term2 camDist./term2; ...
term1./term3 camDist./term3; ...
term1./term3 -camDist./term3; ...
-term1./term2 -camDist./term2];
scaledOutP = bsxfun(#times, outP+1, s([2 1])-1)./2+1;
tform = fitgeotrans(scaledInP, scaledOutP, 'projective');
spinImage = imwarp(img, tform, 'OutputView', outputView);
if (theta == 0)
hImage = image(spinImage);
set(gca, 'Visible', 'off');
else
set(hImage, 'CData', spinImage);
end
drawnow;
end
And here's the animation:
You can perform a projective transformation that can be estimated using the position of the corners in the first and second image.
originalP='peppers.png';
original = imread(originalP);
imshow(original);
s = size(original);
matchedPoints1 = [1 1;1 s(1);s(2) s(1);s(2) 1];
matchedPoints2 = [1 1;1 s(1);s(2) s(1)-100;s(2) 100];
transformType = 'projective';
tform = fitgeotrans(matchedPoints1,matchedPoints2,'projective');
outputView = imref2d(size(original));
Ir = imwarp(original,tform,'OutputView',outputView);
figure; imshow(Ir);
This is the result of the code above:
Original image:
Transformed image:
I would like to fit a MR binary data of 281*398*104 matrix which is not a perfect sphere, and find out the center and radius of sphere and error also. I know LMS or SVD is a good choice to fit for sphere.
I have tried sphereFit from matlab file exchange but got an error,
>> sphereFit(data)
Warning: Matrix is singular to working precision.
> In sphereFit at 33
ans =
NaN NaN NaN
Would you let me know where is the problem, or any others solution?
If you want to use sphere fitting algorithm you should first extract the boundary points of the object you assume to be a sphere. The result should be represented by a N-by-3 array containing coordinates of the points. Then you can apply sphereFit function.
In order to obtain boundary point of a binary object, there are several methods. One method is to apply morphological erosion (you need the "imerode" function from the image processing toolbox) with small structuring element, then compute set difference between the two images, and finally use the "find" function to transform binary image into a coordinate array.
the idea is as follow:
dataIn = imerode(data, ones([3 3 3]));
bnd = data & ~data2;
inds = find(bnd);
[y, x, z] = ind2sub(size(data), inds); % be careful about x y order
points = [x y z];
sphere = sphereFitting(points);
By the way, the link you gave refers to circle fitting, I suppose you wanted to point to a sphere fitting submission?
regards,
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.