I've taken the following image:
PandaNoise.bmp and tried to remove the periodic noise by focusing on its Fourier spectrum. Commented lines are the ones I'm not sure about. I can't get it back to the image plane. What am I doing wrong here?
panda = imread('PandaNoise.bmp');
fpanda = fft2(panda); % 2d fast fourier transform
fpanda = fftshift(fpanda); % center FFT
fpanda = abs(fpanda); % get magnitude
fpanda = log(1 + fpanda); % use log to expand range of dark pixels into bright region
fpanda = mat2gray(fpanda); % scale image from 0 to 1
figure; imshow(fpanda,[]); % show the picture
zpanda = fpanda;
zpanda(fpanda<0.5)=0;
zpanda(fpanda>0.5)=1;
%img = ifft2(zpanda);
%img = ifftshift(img);
%img = exp(1-img);
%img = abs(img);
Here is an example of how to work with the complex Fourier transform. We can take the log modulus for display, but don't change the original Fourier transform matrix, since the phase information that we throw away with abs is very important.
% Load data
panda = imread('https://i.stack.imgur.com/9SlW5.png');
panda = im2double(panda);
% Forward transform
fpanda = fft2(panda);
% Prepare FT for display -- don't change fpanda!
fd = fftshift(fpanda);
fd = log(1 + abs(fd));
figure; imshow(fd,[]); % show the picture
% From here we learn that we should keep the central 1/5th along both axes
% Low-pass filter
sz = size(fpanda);
center = floor(sz/2)+1;
half_width = ceil(sz/10)-1;
filter = zeros(sz);
filter(center(1)+(-half_width(1):half_width(1)),...
center(2)+(-half_width(2):half_width(2))) = 1;
filter = ifftshift(filter); % The origin should be on the top-left, like that of fpanda.
fpanda = fpanda .* filter;
% Inverse transform
newpanda = ifft2(fpanda);
figure; imshow(newpanda);
After computing the ifft2, newpanda is supposed to be purely real-valued if we designed the filter correctly (i.e. perfectly symmetric around the origin). Any imaginary component still present should be purely numerical innacuracy. MATLAB will detect that the input to ifft2 is conjugate symmetric, and return a purely real result. Octave will not, and you would have to do newpanda=real(newpanda) to avoid warnings from imshow.
I'm working on an application to determine from an image the degree of alignment of a fiber network. I've read several papers on this issue and they basically do this:
Find the 2D discrete Fourier transform (DFT = F(u,v)) of the image (gray, range 0-255)
Find the Fourier Spectrum (FS = abs(F(u,v))) and the Power Spectrum (PS = FS^2)
Convert spectrum to polar coordinates and divide it into 1º intervals.
Calculate number-averaged line intensities (FI) for each interval (theta), that is, the average of all the intensities (pixels) forming "theta" degrees with respect to the horizontal axis.
Transform FI(theta) to cartesian coordinates
Cxy(theta) = [FI*cos(theta), FI*sin(theta)]
Find eigenvalues (lambda1 and lambda2) of the matrix Cxy'*Cxy
Find alignment index as alpha = 1 - lamda2/lambda1
I've implemented this in MATLAB (code below), but I'm not sure whether it is ok since point 3 and 4 are not really clear for me (I'm getting similar results to those of the papers, but not in all cases). For instance, in point 3, "spectrum" is referring to FS or to PS?. And in point 4, how should this average be done? are all the pixels considered? (even though there are more pixels in the diagonal).
rgb = imread('network.tif');%513x513 pixels
im = rgb2gray(rgb);
im = imrotate(im,-90);%since FFT space is rotated 90º
FT = fft2(im) ;
FS = abs(FT); %Fourier spectrum
PS = FS.^2; % Power spectrum
FS = fftshift(FS);
PS = fftshift(PS);
xoffset = (513-1)/2;
yoffset = (513-1)/2;
% Avoid low frequency points
x1 = 5;
y1 = 0;
% Maximum high frequency pixels
x2 = 255;
y2 = 0;
for theta = 0:pi/180:pi
% Transposed rotation matrix
Rt = [cos(theta) sin(theta);
-sin(theta) cos(theta)];
% Find radial lines necessary for improfile
xy1_rot = Rt * [x1; y1] + [xoffset; yoffset];
xy2_rot = Rt * [x2; y2] + [xoffset; yoffset];
plot([xy1_rot(1) xy2_rot(1)], ...
