I am brand new to MATLAB but am trying to do some image compression code for grayscale images.
Questions
How can I use SVD to trim off low-valued eigenvalues to reconstruct a compressed image?
Work/Attempts so far
My code so far is:
B=imread('images1.jpeg');
B=rgb2gray(B);
doubleB=double(B);
%read the image and store it as matrix B, convert the image to a grayscale
photo and convert the matrix to a class 'double' for values 0-255
[U,S,V]=svd(doubleB);
This allows me to successfully decompose the image matrix with eigenvalues stored in variable S.
How do I truncate S (which is 167x301, class double)? Let's say of the 167 eigenvalues I want to take only the top 100 (or any n really), how do I do that and reconstruct the compressed image?
Updated code/thoughts
Instead of putting a bunch of code in the comments section, this is the current draft I have. I have been able to successfully create the compressed image by manually changing N, but I would like to do 2 additional things:
1- Show a pannel of images for various compressions (i/e, run a loop for N = 5,10,25, etc.)
2- Somehow calculate the difference (error) between each image and the original and graph it.
I am horrible with understanding loops and output, but this is what I have tried:
B=imread('images1.jpeg');
B=rgb2gray(B);
doubleB=im2double(B);%
%read the image and store it as matrix B, convert the image to a grayscale
%photo and convert the image to a class 'double'
[U,S,V]=svd(doubleB);
C=S;
for N=[5,10,25,50,100]
C(N+1:end,:)=0;
C(:,N+1:end)=0;
D=U*C*V';
%Use singular value decomposition on the image doubleB, create a new matrix
%C (for Compression diagonal) and zero out all entries above N, (which in
%this case is 100). Then construct a new image, D, by using the new
%diagonal matrix C.
imshow(D);
error=C-D;
end
Obviously there are some errors because I don't get multiple pictures or know how to "graph" the error matrix
Although this question is old, it has helped me a lot to understand SVD. I have modified the code you have written in your question to make it work.
I believe you might have solved the problem, however just for the future reference for anyone visiting this page, I am including the complete code here with the output images and graph.
Below is the code:
close all
clear all
clc
%reading and converting the image
inImage=imread('fruits.jpg');
inImage=rgb2gray(inImage);
inImageD=double(inImage);
% decomposing the image using singular value decomposition
[U,S,V]=svd(inImageD);
% Using different number of singular values (diagonal of S) to compress and
% reconstruct the image
dispEr = [];
numSVals = [];
for N=5:25:300
% store the singular values in a temporary var
C = S;
% discard the diagonal values not required for compression
C(N+1:end,:)=0;
C(:,N+1:end)=0;
% Construct an Image using the selected singular values
D=U*C*V';
% display and compute error
figure;
buffer = sprintf('Image output using %d singular values', N)
imshow(uint8(D));
title(buffer);
error=sum(sum((inImageD-D).^2));
% store vals for display
dispEr = [dispEr; error];
numSVals = [numSVals; N];
end
% dislay the error graph
figure;
title('Error in compression');
plot(numSVals, dispEr);
grid on
xlabel('Number of Singular Values used');
ylabel('Error between compress and original image');
Applying this to the following image:
Gives the following result with only first 5 Singular Values,
with first 30 Singular Values,
and the first 55 Singular Values,
The change in error with increasing number of singular values can be seen in the graph below.
Here you can notice the graph is showing that using approximately 200 first singular values yields to approximately zero error.
Just to start, I assume you're aware that the SVD is really not the best tool to decorrelate the pixels in a single image. But it is good practice.
OK, so we know that B = U*S*V'. And we know S is diagonal, and sorted by magnitude. So by using only the top few values of S, you'll get an approximation of your image. Let's say C=U*S2*V', where S2 is your modified S. The sizes of U and V haven't changed, so the easiest thing to do for now is to zero the elements of S that you don't want to use, and run the reconstruction. (Easiest way to do this: S2=S; S2(N+1:end, :) = 0; S2(:, N+1:end) = 0;).
Now for the compression part. U is full, and so is V, so no matter what happens to S2, your data volume doesn't change. But look at what happens to U*S2. (Plot the image). If you kept N singular values in S2, then only the first N rows of S2 are nonzero. Compression! Except you still have to deal with V. You can't use the same trick after you've already done (U*S2), since more of U*S2 is nonzero than S2 was by itself. How can we use S2 on both sides? Well, it's diagonal, so use D=sqrt(S2), and now C=U*D*D*V'. So now U*D has only N nonzero rows, and D*V' has only N nonzero columns. Transmit only those quantities, and you can reconstruct C, which is approximately like B.
For example, here's a 512 x 512 B&W image of Lena:
We compute the SVD of Lena. Choosing the singular values above 1% of the maximum singular value, we are left with just 53 singular values. Reconstructing Lena with these singular values and the corresponding (left and right) singular vectors, we obtain a low-rank approximation of Lena:
Instead of storing 512 * 512 = 262144 values (each taking 8 bits), we can store 2 x (512 x 53) + 53 = 54325 values, which is approximately 20% of the original size. This is one example of how SVD can be used to do lossy image compression.
Here's the MATLAB code:
% open Lena image and convert from uint8 to double
Lena = double(imread('LenaBW.bmp'));
% perform SVD on Lena
[U,S,V] = svd(Lena);
% extract singular values
singvals = diag(S);
% find out where to truncate the U, S, V matrices
indices = find(singvals >= 0.01 * singvals(1));
% reduce SVD matrices
U_red = U(:,indices);
S_red = S(indices,indices);
V_red = V(:,indices);
% construct low-rank approximation of Lena
Lena_red = U_red * S_red * V_red';
% print results to command window
r = num2str(length(indices));
m = num2str(length(singvals));
disp(['Low-rank approximation used ',r,' of ',m,' singular values']);
% save reduced Lena
imwrite(uint8(Lena_red),'Reduced Lena.bmp');
taking the first n max number of eigenvalues and their corresponding eigenvectors may solve your problem.For PCA, the original data multiplied by the first ascending eigenvectors will construct your image by n x d where d represents the number of eigenvectors.
Related
.
I'm to take an image, convert it into a set of 3 matrices using imread(), then calculate a truncated-sum approximation to each matrix using N=1,2,3,4,8,16,32,64,128 terms. I have the matrices, but I'm not really sure about that last part and the reading is a bit vague. What do they mean by a truncated-sum approximation?
Update based on a given answer:
I tried the following:
A = double(imread("image.jpg"))/255;
[U1, S1, V1] = svd(A(:,:,1));
[U2, S2, V2] = svd(A(:,:,2));
[U3, S3, V3] = svd(A(:,:,3));
N = 128;
trunc_image = (U1(1:763,1:N)*S1(1:N,1:N)*V1(1:N,1:691))*255;
imwrite(trunc_image, "truncimg.jpg", "jpg");
...but the resulting image looks like this:
When you perform svd on an image I:
[U,S,V] = svd(I,'econ'); %//you get matrices U, S, V
S will be a diagonal matrix, with decreasing singular values along the diagonals.
Approximation by truncating... means that I can reconstruct I' by zeroing out singular values in S:
I_recon = U(1:256,1:N)*S(1:N,1:N)*V(1:256,1:N).'; %//Reconstruct by keeping the first N singular values in S.
What happens here is that I_recon is an image reconstructed from the N most significant singular values. The purpose of doing this is so that we can remove less significant contributions to the image I, and represent I with less data.
This is an example of reconstructed images with varying N:
So I need to take the derivative of an image in the x-direction for this assignment, with the goal of getting some form of gradient. My thought is to use the diff(command) on each row of the image and then apply a Gaussian filter. I haven't started the second part because the first is giving me trouble. In attempting to get the x-derivative I have:
origImage = imread('TightRope.png');
for h = 1:3 %%h represents color channel
for i = size(origImage,1)
newImage(i,:,h) = diff(origImage(i,:,h)); %%take derivative of row and translate to new row
end
end
The issue is somewhere along the way I get the error 'Subscripted assignment dimension mismatch.'.
Error in Untitled2 (line 14)
newImage(i,:,h) = diff(origImage(i,:,h));
Does anyone have any ideas on why that might be happening and if my approach is correct for getting the gradient/gaussian derivative?
