I have a simple program which replaces a selected region of one image with the corresponding region in another image. I am trying to use imrect() in conjunction with makeConstrainToRectFcn to select a rectangular ROI which cannot be extended beyond the boundaries of the image.
However, when I run the code, the ROI can initially be drawn to include the areas outside the image frame. This leads to the error: Index exceeds matrix dimensions.
Is there any way that the rectangle cannot be drawn outside the image from the outset? Alternatively, is it possible to ensure that the operation does not execute unless the rectangle is constrained within the axes limits?
Any suggestions would be greatly appreciated.
My code:
% Sample images:
X=imread('office_1.jpg');
Y=imread('office_5.jpg');
figure, imshow(X)
h = imrect;
api = iptgetapi(h);
fcn = makeConstrainToRectFcn('imrect',get(gca,'XLim'),...
get(gca,'YLim'));
api.setPositionConstraintFcn(fcn);
wait(h);
rect = getPosition(h);
x1 =rect(1);
x2 = x1 + rect(3);
y1 =rect(2);
y2 = y1 + rect(4);
Z = X; % Initialize
Z(y1:y2, x1:x2, :) = Y(y1:y2, x1:x2, :);
imshow(Z)
This should do the job:
% Sample images:
X = imread('office_1.jpg');
Y = imread('office_5.jpg');
% Show image X:
figure, imshow(X);
% Define the ROI constraint:
h = imrect();
h.setPositionConstraintFcn(#(p) roi_constraint(p,size(X)));
% Wait for the ROI to be confirmed:
roi = round(wait(h));
x1 = roi(1);
x2 = x1 + roi(3);
y1 = roi(2);
y2 = y1 + roi(4);
% Create the final image Z and display it:
Z = X;
Z(y1:y2,x1:x2,:) = Y(y1:y2,x1:x2,:);
imshow(Z);
% Auxiliary function for ROI constraint:
function p_adj = roi_constraint(p,img_size)
p_adj(1) = max([1 p(1)]);
p_adj(2) = max([1 p(2)]);
p_adj(3) = min([(img_size(2) - 1) p(3)]);
p_adj(4) = min([(img_size(1) - 1) p(4)]);
end
The script has been tested under Matlab 2017a and works as expected. As you can see, the main difference is the way the size constraint is being handled: in your case, it wasn't properly applied before wait was hit, thus returning an invalid rectangle. Also, in order to avoid a wrong offsetting, the round function has been applied to the rectangle.
Related
I have an image and I would like to blur it in one specific direction and distance using Matlab.
I found out there is a filter called fspecial('motion',len,theta).
Here there is an example:
I = imread('cameraman.tif');
imshow(I);
H = fspecial('motion',20,45);
MotionBlur = imfilter(I,H,'replicate');
imshow(MotionBlur);
However the blurred picture is blurred in 2 directions! In this case 225 and 45 degrees.
What should it do in order to blur it just in a specific direction (e.g. 45) and not both?
I think you want what's called a "comet" kernel. I'm not sure what kernel is used for the "motion" blur, but I'd guess that it's symmetrical based on the image you provided.
Here is some code to play with that applies the comet kernel in one direction. You'll have to change things around if you want an arbitrary angle. You can see from the output that it's smearing in one direction, since there is a black band on only one side (due to the lack of pixels there).
L = 5; % kernel width
sigma=0.2; % kernel smoothness
I = imread('cameraman.tif');
x = -L:1.0:L;
[X,Y] = meshgrid(x,x);
H1 = exp((-sigma.*X.^2)+(-sigma.*Y.^2));
kernel = H1/sum((H1(:)));
Hflag = double((X>0));
comet_kernel = Hflag.*H1;
comet_kernel=comet_kernel/sum(comet_kernel(:));
smearedImage = conv2(double(I),comet_kernel,'same');
imshow(smearedImage,[]);
Updated code: This will apply an arbitrary rotation to the comet kernel. Note also the difference between sigma in the previous example and sx and sy here, which control the length and width parameters of the kernel, as suggested by Andras in the comments.