[xy1_rot(2) xy2_rot(2)], ...
'linestyle','none', ...
'marker','o', ...
'color','k');
prof = improfile(F,[xy1_rot(1) xy2_rot(1)],[xy1_rot(2) xy2_rot(2)]);
i = i + 1;
FI(i) = sum(prof(:))/length(prof);
Cxy(i,:) = [FI(i)*cos(theta), FI(i)*sin(theta)];
end
C = Cxy'*Cxy;
[V,D] = eig(C)
lambda2 = D(1,1);
lambda1 = D(2,2);
alpha = 1 - lambda2/lambda1
Figure: A) original image, B) plot of log(P+1), C) polar plot of FI.
My main concern is that when I choose an artificial image perfectly aligned (attached figure), I get alpha = 0.91, and it should be exactly 1.
Any help will be greatly appreciated.
PD: those black dots in the middle plot are just the points used by improfile.
I believe that there are a couple sources of potential error here that are leading to you not getting a perfect alpha value.
Discrete Fourier Transform
You have discrete imaging data which forces you to take a discrete Fourier transform which inevitably (depending on the resolution of the input data) have some accuracy issues.
Binning vs. Sampling Along a Line
The way that you have done the binning is that you literally drew a line (rotated by a particular angle) and sampled the image along that line using improfile. Using improfile performs interpolation of your data along that line introducing yet another potential source of error. The default is nearest neighbor interpolation which in the example shown below can cause multiple "profiles" to all pick up the same points.
This was with a rotation of 1-degree off-vertical when technically you'd want those peaks to only appear for a perfectly vertical line. It is clear to see how this sort of interpolation of the Fourier spectrum can lead to a spread around the "correct" answer.
Data Undersampling
Similar to Nyquist sampling in the Fourier domain, sampling in the spatial domain has some requirements as well.
Imagine for a second that you wanted to use 45-degree bin widths instead of the 1-degree. Your approach would still sample along a thin line and use that sample to represent 45-degrees worth or data. Clearly, this is a gross under-sampling of the data and you can imagine that the result wouldn't be very accurate.
It becomes more and more of an issue the further you get from the center of the image since the data in this "bin" is really pie wedge shaped and you're approximating it with a line.
A Potential Solution
A different approach to binning would be to determine the polar coordinates (r, theta) for all pixel centers in the image. Then to bin the theta components into 1-degree bins. Then sum all of the values that fall into that bin.
This has several advantages:
It removes the undersampling that we talked about and draws samples from the entire "pie wedge" regardless of the sampling angle.
It ensures that each pixel belongs to one and only one angular bin
I have implemented this alternate approach in the code below with some false horizontal line data and am able to achieve an alpha value of 0.988 which I'd say is pretty good given the discrete nature of the data.
