I have an image I1=(50x50) with two circles - A (which pixels have an intensity value of 100) and B (intensity values of 500).
I want to resize the image to I2=(100x100). I started by doing a FFT2 of the image, after that I zero padded it, making the FT a matrix of 100x100 and then doing the IFFT2 again.
FT = fftshift(fft2(M));
FT = padarray(FT,[50 50]);
I2 = ifft2(ifftshift(FT));
So now I have a new image with size 100x100 but now the amplitude for the circles is different. How do I correct this scaling problem?
After this I apply a mask to select only the circle A so I have now I3=(100x100) with only the circle A. To this image I am applying a FFT2 again and taking only the 50x50 part in the center of this FT and doing the IFFT2 to get another image PSF=(50x50) where I can see a point spread function (PSF) of this circle.
FT = fftshift(fft2(I3));
FTcenter = FT(25:75,25:75);
I4 = ifft2(ifftshift(FTcenter));
Again, should I apply any scaling here as well?
I love using the built-in interpft for FFT-based sinc-interpolation because it takes all the legwork out of zero-padding, shifting, scaling, etc. (and you can inspect how it does it: edit interpft, it’s all legit). It only does 1D interpolation but you can run it twice in both dimensions for 2D.
I made a 100 by 100 grayscale image containing two circles, with min/max intensities [29, 255]:
Then, in the Matlab REPL:
>> im = double(imread('circles.png'));
>> whos im
Name Size Bytes Class Attributes
im 100x100 80000 double
>> imup = interpft(interpft(im, 2*size(im,1), 1), 2*size(im,2), 2);
>> whos imup
Name Size Bytes Class Attributes
imup 200x200 320000 double
This creates the 2× upsampled image. It’s min and max are [23.059, 271.66], outside the bounds of the original [29, 255], so I rescale the values of the upscaled image to live in the same bounds as the original image, convert to uint8, and save it as PNG:
>> rescaled = interp1([min(imup(:)), max(imup(:))], [min(im(:)), max(im(:))], imup);
>> imwrite(uint8(rescaled), 'circles-up.png')
Looking closely at this result, you can see why the min/max values of the upscaled array are broader than the original: this kind of sinc-interpolation causes Gibbs ringing at sharp transitions in the image like the circle edges, which causes the upscaled image to overshoot peaks and undershoot troughs.
interpft can both interpolate (upsample) as well as decimate (downsample). So here’s a complete script—with one tweak: I “invert” the colors from my original image so the background is black instead of white: this makes masking easier:
clearvars
interpft2 = #(x, sz) interpft(interpft(x, sz(1), 1), sz(2), 2);
im = 255 - double(imread('circles.png'));
imup = interpft2(im, 2*size(im));
mask = double(imread('circles-mask.png')) ~= 255;
psf = interpft2(mask .* imup, round(size(imup) / 2));
imwrite(uint8(255 - round(psf)), 'psf.png')
Here is the upsampled circles-mask.png mask (where black means data to keep):
And here's the resulting psf image (inverted), which is the same as the original circles.png image without the little circle:
It appears that, because of linearity, applying the mask in the 2× upsampled image is equivalent to just applying a smaller mask in the original image.
So I should ask if what I did above match what you’re trying to do with the PSF? I’d like to again emphasize that the interpolating and decimating steps you describe in your question (FFT→zeropad→IFFT and FFT→keep center→IFFT) are exactly what interpft does (plus the bookkeeping to make sure amplitudes and symmetries are correct).
Related
i'm working in matlab and i wanted to apply the Contrast Stretching for grey scale image and also RGB image ,
so for the grey scale i've tried this one and it worked
clear all;
clc;
itemp = imread('cameraman.tif'); %read the image
i = itemp(:,:,1);
rtemp = min(i); % find the min. value of pixels in all the
columns (row vector)
rmin = min(rtemp); % find the min. value of pixel in the image
rtemp = max(i); % find the max. value of pixels in all the
columns (row vector)
rmax = max(rtemp); % find the max. value of pixel in the image
m = 255/(rmax - rmin); % find the slope of line joining point
(0,255) to (rmin,rmax)
c = 255 - m*rmax; % find the intercept of the straight line
with the axis
i_new = m*i + c; % transform the image according to new slope
figure,imshow(i); % display original image
figure,imshow(i_new); % display transformed image
this is for greyscale image ,
the problem is that that i don't know how to do for the RGB image
any idea? how to implement that?
thank you :)
Could the function stretchlim (reference) be useful for your purpose?
Find limits to contrast stretch image.
Low_High = stretchlim(RGB,Tol) returns Low_High, a two-element vector
of pixel values that specify lower and upper limits that can be used
for contrast stretching truecolor image RGB.
img = imread('myimg.png');
lohi = stretchlim(img,[0.2 0.8]);
If you write
rmin = min(i(:));
Then it computes the minimum over all values in i. This will work for RGB images also, which simply are 3D matrices with 3 values along the 3rd dimension.
