I have a picture in MATLAB and decompositioned it to R, G and B channel.
And now I need to figure out resolution, median, upper and lower quartil, interquartil span and modus.
Here is the code I use for it:
close all; clc; clear all
I = imread('image_S006_I0000.jpg');
imshow(I)
[R S T] = size(I);
Red(:,:,1)=I(:,:,1);
Red(:,:,2)=zeros(R,S);
Red(:,:,3)=zeros(R,S);
Green(:,:,1)=zeros(R,S);
Green(:,:,2)=I(:,:,1);
Green(:,:,3)=zeros(R,S);
Blue(:,:,1)=zeros(R,S);
Blue(:,:,2)=zeros(R,S);
Blue(:,:,3)=I(:,:,1);
OR
Red=double(I(:,:,1));
Red=zeros(R,S);
Red=zeros(R,S);
Green=zeros(R,S);
Green=double(I(:,:,1));
Green=zeros(R,S);
Blue=zeros(R,S);
Blue=zeros(R,S);
Blue=double(I(:,:,1));
OR
Red(:,:,1)=double(I(:,:,1));
Green(:,:,1)=double(I(:,:,1));
Blue(:,:,1)=double(I(:,:,1));
cat(3, uint8(Red), zeros(R, S), zeros(R, S));
others = zeros(R, S);
red_plt = cat(3, uint8(Red), others, others);
green_plt = cat(3, others, uint8(Green), others);
blue_plt = cat(3, others, others, uint8(Blue));
figure()
subplot(131);imshow(red_plt)
subplot(132);imshow(green_plt)
subplot(133);imshow(blue_plt)
NOW PLOTTITG (It doesn't print it in Red, Green and Blue color. First two
are just all black, the third one is black and white):
figure()
subplot(131);imshow(uint8(Red))
subplot(132);imshow(uint8(Green))
subplot(133);imshow(uint8(Blue))
[x1 y1 z1] = size(I);
% MEDIAN.
imgmedianR = median (Red(:))
imgmedianG = median (Green(:))
imgmedianB = median (Blue(:))
%QUARTIL
r025 = quantile(Red,0.25)
r075 = quantile(Red,0.75)
g025 = quantile(Green,0.25)
g075 = quantile(Green,0.75)
b025 = quantile(Blue,0.25)
b075 = quantile(Blue,0.75)
%INTERQUARTIL SPAN
r_iqr = iqr(Red)
g_iqr = iqr(Green)
b_iqr = iqr(Blue)
modus_Red = mode(Red(:))
modus_Green = mode(Green(:))
modus_Blue = mode(Blue(:))
MEDIAN: When I try to count MEDIAN, it gives me a bunch of numbers (actually matlab prints out a number for each column separately).
What have I done wrong there?
QUARTIL: The same problem as I have in code for MEDIAN. What have I done wrong there?
INTERQUARTIL SPAN: Matlab prints out subsequent error:
Mixed integer class inputs are not supported.
What have I done wrong?
RESOLUTION: I need to find out a resolution of the image using Matlab. I have tried out funciton imfinfo, but the info isn't included. How can I find such an information using any kind of Matlab function?
Thank you very much in advance!
You save every separate colour in a 3d matrix. This is not nessecary. So save them in a 2d matrix as follows:
Red=double(I(:,:,1));
Green=double(I(:,:,2));
Blue=double(I(:,:,3));
Note there is a conversion to double. That's needed for the iqr. Also use in every function Red(:) instead of Red (same for other colours of course). If you change all that, it should work just fine.
Concerning the resolution, that is just the number of pixels for the height and the width, in your case R and S.
ps you're calling size(I) twice.
In case of plotting the red, green and blue separately, you need to make the same matrix as the original, with exception that only 1 colour is non-zero. The image is a RxSx3 matrix. So if you want to plot the red part you need to make an RxSx3 matrix where only red is non-zero:
cat(3, uint8(Red), zeros(R, S), zeros(R, S));
So in total it should look like this
others = zeros(R, S);
red_plt = cat(3, uint8(Red), others, others);
green_plt = cat(3, others, uint8(Green), others);
blue_plt = cat(3, others, others, uint8(Blue));
figure()
subplot(131);imshow(red_plt)
subplot(132);imshow(green_plt)
subplot(133);imshow(blue_plt)
EDIT:
The whole code:
I = imread('image_S006_I0000.jpg');
imshow(I)
[R S T] = size(I);
Red=double(I(:,:,1));
Green=double(I(:,:,2));
Blue=double(I(:,:,3));
others = zeros(R, S);
red_plt = cat(3, uint8(Red), others, others);
green_plt = cat(3, others, uint8(Green), others);
blue_plt = cat(3, others, others, uint8(Blue));
figure()
subplot(131);imshow(red_plt)
subplot(132);imshow(green_plt)
subplot(133);imshow(blue_plt)
% MEDIAN.
