MATLAB: image AlphaData changes figure size - image

I have two images: im2 needs to be overlayed over im1 with a constant alpha map of 0.5. Whenever I set the 'AlphaData' for im2 (to reveal im1 below), the size of the figure/axes changes slightly. If my axes occupies the whole figure, some border pixels go missing. The following example demonstrates this problem by two example images, where im2 has a yellow border so you can clearly notice this problem:
close all;
frame_sz = [106 777];
% create a white im1
im1 = ones([frame_sz 3]);
% create im2 with a yellow border
im2 = zeros([frame_sz 3]);
im2(:,1,[1 2]) = 1;
im2(:,end,[1 2]) = 1;
im2(1,:,[1 2]) = 1;
im2(end,:,[1 2]) = 1;
% create the figure at a specific location with size equal to image
screen_sz = get(0,'ScreenSize');
fg_h = figure('units', 'pixels', 'position', ...
[screen_sz([3 4])-frame_sz([2 1])-40 frame_sz([2 1])], ...
'paperpositionmode', 'auto');
ax_h = axes('Parent',fg_h);
% show first image
imshow(im1, 'Parent',ax_h, 'InitialMagnification','fit');
hold on;
% display im2 ontop of im1
h = image(im2, 'Parent',ax_h);
% set the axis such that image occupies the whole figure
set(ax_h, 'Units','normalized', ...
'position', [0 0 1 1], 'visible', 'off');
axis off;
%%%%%%%%%%%% PROBLEM HAPPENS HERE %%%%%%%%%%%%%
% as soon as I set the AlphaData couple of pixels
% go missing from the right and bottom side
set(h, 'AlphaData', ones(frame_sz([1 2]))*0.5);
The problem occurs regardless of if I am overlaying using image() or just simply using image() to display an image. As soon as I set the AlphaData, the sizing of the axes slightly changes and chops off some borders.
How do I fix this?