Why not use fspecial along with imfilter instead?
figure;
I = imread('cameraman.tif');
subplot 131; imshow(I); title('original')
h = fspecial('prewitt');
derivative = imfilter(I,h','replicate'); %'
subplot 132; imshow(derivative); title('derivative')
hsize = 5;
sigma = 1;
h = fspecial('gaussian', hsize, sigma) ;
gaussian = imfilter(derivative,h','replicate'); %'
subplot 133; imshow(gaussian); title('derivative + gaussian')
The result is the following one:
If your goal is to use diff to generate the derivative rather than to create a loop, you can just tell diff to give you the derivative in the x-direction (along dimension 2):
newImage = diff(double(origImage), 1, 2);
The 1 is for the first derivative and 2 is for the derivative along the second dimension. See diff.
As #rayryeng mentions in his answer, it's important to cast the image as double.
Given a N element vector, diff returns a N-1 length vector, so the reason why you are getting an alignment mismatch is because you are trying to assign the output of diff into an incorrect number of slots. Concretely, supposing that N is the total number of columns, you are using diff on a 1 X N vector which thus returns a 1 x (N - 1) vector and you are trying to assign this output as a single row into the output image which is expected to be 1 x N. The missing element is causing the alignment mismatch. diff works by taking pairs of elements in the vector and subtracting them to produce new elements, thus the reason why there is one element missing in the final output.
If you want to get your code working, one way is to pad each row of the image or signal vector with an additional zero (for example) as input into diff. Something like this could work. Take note that I'll be converting your image to double to allow the derivative to take on negative values:
origImage = imread('...'); %// Place path to image here and read in
origImage = im2double(origImage); %// Change - Convert to double precision
newImage = zeros(size(origImage)); %// Change - Create blank new image and populate each row per channel manually
for h = 1:3 %%h represents color channel
for ii = 1:size(origImage,1) %// Change - fixed for loop iteration
newImage(ii,:,h) = diff([0 origImage(ii,:,h)]); %// Change
end
end
Take note that your for loop was incorrect since it didn't go over every row... just the last row.
When I use the onion.png image that's part of the image processing toolbox:
...and when I run this code, I get this image using imshow(newImage,[]);:
Take note that the difference filter was applied to each channel individually and I changed the intensities per channel so that the smallest value gets mapped to 0 and the largest value gets mapped to 1. How you can interpret this image is that any areasthat have a non-black colour have some non-zero differences and hence there is some activity going on in those areas and any areas that have a dark / black colour means that there is no activity going on in those areas. Take note that we applied a horizontal filter, so if you wanted to do this vertically, you'd simply repeat the behaviour but apply this column-wise instead of row-wise as you did above.
I have a randomly defined H matrix of size 600 x 10. Each element in this matrix H can be represented as H(k,t). I obtained a speech spectrogram S which is 600 x 597. I obtained it using Mel features, so it should be 40 x 611 but then I used a frame stacking concept in which I stacked 15 frames together. Therefore it gave me (40x15) x (611-15+1) which is 600 x 597.
Now I want to obtain an output matrix Y which is given by the equation based on convolution Y(k,t) = ∑ H(k,τ)S(k,t-τ). The sum goes from τ=0 to τ=Lh-1. Lh in this case would be 597.
I don't know how to obtain Y. Also, my doubt is the indexing into both H and S when computing the convolution. Specifically, for Y(1,1), we have:
Y(1,1) = H(1,0)S(1,1) + H(1,1)S(1,0) + H(1,2)S(1,-1) + H(1,3)S(1,-2) + ...
Now, there is no such thing as negative indices in MATLAB - for example, S(1,-1) S(1,-2) and so on. So, what type of convolution should I use to obtain Y? I tried using conv2 or fftfilt but I think that will not give me Y because Y must also be the size of S.
That's very easy. That's a convolution on a 2D signal only being applied to 1 dimension. If we assume that the variable k is used to access the rows and t is used to access the columns, you can consider each row of H and S as separate signals where each row of S is a 1D signal and each row of H is a convolution kernel.
There are two ways you can approach this problem.
Time domain
If you want to stick with time domain, the easiest thing would be to loop over each row of the output, find the convolution of each pair of rows of S and H and store the output in the corresponding output row. From what I can tell, there is no utility that can convolve in one dimension only given an N-D signal.... unless you go into frequency domain stuff, but let's leave that for later.
Something like:
Y = zeros(size(S));
for idx = 1 : size(Y,1)
Y(idx,:) = conv(S(idx,:), H(idx,:), 'same');
end
For each row of the output, we perform a row-wise convolution with a row of S and a row of H. I use the 'same' flag because the output should be the same size as a row of S... which is the bigger row.
Frequency domain
You can also perform the same computation in frequency domain. If you know anything about the properties of convolution and the Fourier Transform, you know that convolution in time domain is multiplication in the frequency domain. You take the Fourier Transform of both signals, multiply them element-wise, then take the Inverse Fourier Transform back.
However, you need to keep the following intricacies in mind:
Performing a full convolution means that the final length of the output signal is length(A)+length(B)-1, assuming A and B are 1D signals. Therefore, you need to make sure that both A and B are zero-padded so that they both match the same size. The reason why you make sure that the signals are the same size is to allow for the multiplication operation to work.
Once you multiply the signals in the frequency domain then take the inverse, you will see that each row of Y is the full length of the convolution. To ensure that you get an output that is the same size as the input, you need to trim off some points at the beginning and at the end. Specifically, since each kernel / column length of H is 10, you would have to remove the first 5 and last 5 points of each signal in the output to match what you get in the for loop code.
Usually after the inverse Fourier Transform, there are some residual complex coefficients due to the nature of the FFT algorithm. It's good practice to use real to remove the complex valued parts of the results.
Putting all of this theory together, this is what the code would look like:
%// Define zero-padded H and S matrices
%// Rows are the same, but columns must be padded to match point #1
H2 = zeros(size(H,1), size(H,2)+size(S,2)-1);
S2 = zeros(size(S,1), size(H,2)+size(S,2)-1);
%// Place H and S at the beginning and leave the rest of the columns zero
H2(:,1:size(H,2)) = H;
S2(:,1:size(S,2)) = S;
%// Perform Fourier Transform on each row separately of padded matrices
Hfft = fft(H2, [], 2);
Sfft = fft(S2, [], 2);
%// Perform convolution
Yfft = Hfft .* Sfft;
%// Take inverse Fourier Transform and convert to real
Y2 = real(ifft(Yfft, [], 2));
%// Trim off unnecessary values
Y2 = Y2(:,size(H,2)/2 + 1 : end - size(H,2)/2 + 1);
Y2 should be the convolved result and should match Y in the previous for loop code.
Comparison between them both
If you actually want to compare them, we can. What we'll need to do first is define H and S. To reconstruct what I did, I generated random values with a known seed:
rng(123);
H = rand(600,10);
S = rand(600,597);
Once we run the above code for both the time domain version and frequency domain version, let's see how they match up in the command prompt. Let's show the first 5 rows and 5 columns:
>> format long g;
>> Y(1:5,1:5)
ans =
1.63740867892464 1.94924208172753 2.38365646354643 2.05455605619097 2.21772526557861
2.04478411247085 2.15915645246324 2.13672842742653 2.07661341840867 2.61567534623066
0.987777477630861 1.3969752201781 2.46239452105228 3.07699790208937 3.04588738611503
1.36555260994797 1.48506871890027 1.69896157726456 1.82433906982894 1.62526864072424
1.52085236885395 2.53506897420001 2.36780282057747 2.22335617436888 3.04025523335182
>> Y2(1:5,1:5)
ans =
1.63740867892464 1.94924208172753 2.38365646354643 2.05455605619097 2.21772526557861
2.04478411247085 2.15915645246324 2.13672842742653 2.07661341840867 2.61567534623066
0.987777477630861 1.3969752201781 2.46239452105228 3.07699790208937 3.04588738611503
1.36555260994797 1.48506871890027 1.69896157726456 1.82433906982894 1.62526864072424
1.52085236885395 2.53506897420001 2.36780282057747 2.22335617436888 3.04025523335182
Looks good to me! As another measure, let's figure out what the largest difference is between one value in Y and a corresponding value in Y2:
>> max(abs(Y(:) - Y2(:)))
ans =
5.32907051820075e-15
That's saying that the max error seen between both outputs is in the order of 10-15. I'd say that's pretty good.
Data: Say I have a 2000 rows by 500 column matrix (image)
What I need: Compute the FFT of 64 rows by 10 column chunks of above data. In other words, I want to compute the FFT of 64X10 window that is run across the entire data matrix. The FFT result is used to compute a scalar value (say peak amplitude frequency) which is used to create a new "FFT value" image.
Now, I need the final FFT image to be the same size as the original data (2000 X 500).