% Draw a bunch of horizontal lines
data = zeros(101);
data([5:5:end],:) = 1;
fourier = fftshift(fft2(data));
FS = abs(fourier);
PS = FS.^2;
center = fliplr(size(FS)) / 2;
[xx,yy] = meshgrid(1:size(FS,2), 1:size(FS, 1));
coords = [xx(:), yy(:)];
% De-mean coordinates to center at the middle of the image
coords = bsxfun(#minus, coords, center);
[theta, R] = cart2pol(coords(:,1), coords(:,2));
% Convert to degrees and round them to the nearest degree
degrees = mod(round(rad2deg(theta)), 360);
degreeRange = 0:359;
% Band pass to ignore high and low frequency components;
lowfreq = 5;
highfreq = size(FS,1)/2;
% Now average everything with the same degrees (sum over PS and average by the number of pixels)
for k = degreeRange
ps_integral(k+1) = mean(PS(degrees == k & R > lowfreq & R < highfreq));
fs_integral(k+1) = mean(FS(degrees == k & R > lowfreq & R < highfreq));
end
thetas = deg2rad(degreeRange);
Cxy = [ps_integral.*cos(thetas);
ps_integral.*sin(thetas)]';
C = Cxy' * Cxy;
[V,D] = eig(C);
lambda2 = D(1,1);
lambda1 = D(2,2);
alpha = 1 - lambda2/lambda1;
I have a noisy image that I am trying to clean using a lowpass filter (code below, modified from here). The image I get as a result is essentially identical to the one I gave as an input.
I'm not an expert, but my conclusion would be that the input image is so noisy that no patterns are found. Do you agree? Do you have any suggestion on how to interpret the result?
Result from the code:
Input image:
Code:
clear; close all;
frame = 20;
size_y = 512; % This is actually size_x
size_x = 256; % This is actually size_y
roi=5;thresh=100000;
AA = imread('image.png');
A = zeros(size_x, size_y);
A = AA(1:size_x, 1:size_y);
A(isnan(A)) = 0 ;
B = fftshift(fft2(A));
fabs = abs(B);
figure; imshow(B);
local_extr = ordfilt2(fabs, roi^2, ones(roi)); % find local maximum within 3*3 range
result = (fabs == local_extr) & (fabs > thresh);
[r, c] = find(result);
for i=1:length(r)
if (r(i)-128)^2+(c(i)-128)^2>thresh % periodic noise locates in the position outside the 20-pixel-radius circle
B(r(i)-2:r(i)+2,c(i)-2:c(i)+2)=0; % zero the frequency components
end
end
Inew=ifft2(fftshift(B));
figure;
subplot(2,1,1); imagesc(A), colormap(gray); title('Original image');
subplot(2,1,2);imagesc(real(Inew)),colormap(gray); title('Filtered image');
For filtering this kind of signal, you can try to use the median filter. It might be more appropriated than a means or Gaussian filter. The median filter is very effective on "salt and paper" noise when the mean just blur the noise.
As the signal seems very noisy, you need to try to find the good size of kernel for the filter. You can also try to increase the contrast of the image (after filtering) in order to see more the difference between the gray levels.
what i want to do is a histogram for quantized DCT coefficient for an image, to detect Double Quantization effect. when i use hist(x) it will categorize it in to 10s and if i changed it to hist(x,20) or 30 it does not really show the DQ effect. so is there any better way for this??
here is the code: on matlab
im = jpeg_read('image');
% Pull image information - Lum, Cb, Cr
lum = im.coef_arrays{im.comp_info(1).component_id};
cb = im.coef_arrays{im.comp_info(2).component_id};
cr = im.coef_arrays{im.comp_info(3).component_id};
% Pull quantization arrays
lqtable = im.quant_tables{im.comp_info(1).quant_tbl_no};
cqtable = im.quant_tables{im.comp_info(2).quant_tbl_no};
% Quantize above two sets of information
qcof = quantize(lum,lqtable);
bqcof = quantize(cb,cqtable);
rqcof = quantize(cr,cqtable);
hist(qcof,30); %lum quantized dct coefficient histogram
First, no need to quantize the coefficients. Secondly, the effect can be observed by plotting histograms of certain frequencies. You need to go through various positions in the blocks and look for the pattern. Plotting FFT of the histogram helps.
Here is the matlab code:
imJPG2 = jpeg_read('foto2.jpg');
lum = imJPG2.coef_arrays{imJPG2.comp_info(1).component_id};
for i = 1:8
for j = 1:8
r = lum( i:8:end, j:8:end );
histogram(r(:), 'binmethod','integers');
pause();
end
end
More details and background can be found in this paper: http://www.sciencedirect.com/science/article/pii/S0031320309001198
Background:
My question relates to extracting feature from an electrophoresis gel (see below). In this gel, DNA is loaded from the top and allowed to migrate under a voltage gradient. The gel has sieves so smaller molecules migrate further than longer molecules resulting in the separation of DNA by length. So higher up the molecule, the longer it is.