L = 5; % kernel width
sx=3;
sy=10;
theta=0;
I = imread('cameraman.tif');
x = -L:1.0:L;
[X,Y] = meshgrid(x,x);
rX = X.*cos(theta)-Y.*sin(theta);
rY = X.*sin(theta)+Y.*cos(theta);
H1 = exp(-((rX./sx).^2)-((rY./sy).^2));
Hflag = double((0.*rX+rY)>0);
H1 = H1.*Hflag;
comet_kernel = H1/sum((H1(:)))
smearedImage = conv2(double(I),comet_kernel,'same');
imshow(smearedImage,[]);
Based on Anger Density's answer I wrote this code that solves my problem completely:
L = 10; % kernel width
sx=0.1;
sy=100;
THETA = ([0,45,90,135,180,225,270,320,360])*pi/180;
for i=1:length(THETA)
theta=(THETA(i)+pi)*-1;
I = imread('cameraman.tif');
x = -L:1.0:L;
[X,Y] = meshgrid(x,x);
rX = X.*cos(theta)-Y.*sin(theta);
rY = X.*sin(theta)+Y.*cos(theta);
H1 = exp(-((rX./sx).^2)-((rY./sy).^2));
Hflag = double((0.*rX+rY)>0);
H1 = H1.*Hflag;
comet_kernel = H1/sum((H1(:)));
smearedImage = conv2(double(I),comet_kernel,'same');
% Fix edges
smearedImage(:,[1:L, end-L:end]) = I(:,[1:L, end-L:end]); % Left/Right edge
smearedImage([1:L, end-L:end], :) = I([1:L, end-L:end], :); % Top/bottom edge
% Keep only inner blur
smearedImage(L:end-L,L:end-L) = min(smearedImage(L:end-L,L:end-L),double(I(L:end-L,L:end-L)));
figure
imshow(smearedImage,[]);
title(num2str(THETA(i)*180/pi))
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
end
I'm trying to convert an image with many circles with the same center, from Cartesian to Polar (so that the new image will be the circles but lines instead of the circles, see the image below), and that's working out just fine using the following code:
[r, c] = size(img);
r=floor(r/2);
c=floor(c/2);
[X, Y] = meshgrid(-c:c-1,-r:r-1);
[theta, rho] = cart2pol(X, Y);
subplot(221), imshow(img), axis on;
hold on;
subplot(221), plot(xCenter,yCenter, 'r+');
subplot(222), warp(theta, rho, zeros(size(theta)), img);
view(2), axis square;
The problem is, I don't understand why does it even work? (obviously it's not my code), I mean, when I use the function cart2pol I don't even use the image, it's just some vectors x and y generated from the meshgrid function..
and another problem is, I want somehow to have a new image (not just to be able to draw it with the wrap function) which is the original image but by the theta and rho coordinates (meaning the same pixels but rearranged)... I'm not even sure how to ask this, in the end I want to have an image which is a matrix so that I can sum each row and turn the matrix into a column vector...
You can think of your image as being a 2D matrix, where each pixel has an X and Y coordinate
[(1,1) (1,2) (1,3) .... (1,c)]
[(2,1) (2,2) (2,3) .... (2,c)]
[(3,1) (3,2) (3,3) .... (3,c)]
[.... .... .... .... .... ]
[(r,1) (r,2) (r,3) .... (r,c)]
In the code that you posted, it maps each of these (X,Y) coordinates to it's equivalent polar coordinate (R, theta) using the center of the image floor(c/2) and floor(r/2) as the reference point.
% Map pixel value at (1,1) to it's polar equivalent
[r,theta] = cart2pol(1 - floor(r/2),1 - floor(c/2));
So whatever pixel value was used for (1,1) should now appear in your new polar coordinate space at (r,theta). It is important to note that to do this conversion, no information about the actual pixel values in the image matters, rather we just want to perform this transformation for each pixel within the image.
So first we figure out where the center of the image is:
[r, c] = size(img);
r = floor(r / 2);
c = floor(c / 2);
Then we figure out the (X,Y) coordinates for every point in the image (after the center has already been subtracted out
[X, Y] = meshgrid(-c:c-1,-r:r-1);
Now convert all of these cartesian points to polar coordinates
[theta, rho] = cart2pol(X, Y);
All that warp now does, is say "display the value of img at (X,Y) at it's corresponding location in (theta, rho)"
warp(theta, rho, zeros(size(theta)), img);
Now it seems that you want a new 2D image where the dimensions are [nTheta, nRho]. To do this, you could use griddata to interpolate your scattered (theta, rho) image (which is displayed by warp above) to a regular grid.
% These is the spacing of your radius axis (columns)
rhoRange = linspace(0, max(rho(:)), 100);
% This is the spacing of your theta axis (rows)
thetaRange = linspace(-pi, pi, 100);
% Generate a grid of all (theta, rho) coordinates in your destination image
[T,R] = meshgrid(thetaRange, rhoRange);
% Now map the values in img to your new image domain
theta_rho_image = griddata(theta, rho, double(img), T, R);
Take a look at all the interpolation methods for griddata to figure out which is most appropriate for your scenario.