Related

Place image in black pixels of another image

I have an image (white background with 1-5 black dots) that is called main.jpg (main image).
I am trying to place another image (secondary.jpg) in every black dot that is found in main image.
In order to do that:
I found the black pixels in main image
resize the secondary image to specific size that I want
plot the image in every coordinate that I found in step one. (the black pixel should be the center coordinates of the secondary image)
Unfortunately, I don't know how to do the third step.
for example:
main image is:
secondary image is:
output:
(The dots are behind the chairs. They are the image center points)
This is my code:
mainImage=imread('main.jpg')
secondaryImage=imread('secondary.jpg')
secondaryImageResized = resizeImage(secondaryImage)
[m n]=size(mainImage)
for i=1:n
for j=1:m
% if it's black pixel
if (mainImage(i,j)==1)
outputImage = plotImageInCoordinates(secondaryImageResized, i, j)
% save this image
imwrite(outputImage,map,'clown.bmp')
end
end
end
% resize the image to (250,350) width, height
function [ Image ] = resizeImage(img)
image = imresize(img, [250 350]);
end
function [outputImage] = plotImageInCoordinates(image, x, y)
% Do something
end
Any help appreciated!
Here's an alternative without convolution. One intricacy that you must take into account is that if you want to place each image at the centre of each dot, you must determine where the top left corner is and index into your output image so that you draw the desired object from the top left corner to the bottom right corner. You can do this by taking each black dot location and subtracting by half the width horizontally and half the height vertically.
Now onto your actual problem. It's much more efficient if you loop through the set of points that are black, not the entire image. You can do this by using the find command to determine the row and column locations that are 0. Once you do this, loop through each pair of row and column coordinates, do the subtraction of the coordinates and then place it on the output image.
I will impose an additional requirement where the objects may overlap. To accommodate for this, I will accumulate pixels, then find the average of the non-zero locations.
Your code modified to accommodate for this is as follows. Take note that because you are using JPEG compression, you will have compression artifacts so regions that are 0 may not necessarily be 0. I will threshold with an intensity of 128 to ensure that zero regions are actually zero. You will also have the situation where objects may go outside the boundaries of the image. Therefore to accommodate for this, pad the image sufficiently with twice of half the width horizontally and twice of half the height vertically then crop it after you're done placing the objects.
mainImage=imread('https://i.stack.imgur.com/gbhWJ.png');
secondaryImage=imread('https://i.stack.imgur.com/P0meM.png');
secondaryImageResized = imresize(secondaryImage, [250 300]);
% Find half height and width
rows = size(secondaryImageResized, 1);
cols = size(secondaryImageResized, 2);
halfHeight = floor(rows / 2);
halfWidth = floor(cols / 2);
% Create a padded image that contains our main image. Pad with white
% pixels.
rowsMain = size(mainImage, 1);
colsMain = size(mainImage, 2);
outputImage = 255*ones([2*halfHeight + rowsMain, 2*halfWidth + colsMain, size(mainImage, 3)], class(mainImage));
outputImage(halfHeight + 1 : halfHeight + rowsMain, ...
halfWidth + 1 : halfWidth + colsMain, :) = mainImage;
% Find a mask of the black pixels
mask = outputImage(:,:,1) < 128;
% Obtain black pixel locations
[row, col] = find(mask);
% Reset the output image so that they're all zeros now. We use this
% to output our final image. Also cast to ensure accumulation is proper.
outputImage(:) = 0;
outputImage = double(outputImage);
% Keeps track of how many times each pixel was hit by the object
% This is so that we can find the average at each location.
counts = zeros([size(mask), size(mainImage, 3)]);
% For each row and column location in the image
for i = 1 : numel(row)
% Get the row and column locations
r = row(i); c = col(i);
% Offset to get the top left corner
r = r - halfHeight;
c = c - halfWidth;
% Place onto final image
outputImage(r:r+rows-1, c:c+cols-1, :) = outputImage(r:r+rows-1, c:c+cols-1, :) + double(secondaryImageResized);
% Accumulate the counts
counts(r:r+rows-1,c:c+cols-1,:) = counts(r:r+rows-1,c:c+cols-1,:) + 1;
end
% Find average - Any values that were not hit, change to white
outputImage = outputImage ./ counts;
outputImage(counts == 0) = 255;
outputImage = uint8(outputImage);
% Now crop and show
outputImage = outputImage(halfHeight + 1 : halfHeight + rowsMain, ...
halfWidth + 1 : halfWidth + colsMain, :);
close all; imshow(outputImage);
% Write the final output
imwrite(outputImage, 'finalimage.jpg', 'Quality', 100);
We get:
Edit
I wasn't told that your images had transparency. Therefore what you need to do is use imread but ensure that you read in the alpha channel. We then check to see if one exists and if one does, we will ensure that the background of any values with no transparency are set to white. You can do that with the following code. Ensure this gets placed at the very top of your code, replacing the images being loaded in:
mainImage=imread('https://i.stack.imgur.com/gbhWJ.png');
% Change - to accommodate for transparency
[secondaryImage, ~, alpha] = imread('https://i.imgur.com/qYJSzEZ.png');
if ~isempty(alpha)
m = alpha == 0;
for i = 1 : size(secondaryImage,3)
m2 = secondaryImage(:,:,i);
m2(m) = 255;
secondaryImage(:,:,i) = m2;
end
end
secondaryImageResized = imresize(secondaryImage, [250 300]);
% Rest of your code follows...
% ...
The code above has been modified to read in the basketball image. The rest of the code remains the same and we thus get:
You can use convolution to achieve the desired effect. This will place a copy of im everywhere there is a black dot in imz.
% load secondary image
im = double(imread('secondary.jpg'))/255.0;
% create some artificial image with black indicators
imz = ones(500,500,3);
imz(50,50,:) = 0;
imz(400,200,:) = 0;
imz(200,400,:) = 0;
% create output image
imout = zeros(size(imz));
imout(:,:,1) = conv2(1-imz(:,:,1),1-im(:,:,1),'same');
imout(:,:,2) = conv2(1-imz(:,:,2),1-im(:,:,2),'same');
imout(:,:,3) = conv2(1-imz(:,:,3),1-im(:,:,3),'same');
imout = 1-imout;
% output
imshow(imout);
Also, you probably want to avoid saving main.jpg as a .jpg since it results in lossy compression and will likely cause issues with any method that relies on exact pixel values. I would recommend using .png which is lossless and will also likely compress better than .jpg for synthetic images where the same colors repeat many times.