The rest of your code also applies directly to such images.
I'm using the watershed algorithm to try and segment touching nuclei. A typical image may look like:
or this:
I'm trying to apply the watershed algorithm with this code:
show(RGB_img)
%Convert to grayscale image
I = rgb2gray(RGB_img);
%Take structuring element of a disk of size 10, for the morphological transformations
%Attempt to subtract the background from the image: top hat is the
%subtraction of the open image from the original
%Morphological transformation to subtract background noise from the image
%Tophat is the subtraction of an opened image from the original. Remove all
%images smaller than the structuring element of 10
I1 = imtophat(I, strel('disk', 10));
%Increases contrast
I2 = imadjust(I1);
%show(I2,'contrast')
%Assume we have background and foreground and assess thresh as such
level = graythresh(I2);
%Convert to binary image based on graythreshold
BW = im2bw(I2,level);
show(BW,'C');
BW = bwareaopen(BW,8);
show(BW,'C2');
BW = bwdist(BW) <= 1;
show(BW,'joined');
%Complement because we want image to be black and background white
C = ~BW;
%Use distance tranform to find nearest nonzero values from every pixel
D = -bwdist(C);
%Assign Minus infinity values to the values of C inside of the D image
% Modify the image so that the background pixels and the extended maxima
% pixels are forced to be the only local minima in the image (So you could
% hypothetically fill in water on the image
D(C) = -Inf;
%Gets 0 for all watershed lines and integers for each object (basins)
L = watershed(D);
show(L,'L');
%Takes the labels and converts to an RGB (Using hot colormap)
fin = label2rgb(L,'hot','w');
% show(fin,'fin');
im = I;
%Superimpose ridgelines,L has all of them as 0 -> so mark these as 0(black)
im(L==0)=0;
clean_img = L;
show(clean_img)
For whatever reason after C = ~BW; the whole image goes dark. This very same code block has worked on a handful of other images, all of which were more "solid" or not as grainy as these. However, I thought I compensated for this with BW = bwdist(BW) <= 1;. I've experimented a ton and I don't really know what's happening. Any help would be great!
Ps. this is the image after BW = bwareaopen(BW,8);
Before the top-hat, you should perform a closing and an opening in order to reduce the noise.
If you perform an area opening on a noisy image, you may end up with the result on your black and white image.
So it would be:
Closing and opening
Top-Hat
Area opening if still necessary
Thresholding
Erosion and dilation to find the inner and outer markers respectively
Watershed (never use a watershed without markers).
I'm using the watershed algorithm to try and segment touching nuclei. A typical image may look like:
or this:
I'm trying to apply the watershed algorithm with this code:
show(RGB_img)
%Convert to grayscale image
I = rgb2gray(RGB_img);
%Take structuring element of a disk of size 10, for the morphological transformations
%Attempt to subtract the background from the image: top hat is the
%subtraction of the open image from the original
%Morphological transformation to subtract background noise from the image
%Tophat is the subtraction of an opened image from the original. Remove all
%images smaller than the structuring element of 10
I1 = imtophat(I, strel('disk', 10));
%Increases contrast
I2 = imadjust(I1);
%show(I2,'contrast')
%Assume we have background and foreground and assess thresh as such
level = graythresh(I2);
%Convert to binary image based on graythreshold
BW = im2bw(I2,level);
show(BW,'C');
BW = bwareaopen(BW,8);
show(BW,'C2');
BW = bwdist(BW) <= 1;
show(BW,'joined');
%Complement because we want image to be black and background white
C = ~BW;
%Use distance tranform to find nearest nonzero values from every pixel
D = -bwdist(C);
%Assign Minus infinity values to the values of C inside of the D image
% Modify the image so that the background pixels and the extended maxima
% pixels are forced to be the only local minima in the image (So you could
% hypothetically fill in water on the image
D(C) = -Inf;
%Gets 0 for all watershed lines and integers for each object (basins)
L = watershed(D);
show(L,'L');
%Takes the labels and converts to an RGB (Using hot colormap)
fin = label2rgb(L,'hot','w');
% show(fin,'fin');
im = I;
%Superimpose ridgelines,L has all of them as 0 -> so mark these as 0(black)
im(L==0)=0;
clean_img = L;
show(clean_img)
After C = ~BW; the whole image goes dark. I believe this is because the image pixels are all -inf or some smaller negative number. This is there a way around this and if so what could I change in my code to get this algorithm working? I've experimented a ton and I don't really know what's happening. Any help would be great!
The problem is with your show command. As you said in the comments this uses imshow under the hood. If you try imshow directly you'll see you also get a black image. However, if you call it with appropriate limits:
imshow(clean_img,[min(clean_img(:)), max(clean_img(:))])
you'll see everything you expect to see.
In general I usually prefer imagesc for that reason. imshow makes arbitrary judgements as to what range to represent, and I usually can't be bothered to keep up with it. I think in your case, your end image is uint16 so imshow chooses to represent the range [1, 65025]. Since all your pixel values are below 400, they look black to the naked eye for that range.