Question:
In this image there are 9 lanes each with separate source of DNA. I am interested in measuring the mean location (value on the y axis) of each lane.
I am really new to image processing, but I do know MATLAB and I can get by with R with some difficulty. I would really appreciate it if someone can show me how to go about finding the mean of each lane.
Here's my try. It requires that the gels are nice (i.e. straight lanes and the gel should not be rotated), but should otherwise work fairly generically. Note that there are two image-size-dependent parameters that will need to be adjusted to make this work on images of different size.
%# first size-dependent parameter: should be about 1/4th-1/5th
%# of the lane width in pixels.
minFilterWidth = 10;
%# second size-dependent parameter for filtering the
%# lane profiles
gaussWidth = 5;
%# read the image, normalize to 0...1
img = imread('http://img823.imageshack.us/img823/588/gele.png');
img = rgb2gray(img);
img = double(img)/255;
%# Otsu thresholding to (roughly) find lanes
thMsk = img < graythresh(img);
%# count the mask-pixels in each columns. Due to
%# lane separation, there will be fewer pixels
%# between lanes
cts = sum(thMsk,1);
%# widen the local minima, so that we get a nice
%# separation between lanes
ctsEroded = imerode(cts,ones(1,minFilterWidth));
%# use imregionalmin to identify the separation
%# between lanes. Invert to get a positive mask
laneMsk = ~repmat(imregionalmin(ctsEroded),size(img,1),1);
Image with lanes that will be used for analysis
%# for each lane, create an averaged profile
lblMsk = bwlabel(laneMsk);
nLanes = max(lblMsk(:));
profiles = zeros(size(img,1),nLanes);
midLane = zeros(1,nLanes);
for i = 1:nLanes
profiles(:,i) = mean(img.*(lblMsk==i),2);
midLane(:,i) = mean(find(lblMsk(1,:)==i));
end
%# Gauss-filter the profiles (each column is an
%# averaged intensity profile
G = exp(-(-gaussWidth*5:gaussWidth*5).^2/(2*gaussWidth^2));
G=G./sum(G);
profiles = imfilter(profiles,G','replicate'); %'
%# find the minima
[~,idx] = min(profiles,[],1);
%# plot
figure,imshow(img,[])
hold on, plot(midLane,idx,'.r')
Here's my stab at a simple template for an interactive way to do this:
% Load image
img = imread('gel.png');
img = rgb2gray(img);
% Identify lanes
imshow(img)
[x,y] = ginput;
% Invert image
img = max(img(:)) - img;
% Subtract background
[xn,yn] = ginput(1);
noise = img((yn-2):(yn+2), (xn-2):(xn+2));
noise = mean(noise(:));
img = img - noise;
% Calculate means
means = (1:size(img,1)) * double(img(:,round(x))) ./ sum(double(img(:,round(x))), 1);
% Plot
hold on
plot(x, means, 'r.')
The first thing to do to is convert your RGB image to grayscale:
gr = rgb2gray(imread('gelk.png'));
Then, take a look at the image intensity histogram using imhist. Notice anything funny about it? Use imcontrast(imshow(gr)) to pull up the contrast adjustment tool. I found that eliminating the weird stuff after the major intensity peak was beneficial.
The image processing task itself can be divided into several steps.
Separate each lane
Identify ('segment') the band in each lane
Calculate the location of the bands
Step 1 can be done "by hand," if the lane widths are guaranteed. If not, the line detection offered by the Hough transform is probably the way to go. The documentation on the Image Processing Toolbox has a really nice tutorial on this topic. My code recapitulates that tutorial with better parameters for your image. I only spent a few minutes with them, I'm sure you can improve the results by tuning the parameters further.