What is the fastest way to accomplish this in MATLAB? I am currently using for loops which is relatively slow. Also I use interpolation to size up the final image to the original data size.
As #EitanT pointed out, you can use blockproc for batch block processing of an image J. However you should define your function handle as
fun = #(block_struct) fft2(block_struct.data);
B = blockproc(J, [64 10], fun);
For a [2000 x 500] matrix this will give you a [2000 x 500] output of complex Fourier values, evaluated at sub-sampled pixel locations with a local support (size of the input to FFT) of [64 x 10]. Now, to replace those values with a single, e.g. with the peak log-magnitude, you can further specify
fun = #(block_struct) max(max(log(abs(fft2(block_struct.data)))));
B = blockproc(J, [64 10], fun);
The output then is a [2000/64 x 500/10] output of block-patch values, which you can resize by nearest-neighbor interpolation (or something else for smoother versions) to the desired [2000 x 500] original size
C = imresize(B, [2000 500], 'nearest');
I can include a real image example if it will further help.
Update: To get overlapping blocks you can use the 'Bordersize' option of blockproc by setting the overlap [V H] such that the final windows of size [M + 2*V, N + 2*H] will still be [64, 10] in size. Example:
fun = #(block_struct) log(abs(fft2(block_struct.data)));
V = 16; H = 3; % overlap values
overlap = [V H];
M = 32; N = 4; % non-overlapping values
B1 = blockproc(J, [M N], fun, 'BorderSize', overlap); % final windows are 64 x 10
However, this will work with keeping the full Fourier response, not the single-value version with max(max()) above.
See also this post for filtering using blockproc:Dealing with “Really Big” Images: Block Processing.
If you want to apply the same function (in your case, the 2-D Fourier transform) on individual distinct blocks in a larger matrix, you can do that with the blkproc function, which is replaced in newer MATLAB releases by blockproc.
However, I infer that you wish to apply apply fft2 on overlapping blocks in a "sliding window" fashion. For this purpose you can use colfilt with the 'sliding' option. Note that the function that we're applying on each block is the fft:
block_size = [64, 10];
temp_size = 5 * block_size;
col_func = #(x)cellfun(#(y)max(max(abs(fft2(y)))), num2cell(x, 1), 'Un', 0);
B = colfilt(A, block_size, 10 * block_size, 'sliding', col_func);
How does this work? colfilt processes the matrix A by rearranging each "sliding" block into a separate column of a new temporary matrix, and then applying the col_func to this new matrix. col_func in turn restores each column into the original block and applies fft2 on it, returning the largest amplitude value for each column.
Important things to note:
Since this mentioned temporary matrix includes all possible "sliding" blocks, memory could be a limitation. Therefore, in order to use less memory in calculations, colfilt breaks up the original matrix A into sub-matrices of temp_size, and performs calculations separately on each. The resulting matrix B is still the same, of course.
Each element in the resulting matrix B is computed from the corresponding block neighborhood. The larger your image is, the more blocks you will need to process, so the computation time will increase geometrically. I believe that you'll have to wait quite a bit until MATLAB finishes processing all sliding windows on your 2000-by-500 matrix.
i have 100 b&w image of smthing.the probllem is i want to scan each image in 0&1 formatin mby n format and then place each image to one over one and again scan and save them in mbynby100 form.
how i do this and from where i should start
_jaysean
Your question is vague and hard to understand, but my guess is that you want to take 100 M-by-N grayscale intensity images, threshold them to create logical matrices (i.e. containing zeroes and ones), then put them together into one M-by-N-by-100 matrix. You can do the thresholding by simply picking a threshold value yourself, like 0.5, and applying it to an image A as follows:
B = A > 0.5;
The matrix B will now be an M-by-N logical matrix with ones where A is greater than 0.5 and zeroes where A is less than or equal to 0.5.
If you have the Image Processing Toolbox, you could instead use the function GRAYTHRESH to pick a threshold and the function IM2BW to apply it:
B = im2bw(A,graythresh(A));
Once you do this, you can easily put the images into an M-by-N-by-100 logical matrix. Here's an example of how you could do this in a loop, assuming the variables M and N are defined:
allImages = false(M,N,100); %# Initialize the matrix to store all the images
for k = 1:100
%# Here, you would load your image into variable A
allImages(:,:,k) = im2bw(A,graythresh(A)); %# Threshold A and add it to
%# the matrix allImages
end