Merging 2 images by showing one next to the other separated by a diagonal line

I have 2 images ("before" and "after"). I would like to show a final image where the left half is taken from the before image and the right half is taken from the after image.
The images should be separated by a white diagonal line of predefined width (2 or 3 pixels), where the diagonal is specified either by a certain angle or by 2 start and end coordinates. The diagonal should overwrite a part of the final image such that the size is the same as the sources'.
Example:
I know it can be done by looping over all pixels to recombine and create the final image, but is there an efficient way, or better yet, a built-in function that can do this?
Unfortunately I don't believe there is a built-in solution to your problem, but I've developed some code to help you do this but it will unfortunately require the image processing toolbox to play nicely with the code. As mentioned in your comments, you have this already so we should be fine.
The logic behind this is relatively simple. We will assume that your before and after pictures are the same size and also share the same number of channels. The first part is to declare a blank image and we draw a straight line down the middle of a certain thickness. The intricacy behind this is to declare an image that is slightly bigger than the original size of the image. The reason why is because I'm going to draw a line down the middle, then rotate this blank image by a certain angle to achieve the first part of what you desire. I'll be using imrotate to rotate an image by any angle you desire. The first instinct is to declare an image that's the same size as either the originals, draw a line down the middle and rotate it. However, if you do this you'll end up with the line being disconnected and not draw from the top to the bottom of the image. That makes sense because the line being drawn on an angle covers more pixels than if you were to draw this vertically.
Using Pythagorean's theorem, we know that the longest line that can ever be drawn on your image is the diagonal. Therefore we declare an image that is sqrt(rows*rows + cols*cols) in both the rows and columns where rows and cols are the rows and columns of the original image. After, we'll take the ceiling to make sure we've covered as much as possible and we add a bit of extra room to accommodate for the width of the line. We draw a line on this image, rotate it then we'll crop the image after so that it's the same size as the input images. This ensures that the line drawn at whatever angle you wish is fully drawn from top to bottom.
That logic is the hardest part. Once you do that, you declare two logical masks where you use imfill to fill the left side of the mask as one mask and we'll invert the mask to find the other mask. You will also need to use the line image that we created earlier with imrotate to index into the masks and set the values to false so that we ignore these pixels that are on the line.
Finally, you take each mask, index into your image and copy over each portion of the image you desire. You finally use the line image to index into the output and set the values to white.
Without further ado, here's the code:
% Load some example data
load mandrill;
% im is the image before
% im2 is the image after
% Before image is a colour image
im = im2uint8(ind2rgb(X, map));
% After image is a grayscale image
im2 = rgb2gray(im);
im2 = cat(3, im2, im2, im2);
% Declare line image
rows = size(im, 1); cols = size(im, 2);
width = 5;
m = ceil(sqrt(rows*rows + cols*cols + width*width));
ln = false([m m]);
mhalf = floor(m / 2); % Find halfway point width wise and draw the line
ln(:,mhalf - floor(width/2) : mhalf + floor(width/2)) = true;
% Rotate the line image
ang = 20; % 20 degrees
lnrotate = imrotate(ln, ang, 'crop');
% Crop the image so that it's the same dimensions as the originals
mrowstart = mhalf - floor(rows/2);
mcolstart = mhalf - floor(cols/2);
lnfinal = lnrotate(mrowstart : mrowstart + rows - 1, mcolstart : mcolstart + cols - 1);
% Make the masks
mask1 = imfill(lnfinal, [1 1]);
mask2 = ~mask1;
mask1(lnfinal) = false;
mask2(lnfinal) = false;
% Make sure the masks have as many channels as the original
mask1 = repmat(mask1, [1 1 size(im,3)]);
mask2 = repmat(mask2, [1 1 size(im,3)]);
% Do the same for the line
lnfinal = repmat(lnfinal, [1 1 size(im, 3)]);
% Specify output image
out = zeros(size(im), class(im));
out(mask1) = im(mask1);
out(mask2) = im2(mask2);
out(lnfinal) = 255;
% Show the image
figure;
imshow(out);
We get:
If you want the line to go in the other direction, simply make the angle ang negative. In the example script above, I've made the angle 20 degrees counter-clockwise (i.e. positive). To reproduce the example you gave, specify -20 degrees instead. I now get this image:
Here's a solution using polygons:
function q44310306
% Load some image:
I = imread('peppers.png');
B = rgb2gray(I);
lt = I; rt = B;
% Specify the boundaries of the white line:
width = 2; % [px]
offset = 13; % [px]
sz = size(I);
wlb = [floor(sz(2)/2)-offset+[0,width]; ceil(sz(2)/2)+offset-[width,0]];
% [top-left, top-right; bottom-left, bottom-right]
% Configure two polygons:
leftPoly = struct('x',[1 wlb(1,2) wlb(2,2) 1], 'y',[1 1 sz(1) sz(1)]);
rightPoly = struct('x',[sz(2) wlb(1,1) wlb(2,1) sz(2)],'y',[1 1 sz(1) sz(1)]);
% Define a helper grid:
[XX,YY] = meshgrid(1:sz(2),1:sz(1));
rt(inpolygon(XX,YY,leftPoly.x,leftPoly.y)) = intmin('uint8');
lt(repmat(inpolygon(XX,YY,rightPoly.x,rightPoly.y),1,1,3)) = intmin('uint8');
rt(inpolygon(XX,YY,leftPoly.x,leftPoly.y) & ...
inpolygon(XX,YY,rightPoly.x,rightPoly.y)) = intmax('uint8');
final = bsxfun(#plus,lt,rt);
% Plot:
figure(); imshow(final);
The result:
One solution:
im1 = imread('peppers.png');
im2 = repmat(rgb2gray(im1),1,1,3);
imgsplitter(im1,im2,80) %imgsplitter(image1,image2,angle [0-100])
function imgsplitter(im1,im2,p)
s1 = size(im1,1); s2 = size(im1,2);
pix = floor(p*size(im1,2)/100);
val = abs(pix -(s2-pix));
dia = imresize(tril(ones(s1)),[s1 val]);
len = min(abs([0-pix,s2-pix]));
if p>50
ind = [ones(s1,len) fliplr(~dia) zeros(s1,len)];
else
ind = [ones(s1,len) dia zeros(s1,len)];
end
ind = uint8(ind);
imshow(ind.*im1+uint8(~ind).*im2)
hold on
plot([pix,s2-pix],[0,s1],'w','LineWidth',1)
end
OUTPUT:

Crop the largest square inside a circle object - Matlab

I am trying to find a way to crop from a circle object (Image A) the largest square that can fit inside it.
Can someone please explain/show me how to find the biggest square fit parameters of the white space inside the circle (Image I) and based on them crop the square in the original image (Image A).
Script:
A = imread('E:/CirTest/Test.jpg');
%imshow(A)
level = graythresh(A);
BW = im2bw(A,level);
%imshow(BW)
I = imfill(BW, 'holes');
imshow(I)
d = imdistline;
[centers, radii, metric] = imfindcircles(A,[1 500]);
imageCrop=imcrop(A, [BoxBottomX BoxBottomY NewX NewY]);
I have a solution for you but it requires a bit of extra work. What I would do first is use imfill but directly on the grayscale image. This way, noisy pixels in uniform areas get inpainted with the same intensities so that thresholding is easier. You can still use graythresh or Otsu's thresholding and do this on the inpainted image.
Here's some code to get you started:
figure; % Open up a new figure
% Read in image and convert to grayscale
A = rgb2gray(imread('http://i.stack.imgur.com/vNECg.jpg'));
subplot(1,3,1); imshow(A);
title('Original Image');
% Find the optimum threshold via Otsu
level = graythresh(A);
% Inpaint noisy areas
I = imfill(A, 'holes');
subplot(1,3,2); imshow(I);
title('Inpainted image');
% Threshold the image
BW = im2bw(I, level);
subplot(1,3,3); imshow(BW);
title('Thresholded Image');
The above code does the three operations that I mentioned, and we see this figure:
Notice that the thresholded image has border pixels that need to be removed so we can concentrate on the circular object. You can use the imclearborder function to remove the border pixels. When we do that:
% Clear off the border pixels and leave only the circular object
BW2 = imclearborder(BW);
figure; imshow(BW2);
... we now get this image:
Unfortunately, there are some noisy pixels, but we can very easily use morphology, specifically the opening operation with a small circular disk structuring element to remove these noisy pixels. Using strel with the appropriate structuring element in addition to imopen should help do the trick:
% Clear out noisy pixels
SE = strel('disk', 3, 0);
out = imopen(BW2, SE);
figure; imshow(out);
We now get:
This mask contains the locations of the circular object we now need to use to crop our original image. The last part is to determine the row and column locations using this mask to locate the top left and bottom right corner of the original image and we thus crop it:
% Find row and column locations of circular object
[row,col] = find(out);
% Find top left and bottom right corners
top_row = min(row);
top_col = min(col);
bottom_row = max(row);
bottom_col = max(col);
% Crop the image
crop = A(top_row:bottom_row, top_col:bottom_col);
% Show the cropped image
figure; imshow(crop);
We now get:
It's not perfect, but it will of course get you started. If you want to copy and paste this in its entirety and run this on your computer, here we are:
figure; % Open up a new figure
% Read in image and convert to grayscale
A = rgb2gray(imread('http://i.stack.imgur.com/vNECg.jpg'));
subplot(2,3,1); imshow(A);
title('Original Image');
% Find the optimum threshold via Otsu
level = graythresh(A);
% Inpaint noisy areas
I = imfill(A, 'holes');
subplot(2,3,2); imshow(I);
title('Inpainted image');
% Threshold the image
BW = im2bw(I, level);
subplot(2,3,3); imshow(BW);
title('Thresholded Image');