Step 2 can be done in a few ways. The easiest technique to use is Otsu's method for thresholding grayscale images. This method works by determining a threshold that minimizes the intra-class variance, or, equivalently, maximizes the inter-class variance. Otsu's method is present in MATLAB as the graythresh function. If Otsu's method isn't working well you can try multi-level Otsu or a number of other histogram based threshold determination methods.
Step 3 can be done as you suggest, by calculating the mean y value of the segmented band pixels. This is what my code does, though I've restricted the check to just the center column of each lane, in case the separation was off. I'm worried that the result may not be as good as calculating the band centroid and using its location.
Here is my solution:
function [locations, lanesBW, lanes, cols] = segmentGel(gr)
%%# Detect lane boundaries
unsharp = fspecial('unsharp'); %# Sharpening filter
I = imfilter(gr,unsharp); %# Apply filter
bw = edge(I,'canny',[0.01 0.3],0.5); %# Canny edges with parameters
[H,T,R] = hough(bw); %# Hough transform of edges
P = houghpeaks(H,20,'threshold',ceil(0.5*max(H(:)))); %# Find peaks of Hough transform
lines = houghlines(bw,T,R,P,'FillGap',30,'MinLength',20); %# Use peaks to identify lines
%%# Plot detected lines above image, for quality control
max_len = 0;
imshow(I);
hold on;
for k = 1:length(lines)
xy = [lines(k).point1; lines(k).point2];
plot(xy(:,1),xy(:,2),'LineWidth',2,'Color','green');
%# Plot beginnings and ends of lines
plot(xy(1,1),xy(1,2),'x','LineWidth',2,'Color','yellow');
plot(xy(2,1),xy(2,2),'x','LineWidth',2,'Color','red');
%# Determine the endpoints of the longest line segment
len = norm(lines(k).point1 - lines(k).point2);
if ( len > max_len)
max_len = len;
end
end
hold off;
%%# Use first endpoint of each line to separate lanes
cols = zeros(length(lines),1);
for k = 1:length(lines)
cols(k) = lines(k).point1(1);
end
cols = sort(cols); %# The lines are in no particular order
lanes = cell(length(cols)-1,1);
for k = 2:length(cols)
lanes{k-1} = im2double( gr(:,cols(k-1):cols(k)) ); %# im2double for compatibility with greythresh
end
otsu = cellfun(#graythresh,lanes); %# Calculate threshold for each lane
lanesBW = cell(size(lanes));
for k = 1:length(lanes)
lanesBW{k} = lanes{k} < otsu(k); %# Apply thresholds
end
%%# Use segmented bands to determine migration distance
locations = zeros(size(lanesBW));
for k = 1:length(lanesBW)
width = size(lanesBW{k},2);
[y,~] = find(lanesBW{k}(:,round(width/2))); %# Only use center of lane
locations(k) = mean(y);
end
I suggest you carefully examine not only each output value, but the results from each step of the function, before using it for actual research purposes. In order to get really good results, you will have to read a bit about Hough transforms, Canny edge detection and Otsu's method, and then tune the parameters. You may also have to alter how the lanes are split; this code assumes that there will be lines detected on either side of the image.
Let me add another implementation similar in concept to that of #JohnColby's, only without the manual user-interaction:
%# read image
I = rgb2gray(imread('gele.png'));
%# middle position of each lane
%# (assuming lanes are somewhat evenly spread and of similar width)
x = linspace(1,size(I,2),10);
x = round( (x(1:end-1)+x(2:end))./2 );
%# compute the mean value across those columns
m = mean(I(:,x));
%# find the y-indices of the mean values
[~,idx] = min( bsxfun(#minus, double(I(:,x)), m) );
%# show the result
figure(1)
imshow(I, 'InitialMagnification',100, 'Border','tight')
hold on, plot(x, idx, ...
'Color','r', 'LineStyle','none', 'Marker','.', 'MarkerSize',10)
and applied on the smaller image: