I have a question regarding the parameters in the edge function.
edge(img,'sobel',threshold);
edge(img,'prewitt',threshold) ;
edge(img,'roberts',threshold);
edge(img,'canny',thresh_canny,sigma);
How should the threshold for the first 3 types be chosen? Is there an aspect that can help choosing this threshold (like histogram for instance)? I am aware of the function graythresh but I want to set it manually. So far I know it's a value between 0-1, but I don't know how to interpret it.
Same thing for Canny. I`m trying to input an array for thresh_canny = [low_limit, high_limit]. but don't know how to look at these values. How does the sigma value influence the image?
It really depends on the type of edge you want to see in the output. If you want to see really powerful edges, use a smaller interval in the higher end of threshold (say 0.9-1) and this is relative to highest gradient magnitude of the image.
As far as the sigma is concerned, it is used in filtering the input image before passing to edge. This is to reduce noise in the input image.
Related
I took 200 projections at a step angle of 1.8 degrees using LabVIEW software. The size of the image is 2748 x 2748 pixels, uint16. Then using Matlab, I load the projection images, do the flat field correction, resize the image by 1/3 and save the images as .mat file. Then I run the code below for the filtered backprojection.
interp='linear'; %set interpolation: nearest, linear, spline, pchip, v5cubic
filter='Hann'; %set filter: Ram-Lak, Shepp-Logan, Cosine, Hamming, Hann, None
for s=1:916
for i=1:200
a(i,:)=proj065(:,s,i);
end
a=a';
%figure(3), imagesc(a)
b=iradon(a,1.8,interp,filter);
imagesc(b);
recon(:,:,s)=b;
s
clear a
end
If I used a filter in this code, I will get negative pixel values.
But, if I run the code without the filter, I will get positive pixel values.
Any idea why iradon returns negative pixel values in filtered back projection?
Thank you.
Nurul
Yes, the FBP (filtered back-projection) algorithm will do that. It can wrongly reconstruct voxels as having negative values, due to noise and discretization on the data. Nothing you can do about it than just crop those values generally.
As my PhD is about tomography reconstruction algorithms I feel contractually obligated (joking) to suggest the use of iterative algorithms to possibly obtain better images (never worse, often considerably better). Check SART/SIRT or CGLS for this problem.
However, you are calling your function wrong! In tomography, the step size is not enough to reconstruct an image, you generally need the exact angles, thus iradon doesnt accept a step size as an input, it accepts an array of angles.
in your case, theta should be theta=linspace(0,360-200/360,200), and you should call iradon(a,theta,...)
Due to the nature of my problem, I want to evaluate the numerical implementations of the Radon transform in Matlab (i.e. different interpolation methods give different numerical values).
while trying to code my own Radon, and compare it to Matlab's output, I found out that my radon projection sizes are different than Matlab's.
So a bit of intuition of how I compute the amount if radon samples needed. Let's do the 2D case.
The idea is that the maximum size would be when the diagonal (in a rectangular shape at least) part is proyected in the radon transform, so diago=sqrt(size(I,1),size(I,2)). As we dont wan nothing out, n_r=ceil(diago). n_r should be the amount of discrete samples of the radon transform should be to ensure no data is left out.
I noticed that Matlab's radon output is always even, which makes sense as you would want a "ray" through the rotation center always. And I noticed that there are 2 zeros in the endpoints of the array in all cases.
So in that case, n_r=ceil(diago)+mod(ceil(diago)+1,2)+2;
However, it seems that I get small discrepancies with Matlab.
A MWE:
% Try: 255,256
pixels=256;
I=phantom('Modified Shepp-Logan',pixels);
rd=radon(I,pi/4);
size(rd,1)
s=size(I);
diagsize=sqrt(sum(s.^2));
n_r=ceil(diagsize)+mod(ceil(diagsize)+1,2)+2
rd=
367
n_r =
365
As Matlab's Radon transform is a function I can not look into, I wonder why could it be this discrepancy.
I took another look at the problem and I believe this is actually the right answer. From the "hidden documentation" of radon.m (type in edit radon.m and scroll to the bottom)
Grandfathered syntax
R = RADON(I,THETA,N) returns a Radon transform with the
projection computed at N points. R has N rows. If you do not
specify N, the number of points the projection is computed at
is:
2*ceil(norm(size(I)-floor((size(I)-1)/2)-1))+3
This number is sufficient to compute the projection at unit
intervals, even along the diagonal.
I did not try to rederive this formula, but I think this is what you're looking for.
This is a fairly specialized question, so I'll offer up an idea without being completely sure it is the answer to your specific question (normally I would pass and let someone else answer, but I'm not sure how many readers of stackoverflow have studied radon). I think what you might be overlooking is the floor function in the documentation for the radon function call. From the doc:
The radial coordinates returned in xp are the values along the x'-axis, which is
oriented at theta degrees counterclockwise from the x-axis. The origin of both
axes is the center pixel of the image, which is defined as
floor((size(I)+1)/2)
For example, in a 20-by-30 image, the center pixel is (10,15).
This gives different behavior for odd- or even-sized problems that you pass in. Hence, in your example ("Try: 255, 256"), you would need a different case for odd versus even, and this might involve (in effect) padding with a row and column of zeros.
I have an array named Area, which contains a set of values.
The histogram of the array looks like this
The bin width is 60 in this case. I'd like to fit two gaussians to the two peaks here (even if it won't be a great fit).