imgmedianR = median (Red(:))
imgmedianG = median (Green(:))
imgmedianB = median (Blue(:))
%QUARTIL
r025 = quantile(Red(:),0.25)
r075 = quantile(Red(:),0.75)
g025 = quantile(Green(:),0.25)
g075 = quantile(Green(:),0.75)
b025 = quantile(Blue(:),0.25)
b075 = quantile(Blue(:),0.75)
%INTERQUARTIL SPAN
r_iqr = iqr(Red(:))
g_iqr = iqr(Green(:))
b_iqr = iqr(Blue(:))
modus_Red = mode(Red(:))
modus_Green = mode(Green(:))
modus_Blue = mode(Blue(:))
Related
I have a clean image, and a noisy image. I created a denoiser and applied it to the noisy image, that was my final output. Now to compare how much this image is close to a clean image I need to compare it using PSNR and SSIM, but due to different positions of the image I am unable to compare.
Now I am getting SSIM as 0.5, which is very low, due to the improper placement of both the images. If the images are registered properly, then I guess SSIM should come around 0.80+. But I have not been able to accomplish this.
How can I align these two images to obtain a good SSIM value?
I have two coin images, 1st image (CLEAN), 2nd image (IMPROVED a NOISY IMG), for comparison.
Clean Img:
Noisy Img:
Due to positions of images at different positions ssim(img1,img2) is giving incorrect output. I tried cropping but that did not work.
Here is what I have tried so far:
Attempt 1:
function [valPSNR,valSSIM,badpict]=getSSIM(clean_img,img2)
% pad reference image since object is so close to edges
refpict = padarray(mat2gray(clean_img),[20 20],'replicate','both');
% crop test image down to extract the object alone
badpict = imcrop(mat2gray(img2),[2.5 61.5 357 363]);
% maximize normalized cross-correlation to find offset
szb = size(badpict);
c = normxcorr2(badpict,refpict);
[idxy idxx] = find(c == max(c(:)));
osy = idxy-szb(1);
osx = idxx-szb(2);
% crop the reference pict to the ROI
refpict = refpict(osy:idxy-1,osx:idxx-1);
%imshow(imfuse(badpict,refpict,'checkerboard'));
%imagesc(badpict);
valSSIM=ssim(badpict,refpict);
valPSNR=getPSNR(badpict,refpict);
img2=badpict;
clean_img=refpict;
figure; imshowpair(clean_img,img2);
figure; montage({mat2gray(clean_img),mat2gray(img2)}, 'Size', [1 2], 'BackgroundColor', 'w', 'BorderSize', [2 2]);
end
Attempt 2:
function [valPSNR,valSSIM,badpict]=getSSIM2(clean_img,img2)
% pad reference image since object is so close to edges
bw1 = im2bw(mat2gray(clean_img));
bw2 = imclose(im2bw(mat2gray(img2),0.3),strel('disk',9));
bw2 = bwareafilt(bw2,1);
% make same size
[r,c] = find(bw1);
clean_img = clean_img(min(r):max(r),min(c):max(c));
[r,c] = find(bw2);
img2 = img2(min(r):max(r),min(c):max(c));
img2= imresize(img2, size(clean_img),'bilinear');
valPSNR=getPSNR(mat2gray(clean_img),mat2gray(img2));
valSSIM=ssim(mat2gray(clean_img),mat2gray(img2));
badpict=img2;
figure; imshowpair(clean_img,img2);
figure; montage({mat2gray(clean_img),mat2gray(img2)}, 'Size', [1 2], 'BackgroundColor', 'w', 'BorderSize', [2 2]);
end
As others have pointed out, the resampling required by registration will have some non-zero error. But, here is some sample code that will take you through the registration part that is the crux of your question.