I need a little help guys in Matlab in Matrix Dimensions,
I Have two images imported by imread function:
im1 = imread('1.jpg');
im2 = imread('2.jpg');
im1 is the reference image, while im2 is the Noisy image.
In the workspace window, Matlab shows the im2 Dimensions like this: 768x1024x3
while im2 displayed as: 768x1024
They are both RGB, there's no greyscale images,
In fact the second image is the a compressed image (performed compression algorithm on it ) while the first image is natural JPEG Image, untouched
and for calculating MSE/PNSR for both images, the matrix dimensions must be the same.
I Will need to transform im1 dimensions to be 3d like the first image (768x1024)
I tried this functions (squeeze, reshape) and with no success
You were on the right track with repmat. Here's the correct syntax:
im2 = repmat(im2, [1 1 3]);
This says you want 1 replicate along the first dimension, 1 replicate along the second dimension, and 3 replicates along the third dimension.
Are you sure that both are RGB images because im2 has only one channel and it looks grayscale but it can also be a colormap image in that case try
[im2, map] = imread('im2.jpg');
and see if anything is appearing in map variable. If the image is indeed colormap image, the map variable should be of size 256 X 3.
What donda has suggested is repeating the grayscale channel 3 times to make it of size 768x1024x3. Another possibility is that noisy image was created by converting RGB image to grayscale or by taking green channel of RGB image. Verify the source of the image in that case.
About PSNR computation I have a feeling that there is some problem with your code. I have given my code below use this and see if it works. Get back to me if you face any problem.
function [Psnr_DB] = psnr(I,I_out)
I = double(I);
I_out = double(I_out);
total_error = 0;
for iterz = 1:size(I,3)
for iterx = 1:size(I,1)
for itery = 1:size(I,2)
total_error = total_error + (I(iterx,itery,iterz)-I_out(iterx,itery,iterz))^2;
end
end
end
MSE = total_error/numel(I);
Psnr = (255^2)/MSE;
Psnr_DB = 10*log10(Psnr) %#ok<NOPRT>
can any one please help me in filling these black holes by values taken from neighboring non-zero pixels.
thanks
One nice way to do this is to is to solve the linear heat equation. What you do is fix the "temperature" (intensity) of the pixels in the good area and let the heat flow into the bad pixels. A passable, but somewhat slow, was to do this is repeatedly average the image then set the good pixels back to their original value with newImage(~badPixels) = myData(~badPixels);.
I do the following steps:
Find the bad pixels where the image is zero, then dilate to be sure we get everything
Apply a big blur to get us started faster
Average the image, then set the good pixels back to their original
Repeat step 3
Display
You could repeat averaging until the image stops changing, and you could use a smaller averaging kernel for higher precision---but this gives good results:
The code is as follows:
numIterations = 30;
avgPrecisionSize = 16; % smaller is better, but takes longer
% Read in the image grayscale:
originalImage = double(rgb2gray(imread('c:\temp\testimage.jpg')));
% get the bad pixels where = 0 and dilate to make sure they get everything:
badPixels = (originalImage == 0);
badPixels = imdilate(badPixels, ones(12));
%# Create a big gaussian and an averaging kernel to use:
G = fspecial('gaussian',[1 1]*100,50);
H = fspecial('average', [1,1]*avgPrecisionSize);
%# User a big filter to get started:
newImage = imfilter(originalImage,G,'same');
newImage(~badPixels) = originalImage(~badPixels);
% Now average to
for count = 1:numIterations
newImage = imfilter(newImage, H, 'same');
newImage(~badPixels) = originalImage(~badPixels);
end
%% Plot the results
figure(123);
clf;
% Display the mask:
subplot(1,2,1);
imagesc(badPixels);
axis image
title('Region Of the Bad Pixels');
% Display the result:
subplot(1,2,2);
imagesc(newImage);
axis image
set(gca,'clim', [0 255])
title('Infilled Image');
colormap gray
But you can get a similar solution using roifill from the image processing toolbox like so:
newImage2 = roifill(originalImage, badPixels);
figure(44);
clf;
imagesc(newImage2);
colormap gray
notice I'm using the same badPixels defined from before.
There is a file on Matlab file exchange, - inpaint_nans that does exactly what you want. The author explains why and in which cases it is better than Delaunay triangulation.
To fill one black area, do the following:
1) Identify a sub-region containing the black area, the smaller the better. The best case is just the boundary points of the black hole.
2) Create a Delaunay triangulation of the non-black points in inside the sub-region by:
tri = DelaunayTri(x,y); %# x, y (column vectors) are coordinates of the non-black points.
3) Determine the black points in which Delaunay triangle by:
[t, bc] = pointLocation(tri, [x_b, y_b]); %# x_b, y_b (column vectors) are coordinates of the black points
tri = tri(t,:);
4) Interpolate:
v_b = sum(v(tri).*bc,2); %# v contains the pixel values at the non-black points, and v_b are the interpolated values at the black points.