So I used:
options = statset('Display','final');
obj = gmdistribution.fit(area,2,'Options',options);
gausspdf = pdf(obj, xaxis);
A = sum(gausspdf);
gausspdf = gausspdf/A;
But when I try to plot the two fitted Gaussians, the resulting curve looks like this:
I'm quite confused, as there should be two peaks appearing in the plot?
The gmdistribution.fit method fits data according to maximum-likelihood criterion; that is, it tries to find parameters which maximize the likelihood given the data. It will not necessarily fit what you see or expect visually. Still, there is the possibility that the algorithm converged to a "bad" local minimum. You can try and set the initial conditions according to what you want to get, practically 'helping' the algorithm to converge to the desired result. You do this using the Start option to the fit method, which enables you to give it either an initial guess, in which case you should try and estimate the parameters from the histogram, or an initial component index for each data sample. See the documentation for more details.
I think that your peaks are too close and the function can't distinguish them. so maybe you should change the options for gmdistribution or apply a non-linear function to your data first to get more separate peaks in histogram.
My objective is to handle illumination and expression variations on an image. So I tried to implement a MATLAB code in order to work with only the important information within the image. In other words, to work with only the "USEFUL" information on an image. To do that, it is necessary to delete all unimportant information from the image.
Reference: this paper
Lets see my steps:
1) Apply the Histogram Equalization in order to get an histo_equalized_image=histeq(MyGrayImage). so that large intensity variations
can be handled to some extent.
2) Apply svd approximations on the histo_equalized_image. But before to do that, I applied the svd decomposition ([L D R]=svd(histo_equalized_image)), then these singular values are used to make the derived image J=L*power(D, i)*R where i varies between 1 and 2.
3) Finally, the derived image is combined with the original image to: C=(MyGrayImage+(a*J))/1+a. Where a varies from 0 to 1.
4) But all the steps above are not able to perform well under varying conditions. So finally, wavelet transform should be used to handle those variations(we use only the LL image bloc). Low frequency component contains the useful information, also, unimportant
information gets lost in this component. The (LL) component is ineffective with illumination changes and expression variations.
I wrote a matlab code for that, and I would like to know if my code is correct or no (if no, so how to correct it). Furthermore, I am interested to know if I can optimize these steps. Can we improve this method? if yes, so how? Please I need help.
Lets see now my Matlab code:
%Read the RGB image
image=imread('img.jpg');
%convert it to grayscale
image_gray=rgb2gray(image);
%convert it to double
image_double=im2double(image_gray);
%Apply histogram equalization
histo_equalized_image=histeq(image_double);
%Apply the svd decomposition
[U S V] = svd(histo_equalized_image);
%calculate the derived image
P=U * power(S, 5/4) * V';
%Linearly combine both images
J=(single(histo_equalized_image) + (0.25 * P)) / (1 + 0.25);
%Apply DWT
[c,s]=wavedec2(J,2,'haar');
a1=appcoef2(c,s,'haar',1); % I need only the LL bloc.
You need to define, what do you mean by "USEFUL" or "important" information. And only then do some steps.
Histogram equalization is global transformation, which gives different results on different images. You can make an experiment - do histeq on image, that benefits from it. Then make two copies of the original image and draw in one black square (30% of image area) and white square on second. Then apply histeq and compare results.
Low frequency component contains the useful information, also,
unimportant information gets lost in this component.
Really? Edges and shapes - which are (at least for me) quite important are in high frequencies. Again we need definition of "useful" information.
I cannot see theoretical background why and how your approach would work. Could you a little bit explain, why do you choose this method?
P.S. I`m not sure if this papers are relevant to you, but recommend "Which Edges Matter?" by Bansal et al. and "Multi-Scale Image Contrast Enhancement" by V. Vonikakis and I. Andreadis.
I have real world 3D points which I want to project on a plane. The most of intensity [0-1] values fall in lower region (near zero).
Please see image 'before' his attched below.
I tried to normalize values
Col_=Intensity; % before
max(Col_)=0.46;min(Col_)=0.06;
Col=(Col_-min(Col_))/(max(Col_)-min(Col_));% after
max(Col)=1;min(Col)=0;
But still i have maximum values falling in lower region (near zero).
Please see second fig after normalization.
Result is still most of black region.Any suggestion. How can I strech my intensity information.
regards,!
It looks like you have already normalized as much as you can with linear scaling. If you want to get more contrast, you will have to give up preserving the original scaling and use a non-linear equalization.
For example: http://en.wikipedia.org/wiki/Histogram_equalization
If you have the image processing toolbox, matlab will do it for you:
http://www.mathworks.com/help/toolbox/images/ref/histeq.html
It looks like you have very few values outside the first bin, if you don't need to preserve the uniqueness of the intensities, you could just scale by a larger amount and clip the few that exceed 1.
When I normalize intensities I do something like this:
Col = Col - min(Col(:));
Col = Col/max(Col(:));
This will normalize your data points to the range [0,1].
Now, since you have many small values, you might be able to make out small changes better through log scaling.
Col_scaled = log(1+Col);
Linear scaling with such data rarely works for me. Using the log function is akin to tweaking gamma for visualization purposes.
I think the only thing you can do here is reduce the range.
After normalization do the following:
t = 0.1;
Col(Col > t) = t;
This will simply truncate the range of the data, which may be sufficient for what you are doing. Then you can re-normalize again if you wish.