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.
Related
i'm trying to extract HOG_features for mathematical symbols classification (i will use SVM classifier). I get a 1xn vector then i have to put all the vectors in a single matrix. The problem is that the size of the feature vector is different for each image so I can't concatenate them.
Is there a way to make all vectors having the same size ?
Thank you in advance.
Here is the code:
rep1 = 'D:\mémoire MASTER\data';
ext = '*.tif' ;
chemin = fullfile(rep1, ext);
list = dir(chemin);
for i=1:length(list)
I = imread(fullfile(rep1, list(i).name), ext(3:end));
if size(I,3)==3 % RGB image
I = rgb2gray(I);
end
I1 = imbinarize(I);
% Extract HOG features data
HOG_feat = extractHOGFeatures(I1,'CellSize', [2 2]);
HOG_feat1 = HOG_feat';
end
You can pad each one with zeros to be as long as the longest one:
e.g. to put two vectors, v1 and v2, into a matrix M:
M = zeros(2,max(length(v1),length(v2)));
M(1,1:length(v1)) = v1;
M(2,1:length(v2)) = v2;
You have the problem that all your vectors are a different size. Instead of trying to coerce them to be the sesame size by zero-padding or interpolating (both I think are bad ideas), change your computation so that the length of the output vector does not depend on the size of the image.
This is your current code:
HOG_feat = extractHOGFeatures(I1,'CellSize', [2 2]);
% ^^^
% the image is split in cells of 2x2 pixels
2x2 cells are way too small for this method anyway. You could instead divide your image into a set number of cells, say 100 cells:
cellSize = ceil(size(I1)/10);
HOG_feat = extractHOGFeatures(I1,'CellSize', cellSize);
(I’m using ceil in the division because I figure it’s necessary to have an integer size. But I’m not sure whether ceil or floor or round is needed here, and I don’t have access to this function to test it. A bit of trial and error should show which method gives consistent output size.)
I need to compute a moving average over a data series, within a for loop. I have to get the moving average over N=9 days. The array I'm computing in is 4 series of 365 values (M), which itself are mean values of another set of data. I want to plot the mean values of my data with the moving average in one plot.
I googled a bit about moving averages and the "conv" command and found something which i tried implementing in my code.:
hold on
for ii=1:4;
M=mean(C{ii},2)
wts = [1/24;repmat(1/12,11,1);1/24];
Ms=conv(M,wts,'valid')
plot(M)
plot(Ms,'r')
end
hold off
So basically, I compute my mean and plot it with a (wrong) moving average. I picked the "wts" value right off the mathworks site, so that is incorrect. (source: http://www.mathworks.nl/help/econ/moving-average-trend-estimation.html) My problem though, is that I do not understand what this "wts" is. Could anyone explain? If it has something to do with the weights of the values: that is invalid in this case. All values are weighted the same.
And if I am doing this entirely wrong, could I get some help with it?
My sincerest thanks.
There are two more alternatives:
1) filter
From the doc:
You can use filter to find a running average without using a for loop.
This example finds the running average of a 16-element vector, using a
window size of 5.
data = [1:0.2:4]'; %'
windowSize = 5;
filter(ones(1,windowSize)/windowSize,1,data)
2) smooth as part of the Curve Fitting Toolbox (which is available in most cases)
From the doc:
yy = smooth(y) smooths the data in the column vector y using a moving
average filter. Results are returned in the column vector yy. The
default span for the moving average is 5.
%// Create noisy data with outliers:
x = 15*rand(150,1);
y = sin(x) + 0.5*(rand(size(x))-0.5);
y(ceil(length(x)*rand(2,1))) = 3;
%// Smooth the data using the loess and rloess methods with a span of 10%:
yy1 = smooth(x,y,0.1,'loess');
yy2 = smooth(x,y,0.1,'rloess');
In 2016 MATLAB added the movmean function that calculates a moving average:
N = 9;
M_moving_average = movmean(M,N)
Using conv is an excellent way to implement a moving average. In the code you are using, wts is how much you are weighing each value (as you guessed). the sum of that vector should always be equal to one. If you wish to weight each value evenly and do a size N moving filter then you would want to do
N = 7;
wts = ones(N,1)/N;
sum(wts) % result = 1
Using the 'valid' argument in conv will result in having fewer values in Ms than you have in M. Use 'same' if you don't mind the effects of zero padding. If you have the signal processing toolbox you can use cconv if you want to try a circular moving average. Something like
N = 7;
wts = ones(N,1)/N;
cconv(x,wts,N);
should work.
You should read the conv and cconv documentation for more information if you haven't already.
I would use this:
% does moving average on signal x, window size is w
function y = movingAverage(x, w)
k = ones(1, w) / w
y = conv(x, k, 'same');
end
ripped straight from here.
To comment on your current implementation. wts is the weighting vector, which from the Mathworks, is a 13 point average, with special attention on the first and last point of weightings half of the rest.
I have a two-dimensional matrix whose elements represent data that can be colormapped and represented as an image. I want to interpolate them, but I get strange behavior at some of the boundaries that I can't explain.