% SSIM isn't defined on RGB images, convert to grayscale.
ref = rgb2gray(imread('https://i.stack.imgur.com/tPKEJ.png'));
X = rgb2gray(imread('https://i.stack.imgur.com/KmU4y.png'));
% The input image data has bright borders at the edges that create
% artifacts in resampling, best to just crop those or maybe there are
% aquisitions that don't have these borders?
X = X(3:end-2,3:end-2);
ref = ref(4:end-3,4:end-3);
figure
montage({X,ref});
tform = imregcorr(X,ref,"translation");
Xreg = imwarp(X,tform,OutputView=imref2d(size(ref)),SmoothEdges=true);
figure
imshowpair(Xreg,ref)
ssim(Xreg,ref)
Maybe you can refer to my github.
I implemented a template matching algorithm by OpenCV which you can use NCC-Based Pattern Matching to find targets, and then get a score (similarity).
You can then use this score to decide if it is clean.
Besides, tranforming c++ code may be an issue for you, but just find the all corresponded function in matlab version.
Here are effects (red blocks are areas with similarity higher than threshold 0.85 in comparison with golden sample):
The whole function is too long to be posted here.
Part of the function:
for (int i = 0; i < iSize; i++)
{
Mat matRotatedSrc, matR = getRotationMatrix2D (ptCenter, vecAngles[i], 1);
Mat matResult;
Point ptMaxLoc;
double dValue, dMaxVal;
double dRotate = clock ();
Size sizeBest = GetBestRotationSize (vecMatSrcPyr[iTopLayer].size (), pTemplData->vecPyramid[iTopLayer].size (), vecAngles[i]);
float fTranslationX = (sizeBest.width - 1) / 2.0f - ptCenter.x;
float fTranslationY = (sizeBest.height - 1) / 2.0f - ptCenter.y;
matR.at<double> (0, 2) += fTranslationX;
matR.at<double> (1, 2) += fTranslationY;
warpAffine (vecMatSrcPyr[iTopLayer], matRotatedSrc, matR, sizeBest);
MatchTemplate (matRotatedSrc, pTemplData, matResult, iTopLayer);
minMaxLoc (matResult, 0, &dMaxVal, 0, &ptMaxLoc);
vecMatchParameter[i * (m_iMaxPos + MATCH_CANDIDATE_NUM)] = s_MatchParameter (Point2f (ptMaxLoc.x - fTranslationX, ptMaxLoc.y - fTranslationY), dMaxVal, vecAngles[i]);
for (int j = 0; j < m_iMaxPos + MATCH_CANDIDATE_NUM - 1; j++)
{
ptMaxLoc = GetNextMaxLoc (matResult, ptMaxLoc, -1, pTemplData->vecPyramid[iTopLayer].cols, pTemplData->vecPyramid[iTopLayer].rows, dValue, m_dMaxOverlap);
vecMatchParameter[i * (m_iMaxPos + MATCH_CANDIDATE_NUM) + j + 1] = s_MatchParameter (Point2f (ptMaxLoc.x - fTranslationX, ptMaxLoc.y - fTranslationY), dValue, vecAngles[i]);
}
}
FilterWithScore (&vecMatchParameter, m_dScore-0.05*iTopLayer);
How can I plot a bar out of a
data = 1x10 cell
, where each value in the cell has a different dimension like 3x100, 3x40, 66x2 etc.
My goal is to get a bar plot, where I would have 10 group of bars and in every group three bars for each of the values. On the bar, I want it to be shown the median of the values, and I want to calculate the confidence interval and show it additionally.
On this example there are not group of bars, but my point is to show you how I want the confidence intervals shown. On the site, where I found this example they offer a solution where they have this command line
e1 = errorbar(mean(data), ci95);
but I have the problem that it can't find any ci95
So, are there any other effective ways to do it, without installing or downloading additional services?
I've found Patrick Happel's answer to not work because the figure window (and therefore the variable b) gets cleared out by subsequent calls to errorbar. Simply adding a hold on command takes care of this. To avoid confusion, here's a new answer that reproduces all of Patrick's original code, plus my small tweak:
%% Old answer
%Just to be safe, let's clear everything
clear all
data = cell(1,10);
% Random length of the data
l = randi(500, 10, 1) + 50;
% Random "width" of the data, with 3 more likely
w = randi(4, 10, 1);
w(w==4) = 3;
% random "direction" of the data
d = randi(2, 10, 1);
% sigma of the data (in fraction of mean)
sigma = rand(10,1) / 3;
% means of the data
dmean = randi(150,10,1);
dsigma = dmean.*sigma;
for c = 1 : 10
if d(c) == 1
data{c} = randn(l(c), w(c)) .* dsigma(c) + dmean(c);
else
data{c} = randn(w(c), l(c)) .* dsigma(c) + dmean(c);
end
end
%============================================
%Next thing is
% On the bar, I want it to be shown the median of the values, and I
% want to calculate the confidence interval and show it additionally.
%
%Are you really sure you want to plot the median? The median of some data
%is not connected to the variance of the data, and hus no type of error
%bars are required. I guess you want to show the mean. If you really want
%to show the median, a box plot might be a better alternative.
%
%The following code computes and plots the mean in a bar plot:
%============================================
means = zeros(numel(data),3);
stds = zeros(numel(data),3);
n = zeros(numel(data),3);
for c = 1:numel(data)
d = data{c};
if size(d,1) < size(d,2)
d = d';
end
cols = size(d,2);
means(c, 1:cols) = nanmean(d);
stds(c, 1:cols) = nanstd(d);
n(c, 1:cols) = sum(~isnan((d)));
end
b = bar(means);
%% New code
%This ensures that b continues to reference existing data in the next for
%loop, as the graphics objects can otherwise be deleted.
hold on
%% Continuing Patrick Happel's answer
%============================================
%Now, we need to compute the length of the error bars. Typical choices are
%the standard deviation of the data (already computed by the code above,
%stored in stds), the standard error or the 95% confidence interval (which
%is the 1.96fold of the standard error, assuming the underlying data
%follows a normal distribution).
%============================================
% for standard deviation use stds
% for standard error
ste = stds./sqrt(n);
% for 95% confidence interval
ci95 = 1.96 * ste;
%============================================
%Last thing is to plot the error bars. Here I chose the ci95 as you asked
%in your question, if you want to change that, simply change the variable
%in the call to errorbar:
%============================================
for c = 1:3
size(means(:, c))
size(b(c).XData)
e = errorbar(b(c).XData + b(c).XOffset, means(:,c), ci95(:, c));
e.LineStyle = 'none';
end
Since I am not sure how your data looks like, since in your question you stated that the elements of the cell contain data with different dimension like
3x100, 3x40, 66x2
I assume that your data can be arranged in columns or rows and that not all data requires three bars.
Since you did not provide a short piece of your data for us to test, I generate some artificial data:
data = cell(1,10);
% Random length of the data
l = randi(500, 10, 1) + 50;
% Random "width" of the data, with 3 more likely
w = randi(4, 10, 1);
w(w==4) = 3;
% random "direction" of the data
d = randi(2, 10, 1);
% sigma of the data (in fraction of mean)
sigma = rand(10,1) / 3;
% means of the data
dmean = randi(150,10,1);
dsigma = dmean.*sigma;
for c = 1 : 10
if d(c) == 1
data{c} = randn(l(c), w(c)) .* dsigma(c) + dmean(c);
else
data{c} = randn(w(c), l(c)) .* dsigma(c) + dmean(c);
end
end
Next thing is
On the bar, I want it to be shown the median of the values, and I want to calculate the confidence interval and show it additionally.
Are you really sure you want to plot the median? The median of some data is not connected to the variance of the data, and hus no type of error bars are required. I guess you want to show the mean. If you really want to show the median, a box plot might be a better alternative.
The following code computes and plots the mean in a bar plot:
means = zeros(numel(data),3);
stds = zeros(numel(data),3);
n = zeros(numel(data),3);
for c = 1:numel(data)
d = data{c};
if size(d,1) < size(d,2)
d = d';
end
cols = size(d,2);
means(c, 1:cols) = nanmean(d);
stds(c, 1:cols) = nanstd(d);
n(c, 1:cols) = sum(~isnan((d)));
end
b = bar(means);
Now, we need to compute the length of the error bars. Typical choices are the standard deviation of the data (already computed by the code above, stored in stds), the standard error or the 95% confidence interval (which is the 1.96fold of the standard error, assuming the underlying data follows a normal distribution).
% for standard deviation use stds
% for standard error
ste = stds./sqrt(n);
% for 95% confidence interval
ci95 = 1.96 * ste;
Last thing is to plot the error bars. Here I chose the ci95 as you asked in your question, if you want to change that, simply change the variable in the call to errorbar:
for c = 1:3
size(means(:, c))
size(b(c).XData)
e = errorbar(b(c).XData + b(c).XOffset, means(:,c), ci95(:, c));
e.LineStyle = 'none';
end
I'm relatively new to Matlab, and trying to understand why a piece of code isn't working.
I have a 512x512 image that needs to be downsized to 256, and then resized back up to 512.
How I understand the mathematics, is that I would need to mean the pixels in the image to get the 256, and then sum them back to get the 512. Is that correct ? Following is the code that I'm looking at, and if someone can explain me whats wrong(its giving a blank white image), I would appreciate it:
w = double(imread('walkbridge.tif'));
%read the image
w = w(:,:,1);
for x = 1:256
for y = 1:256
s256(x,y) = (w(2*x,2*y)+ w(2*x,(2*y)-1) + w((2*x)-1,2*y)+ w((2*x)-1,(2*y)-1))/4;
end
end
for x = 1 : 256
for y = 1 : 256
for x1 = 0:1
for y1 = 0:1
R1((2*x)-x1,((2*y)-y1)) = s256(x,y);
end
end
end
end
imshow(R1)
I got your code to work, so you might have some bad values in your image data. Namely, if your image has values in range 0..127 or something similar, it will most likely show as all white. By default, imshow expects color channels to be in range 0..1.
You might also want to simplify your code a bit by indexing the original array instead of accessing individual elements. That way the code is even easy to change:
half_size = 256;
w = magic(2*half_size);
w = w / max(w(:));
figure()
imshow(w)
s = zeros(half_size);
for x = 1:half_size
for y = 1:half_size
ix = w(2*x-1:2*x, 2*y-1:2*y);
s(x,y) = sum(ix(:))/4;
end
end
for x = 1:half_size
for y = 1:half_size
R1(2*x-1:2*x, 2*y-1:2*y) = s(x,y);
end
end
figure()
imshow(R1)
I imagine the calculations could even be vectorised in some way instead of looping, but I didn't bother.
I have a set of three vectors (stored into a 3xN matrix) which are 'entangled' (e.g. some value in the second row should be in the third row and vice versa). This 'entanglement' is based on looking at the figure in which alpha2 is plotted. To separate the vector I use a difference based approach where I calculate the difference of one value with respect the three next values (e.g. comparing (1,i) with (:,i+1)). Then I take the minimum and store that. The method works to separate two of the three vectors, but not for the last.
I was wondering if you guys can share your ideas with me how to solve this problem (if possible). I have added my coded below.
Thanks in advance!
Problem in figures:
clear all; close all; clc;
%%
alpha2 = [-23.32 -23.05 -22.24 -20.91 -19.06 -16.70 -13.83 -10.49 -6.70;
-0.46 -0.33 0.19 2.38 5.44 9.36 14.15 19.80 26.32;
-1.58 -1.13 0.06 0.70 1.61 2.78 4.23 5.99 8.09];
%%% Original
figure()
hold on
plot(alpha2(1,:))
plot(alpha2(2,:))
plot(alpha2(3,:))
%%% Store start values
store1(1,1) = alpha2(1,1);
store2(1,1) = alpha2(2,1);
store3(1,1) = alpha2(3,1);
for i=1:size(alpha2,2)-1
for j=1:size(alpha2,1)
Alpha1(j,i) = abs(store1(1,i)-alpha2(j,i+1));
Alpha2(j,i) = abs(store2(1,i)-alpha2(j,i+1));
Alpha3(j,i) = abs(store3(1,i)-alpha2(j,i+1));
[~, I] = min(Alpha1(:,i));
store1(1,i+1) = alpha2(I,i+1);
[~, I] = min(Alpha2(:,i));
store2(1,i+1) = alpha2(I,i+1);
[~, I] = min(Alpha3(:,i));
store3(1,i+1) = alpha2(I,i+1);
end
end
%%% Plot to see if separation worked
figure()
hold on
plot(store1)
plot(store2)
plot(store3)
Solution using extrapolation via polyfit:
The idea is pretty simple: Iterate over all positions i and use polyfit to fit polynomials of degree d to the d+1 values from F(:,i-(d+1)) up to F(:,i). Use those polynomials to extrapolate the function values F(:,i+1). Then compute the permutation of the real values F(:,i+1) that fits those extrapolations best. This should work quite well, if there are only a few functions involved. There is certainly some room for improvement, but for your simple setting it should suffice.