There were a couple other issues (like the rounding of the center) which caused the result to be slightly incorrect. A fully working example is provided below
% Create an image of circles
radii = linspace(0, 40, 10);
rows = 100;
cols = 100;
img = zeros(rows, cols);
for k = 1:numel(radii)
t = linspace(0, 2*pi, 1000);
xx = round((cos(t) * radii(k)) + (cols / 2));
yy = round((sin(t) * radii(k)) + (rows / 2));
toremove = xx > cols | xx < 1 | yy > rows | yy < 1;
inds = sub2ind(size(img), xx(~toremove), yy(~toremove));
img(inds) = 1;
end
[r,c] = size(img);
center_row = r / 2;
center_col = c / 2;
[X,Y] = meshgrid((1:c) - center_col, (1:r) - center_row);
[theta, rho] = cart2pol(X, Y);
rhoRange = linspace(0, max(rho(:)), 1000);
thetaRange = linspace(-pi, pi, 1000);
[T, R] = meshgrid(thetaRange, rhoRange);
theta_rho_image = griddata(theta, rho, double(img), T, R);
figure
subplot(1,2,1);
imshow(img);
title('Original Image')
subplot(1,2,2);
imshow(theta_rho_image);
title('Polar Image')
And the result
I have the following function that is successful in creating a grey circular mask over the image input, such that the new image is a grey border around a circular image. Example: Grey circular mask.
All I want to do is make the mask a very specific green, but I haven't been successful.
Here is the code:
function [newIm] = myCircularMask(im)
%Setting variables
rad = size(im,1)/2.1; %Radius of the circle window
im = double(im);
[rows, cols, planes]= size(im);
newIm = zeros(rows, cols, planes);
%Generating hard-edged circular mask with 1 inside and 0 outside
M = rows;
[X,Y] = meshgrid(-M/2:1:(M-1)/2, -M/2:1:(M-1)/2);
mask = double(zeros(M,M));
mask(X.^2 + Y.^2 < rad^2) = 1;
% Soften edge of mask
gauss = fspecial('gaussian',[12 12],0.1);
mask = conv2(mask,gauss,'same');
% Multiply image by mask, i.e. x1 inside x0 outside
for k=1:planes
newIm(:,:,k) = im(:,:,k).*mask;
end
% Make mask either 0 inside or -127 outside
mask = (abs(mask-1)*127);
% now add mask to image
for k=1:planes
newIm(:,:,k) = newIm(:,:,k)+mask;
end
newIm = floor(newIm)/255;
The type of green I would like to use is of RGB values [59 178 74].
I'm a beginner with MATLAB, so any help would be greatly appreciated.
Cheers!
Steve
After masking your image, create a color version of your mask:
% test with simple mask
mask = ones(10,10);
mask(5:7,5:7)=0;
% invert mask, multiply with rgb-values, make rgb-matrix:
r_green=59/255; g_green=178/255; b_green=74/255;
invmask=(1-mask); % use mask with ones/zeroes
rgbmask=cat(3,invmask*r_green,invmask*g_green,invmask*b_green);
Add this to your masked image.
Edit:
function [newIm] = myCircularMask(im)
%Setting variables
rad = size(im,1)/2.1; %Radius of the circle window
im = double(im);
[rows, cols, planes]= size(im);
newIm = zeros(rows, cols, planes);
%Generating hard-edged circular mask with 1 inside and 0 outside
M = rows;
[X,Y] = meshgrid(-M/2:1:(M-1)/2, -M/2:1:(M-1)/2);
mask = double(zeros(M,M));
mask(X.^2 + Y.^2 < rad^2) = 1;
% Soften edge of mask
gauss = fspecial('gaussian',[12 12],0.1);
mask = conv2(mask,gauss,'same');
% Multiply image by mask, i.e. x1 inside x0 outside
for k=1:planes
newIm(:,:,k) = im(:,:,k).*mask;
end
% Here follows the new code:
% invert mask, multiply with rgb-values, make rgb-matrix:
r_green=59/255; g_green=178/255; b_green=74/255;
invmask=(1-mask); % use mask with ones/zeroes
rgbmask=cat(3,invmask*r_green,invmask*g_green,invmask*b_green);
newIm=newIm+rgbmask;
Note that I haven't been able to test my suggestion, so there might be errors.
I'm trying to plot small images on a larger plot... Actually its isomap algorithm, I got many points, now each point correspond to some image, I know which image is it... The porblem is how to load that image and plot on the graph?
One more thing I have to plot both image and the points, so, basically the images will overlap the points.
Certainly, the type of image given here
Something like this should get you started. You can use the low-level version of the image function to draw onto a set of axes.
% Define some random data
N = 5;
x = rand(N, 1);
y = rand(N, 1);
% Load an image
rgb = imread('ngc6543a.jpg');
% Draw a scatter plot
scatter(x, y);
axis([0 1 0 1]);
% Offsets of image from associated point
dx = 0.02;
dy = 0.02;
width = 0.1;
height = size(rgb, 1) / size(rgb, 2) * width;
for i = 1:N
image('CData', rgb,...