Here is the original image, the image after 3 iterations of the interpolation routine, and the image after 10 interpolation iterations.
Here's my code:
close all
ifactor = 0.9; % Interpolation factor
% Cut out the meaningless stuff at the edges
bshift_i = bshift(1:16, 1:50);
[m, n] = size(bshift_i);
% Plot the initial data using colormapping
figure()
imagesc(bshift_i, [7.5 10.5])
colorbar()
% Main processing loop
for ii = 1:10
% Every iteration, we generate a grid that has the same size as the
% existing data, and another one that whose axis step sizes are
% (ifactor) times smaller than the existing axes.
[b_x, b_y] = meshgrid(1:ifactor^(ii - 1):n, 1:ifactor^(ii - 1):m);
[b_xi, b_yi] = meshgrid(1:ifactor^ii:n, 1:ifactor^ii:m);
% Interpolate our data and reassign to bshift_i so that we can use it
% in the next iteration
bshift_i = interp2(b_x, b_y, bshift_i, b_xi, b_yi);
end
% Plot the interpolated image
figure()
imagesc(bshift_i, [7.5 10.5])
colorbar()
I'm mainly wondering why the blue artifacts at the bottom and right edges occur, and if so, how I can work around/avoid them.
The problem is how you define the x and y ranges for the interpolation.
Let's take a look at 1:ifactor^(ii - 1):m:
For the first iteration, you get 16 values, from 1 to 16.
For the second iteration, you get 17 values, from 1 to 15.4
For the third iteration, you get 19 values, from 1 to 15.58
An this is enough to illustrate the problem. With the second iteration, everything is fine, because your upper interpolation bound is inside the value range. However, for the third iteration, your upper bound is now outside the value range (15.58 > 15.4).
interp2 does not extrapolate, it returns a NaN (after the 3rd iteration, bshift_i(end,end) is NaN). These NaNs get plotted as 0, hence blue.
To fix this, you have to ensure that your x and y range always include the last value. One way to do this could be:
[b_x, b_y] = meshgrid(linspace(1,n,n./(ifactor^(ii - 1))), ...
linspace(1,m,m./(ifactor^(ii - 1))));
[b_xi, b_yi] = meshgrid(linspace(1,n,n./(ifactor^(ii))), ...
linspace(1,m,m./(ifactor^(ii))));
linspace always includes the first and the last element. However, the third input is not the increment, but the number of elements. That's why you'd have to adapt that condition to resemble your approach.
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.
I would like to write a program which plots the points on top of a semicircle on a certain interval and a straight line everywhere else. Something like this: __n__.
I defined a time domain, which was stored as a vector (t = 0:0.01:5). I assumed that I could define the points on the top of the semicircle using elements of the time vector:
if t>=2|t<=2.3
y = sqrt(.15^2-(t-2.15)^2);
but MATLAB produced an error message saying only square matrices can be squared.
I tried to utilize indices to show that I wanted to square an element of the t vector and not the whole vector:
i = [200:230];
for t(200:230)
y = sqrt(.15^2-(t(i)-2.15)^2);
After these failures, I noticed that squaring a square matrix with one column of non-zero elements would produce a new square matrix with a column of the first matrix's elements squared. If there is some way to eliminate the extra columns of zeros after squaring the matrix, I could use that property of matrices to square the values of the t vector.
What is the simplest and most effective way to address this problem?
It sounds like you want to draw a horizontal line with a semicircular "bump" on it. Here's how you can do this:
t = 0:0.01:5; % Create the time vector
y = zeros(size(t)); % Create a zero vector the same size as t
index = find((t >= 2) & (t <= 2.3)); % Find a set of indices into t
y(index) = sqrt(.15^2-(t(index)-2.15).^2); % Add the "bump" to y
y(1:index(1)) = y(index(1)); % Add the line before the "bump"
y(index(end):end) = y(index(end)); % Add the line after the "bump"
In the above solution, the lines before and after the "bump" could be slightly higher or lower than one another (depending on where your samples in t fall). If you want to make sure they are at the same height, you can instead do the following:
index = (t >= 2) & (t <= 2.3); % Find a set of logical indices
y(index) = sqrt(.15^2-(t(index)-2.15).^2); % Add the "bump" to y
% OPTION #1:
y(~index) = y(find(index,1,'first')); % Use the first circle point as the height
% OPTION #2:
y(~index) = y(find(index,1,'last')); % Use the last circle point as the height
Finally, you can plot the line:
plot(t,y);
Hold on, so your question is, you want to square each element of a vector? All you have to do is:
t.^2
The . represents an element-wise operation in MATLAB on a vector or an array.
And secondly, if I understood your problem currently, you want to create a vector y, which contains a function of elements of t such that t>=2 | t <=2.3?
If so, all you have to do is this:
y = sqrt(0.15^2-(t( (t>=2|t<=2.3) )-2.15).^2));
Essentially, I created a logical index (t>=2 | t<=2.3) and used to access only those elements (which I wanted) in t.
Also, I didn't fully understand what you wanted to achieve. Do you want to plot the topmost point (maxima) of a semicircular curve?