function F = untangle(F, maxExtrapolationDegree)
%// UNTANGLE(F) untangles the functions F(i,:) via extrapolation.
if nargin<2
maxExtrapolationDegree = 4;
end
extrapolate = #(f) polyval(polyfit(1:length(f),f,length(f)-1),length(f)+1);
extrapolateAll = #(F) cellfun(extrapolate, num2cell(F,2));
fitCriterion = #(X,Y) norm(X(:)-Y(:),1);
nFuncs = size(F,1);
nPoints = size(F,2);
swaps = perms(1:nFuncs);
errorOfFit = zeros(1,size(swaps,1));
for i = 1:nPoints-1
nextValues = extrapolateAll(F(:,max(1,i-(maxExtrapolationDegree+1)):i));
for j = 1:size(swaps,1)
errorOfFit(j) = fitCriterion(nextValues, F(swaps(j,:),i+1));
end
[~,j_bestSwap] = min(errorOfFit);
F(:,i+1) = F(swaps(j_bestSwap,:),i+1);
end
Initial solution: (not that pretty - Skip this part)
This is a similar solution that tries to minimize the sum of the derivatives up to some degree of the vector valued function F = #(j) alpha2(:,j). It does so by stepping through the positions i and checks all possible permutations of the coordinates of i to get a minimal seminorm of the function F(1:i).
(I'm actually wondering right now if there is any canonical mathematical way to define the seminorm so we get our expected results... I initially was going for the H^1 and H^2 seminorms, but they didn't quite work...)
function F = untangle(F)
nFuncs = size(F,1);
nPoints = size(F,2);
seminorm = #(x,i) sum(sum(abs(diff(x(:,1:i),1,2)))) + ...
sum(sum(abs(diff(x(:,1:i),2,2)))) + ...
sum(sum(abs(diff(x(:,1:i),3,2)))) + ...
sum(sum(abs(diff(x(:,1:i),4,2))));
doSwap = #(x,swap,i) [x(:,1:i-1), x(swap,i:end)];
swaps = perms(1:nFuncs);
normOfSwap = zeros(1,size(swaps,1));
for i = 2:nPoints
for j = 1:size(swaps,1)
normOfSwap(j) = seminorm(doSwap(F,swaps(j,:),i),i);
end
[~,j_bestSwap] = min(normOfSwap);
F = doSwap(F,swaps(j_bestSwap,:),i);
end
Usage:
The command alpha2 = untangle(alpha2); will untangle your functions:
It should even work for more complicated data, like these shuffled sine-waves:
nPoints = 100;
nFuncs = 5;
t = linspace(0, 2*pi, nPoints);
F = bsxfun(#(a,b) sin(a*b), (1:nFuncs).', t);
for i = 1:nPoints
F(:,i) = F(randperm(nFuncs),i);
end
Remark: I guess if you already know that your functions will be quadratic or some other special form, RANSAC would be a better idea for larger number of functions. This could also be useful if the functions are not given with the same x-value spacing.
I'm taking images using a tunneling microscope. However, the scope is drifting between successive images. I'm trying to use MatLab to calculate the offset between images. The code below calculates in seconds for small images (e.g. 64x64 pixels), but takes >2 hrs to handle the 512x512 pixel images I'm dealing with. Do you have any suggestions for speeding up this code? Or do you know of better ways to track images in MatLab? Thanks for your help!