'XData', [x(i)-dx x(i)-(dx+width)],...
'YData', [y(i)-dy y(i)-(dy+height)]);
end
I have read in an image file to MATLAB and I am trying to stretch it in one direction, but a variable amount (sinusoidal). This would create an accordion affect on the image. I have toyed around with imresize, however that only resizes the image linearly. I would like the amount of "stretch" to vary for each image line. I tried to convey this with the following code:
periods = 10; % Number of "stretch" cycles
sz = size(original_image,2)/periods;
s = 0;
x = 0;
for index = 1:periods
B = original_image(:,round(s+1:s+sz));
if mod(index,2) == 0
amp = 1.5;
else
amp = 0.75;
end
xi = size(B,2)*amp;
new_image(:,x+1:x+xi) = imresize(B, [size(B,1) size(B,2)*amp]);
s = s + sz;
x = x+xi;
end
You can see that segments of the image are stretched, then compressed, then stretched, etc, like an accordion. However, each segment has a uniform amount of stretch, whereas I'd like it to be increasing then decreasing as you move along the image.
I have also looked at MATLAB's example of Applying a Sinusoidal Transformation to a Checkerboard which seems very applicable to my problem, however I have been trying and I cannot get this to produce the desired result for my image.
Any help is much appreciated.
UPDATE:
Thank you for Answer #1. I was unable to get it to work for me, but also realized it would resulted in loss of data, as the code only called for certian lines in the original image, and other lines would have been ignored.
After experimenting further, I developed the code below. I used a checkerboard as an example. While combersome, it does get the job done. However, upon trying the script with an actual high-resolution image, it was extremely slow and ended up failing due to running out of memory. I believe this is because of the excessive number of "imresize" commands that are used in loop.
I = checkerboard(10,50);
I = imrotate(I,90);
[X Y] = size(I);
k = 4; % Number of "cycles"
k = k*2;
x = 1;
y = 2;
z = 2;
F = [];
i = 1;
t = 0;
s = 0;
for j = 1:k/2
t = t + 1;
for inc = round(s+1):round(Y/k*t)
Yi = i + 1;
F(:,(x:y)) = imresize(I(:,(inc:inc)),[X Yi]);
x = y + 1;
y = x + z;
z = z + 1;
i = i + 1;
end
y = y - 2;
z = z - 4;
for inc = round(Y/k*t+1):round(Y/k*(t+1))
Yi = i - 1;
F(:,(x:y)) = imresize(I(:,(inc:inc)),[X Yi]);
x = y + 1;
y = x + z;
z = z - 1;
i = i - 1;
end
y = y + 2;
z = z + 4;
s = Y/k*(t+1);
t = t + 1;
end
Fn = imresize(F, [X Y]);
imshow(Fn);
Does anyone know of a simpler way to achieve this? If you run the code above, you can see the effect I am trying to achieve. Unfortunately, my method above does not allow me to adjust the amplitude of the "stretch" either, only the number of "cycles," or frequency. Help on this would also be appreciated. Much thanks!
Here is how I would approach it:
Determine how the coordinate of each point in your Final image F maps into your Initial image I of size (M,N)
Since you want to stretch horizontally only, given a point (xF,yF) in your final image, that point would be (xI,yI) in your initial image where xI and yI can be obtained as follows:
yI = yF;
xI = xF + Lsin(xFK);
Notes:
these equations do not guarantee that xI remains within the range [1:N] so cropping needs to be added
K controls the how many wrinkles you want to have in your accordion effect. For example, if you only want one wrinkle, K would be 2*pi/N
L controls how much stretching you want to apply
Then simply express your image F from image I with the transforms you have in 1.
Putting it all together, the code below creates a sample image I and generates the image F as follows:
% Generate a sample input image
N=500;
xF=1:N;
I=(1:4)'*xF/N*50;
% Set the parameters for your accordion transform
K=2*pi/N;
L=100;
% Apply the transform
F=I(:, round(min(N*ones(1,N), max(ones(1,N), (xF + L*sin(xF*K))))) );
% Display the input and output images side by side
image(I);
figure;
image(F);
If you run this exact code you get:
As you can see, the final image on the right stretches the center part of the image on the left, giving you an accordion effect with one wrinkle.
You can fiddle with K and L and adjust the formula to get the exact effect you want, but note how by expressing the transform in a matrix form MATLAB executes the code in a fraction of second. If there is one take away for you is that you should stay away from for loops and complex processing whenever you can.
Have fun!