%Test templates
template = .5*ones(32);
template(25:32,:) = 0;
template(:,25:64) = 0;
data_A = template;
close all
imshow(data_A);
template(9:32,41:64) = .5;
template(:,1:24) = 0;
data_B = template;
figure, imshow(data_B);
tic
[m n] = size(data_B);
z = [];
% Loop over all possible displacements
for x = -n:n
for y = -m:m
paddata_B = data_B;
ax = abs(x);
zerocols = zeros(m,ax);
if x > 0
paddata_B(:,1:ax) = [];
paddata_B = [paddata_B zerocols];
else
paddata_B(:,(n-ax+1):n) = [];
paddata_B = [zerocols paddata_B];
end
ay = abs(y);
zerorows = zeros(ay,n);
if y < 0
paddata_B(1:ay,:) = [];
paddata_B = vertcat(paddata_B, zerorows);
else
paddata_B((m-ay+1):m,:) = [];
paddata_B = vertcat(zerorows, paddata_B);
end
% Full matrix sum after array multiplication
C = paddata_B.*data_A;
matsum = sum(sum(C));
% Populate array of matrix sums for each displacement
z(x+n+1, y+m+1) = matsum;
end
end
toc
% Plot matrix sums
figure, surf(z), shading flat
% Find maximum value of z matrix
[max_z, imax] = max(abs(z(:)));
[xpeak, ypeak] = ind2sub(size(z),imax(1))
% Calculate displacement in pixels
corr_offset = [(xpeak-n-1) (ypeak-m-1)];
xoffset = corr_offset(1)
yoffset = corr_offset(2)
What you're calculating is known as the cross-correlation of the two images. You can calculate the cross-correlation of all offsets at once using Discrete Fourier Transforms (DFT or FFT). So try something like
z = ifft2( fft2(dataA) .* fft2(dataB).' );
If you pad with zeros in the Fourier domain, you can even use this sort of math to get offsets in fractions of a pixel, and apply offsets of fractions of a pixel to an image.
A typical approach to this kind of problem is to use the fact that it works quickly for small images to your advantage. When you have large images, decimate them to make small images. Register the small images quickly and use the computed offset as your initial value for the next iteration. In the next iteration, you don't decimate the images as much, but you're starting with a good initial estimate of the offset so you can constrain your search for solutions to a small neighborhood near your initial estimate.
Although not written with tunneling microscopes in mind, a review paper that may be of some assistance is: "Mutual Information-Based Registration of Medical Images: A Survey" by Pluim, Maintz, and Viergever published in IEEE Transactions on Medical Imaging, Vol. 22, No. 8, p. 986.
below link will help you find transformation between 2 images and correct/recover the distorted (in your case, image with offset)
http://in.mathworks.com/help/vision/ref/estimategeometrictransform.html
index_pairs = matchFeatures(featuresOriginal,featuresDistorted, 'unique', true);
matchedPtsOriginal = validPtsOriginal(index_pairs(:,1));
matchedPtsDistorted = validPtsDistorted(index_pairs(:,2));
[tform,inlierPtsDistorted,inlierPtsOriginal] = estimateGeometricTransform(matchedPtsDistorted,matchedPtsOriginal,'similarity');
figure; showMatchedFeatures(original,distorted,inlierPtsOriginal,inlierPtsDistorted);
The inlierPtsDistored, inlierPtsOriginal have attributes called locations.
These are nothing but matching locations of one image on another. I think from that point it is very easy to calculate offset.
The function below was my attempt to compute the cross-correlation of the two images manually. Something's not quite right though. Will look at it again this weekend if I have time. You can call the function with something like:
>> oldImage = rand(64);
>> newImage = circshift(oldImage, floor(64/2)*[1 1]);
>> offset = detectOffset(oldImage, newImage, 10)
offset =
32 -1
function offset = detectOffset(oldImage, newImage, margin)
if size(oldImage) ~= size(newImage)
offset = [];
error('Test images must be the same size.');
end
[imageHeight, imageWidth] = size(oldImage);
corr = zeros(2 * imageHeight - 1, 2 * imageWidth - 1);
for yIndex = [1:2*imageHeight-1; ...
imageHeight:-1:1 ones(1, imageHeight-1); ...
imageHeight*ones(1, imageHeight) imageHeight-1:-1:1];
oldImage = circshift(oldImage, [1 0]);
for xIndex = [1:2*imageWidth-1; ...
imageWidth:-1:1 ones(1, imageWidth-1); ...
imageWidth*ones(1, imageWidth) imageWidth-1:-1:1];
oldImage = circshift(oldImage, [0 1]);
numPoint = abs(yIndex(3) - yIndex(2) + 1) * abs(xIndex(3) - xIndex(2) + 1);
corr(yIndex(1),xIndex(1)) = sum(sum(oldImage(yIndex(2):yIndex(3),xIndex(2):xIndex(3)) .* newImage(yIndex(2):yIndex(3),xIndex(2):xIndex(3)))) * imageHeight * imageWidth / numPoint;
end
end
[value, yOffset] = max(corr(margin+1:end-margin,margin+1:end-margin));
[dummy, xOffset] = max(value);
offset = [yOffset(xOffset)+margin-imageHeight xOffset+margin-imageWidth];