% Clear off the border pixels and leave only the circular object
BW2 = imclearborder(BW);
subplot(2,3,4); imshow(BW2);
title('Cleared Border Pixels');
% Clear out noisy pixels
SE = strel('disk', 3, 0);
out = imopen(BW2, SE);
% Show the final mask
subplot(2,3,5); imshow(out);
title('Final Mask');
% Find row and column locations of circular object
[row,col] = find(out);
% Find top left and bottom right corners
top_row = min(row);
top_col = min(col);
bottom_row = max(row);
bottom_col = max(col);
% Crop the image
crop = A(top_row:bottom_row, top_col:bottom_col);
% Show the cropped image
subplot(2,3,6);
imshow(crop);
title('Cropped Image');
... and our final figure is:
You can use bwdist with L_inf distance (aka 'chessboard') to get the axis-aligned distance to the edges of the region, thus concluding the dimensions of the largest bounded box:
bw = imread('http://i.stack.imgur.com/7yCaD.png');
lb = bwlabel(bw);
reg = lb==2; %// pick largest area
d = bwdist(~reg,'chessboard'); %// compute the axis aligned distance from boundary inward
r = max(d(:)); %// find the largest distance to boundary
[cy cx] = find(d==r,1); %// find the location most distant
boundedBox = [cx-r, cy-r, 2*r, 2*r];
And the result is
figure;
imshow(bw,'border','tight');
hold on;
rectangle('Position', boundedBox, 'EdgeColor','r');
Once you have the bounding box, you can use imcrop to crop the original image
imageCrop = imcrop(A, boundedBox);
Alternatively, you can
imageCrop = A(cy + (-r:r-1), cx + (-r:r-1) );

Re-sizing a rectangular image to a square image

I have an image, size 213 x 145 pixels. I want to resize it to 128 x 128 pixels for example. I've already tried the code below:
i = imread ('alif1.png');
I = imresize (i, [128 128], 'bilinear');
OR
i = imread ('alif1.png');
I = imresize (i, [128 128], 'lanczos3');
it gave me a square image, but the image became disproportionate. However, I believe the aspect ratio was preserved.
I want to resize the image to a square shape without distorting or stretching the image, rather to pad/crop the white background instead. I still can't figure out the right code. I hope anyone could help.
any help will be very much appreciated :)
I = imread('alifi.png');
Crop image, specifying crop rectangle.
I2 = imcrop(I,[75 68 128 128]);
Size and position of the crop rectangle, specified as a four-element position vector of the form [xmin ymin width height].
for more understanding follow this(matlab ) and this(blog) links.
If you want to resize (not crop) the image and keep the aspect ratio (so you don't loose any part of the image AND it doesn't get distorted), you can first add margins to make the image squared.
You can achieve this using the function padarray, or just creating a new image of zeros and then adding your image in the appropiate coordinates.
Once your image is squared, you can resize it to 128x128 using imresize.
In order to add margins, you will have to see where to add them (top&bottom OR left&right).
Also since padarray adds the same amount of margins in both sides, you have to check if the number you need is even. If it's odd add first a last row (or column) of zeros to your image.
So basically you have three options:
Make the image squared by not preserving aspect ratio (which is what you already tried)
Cropping the image as suggested by #ShvetChakra and #bla (but you will loose some image info)
Add margins to the image and resize (but you will end up with a squared image with margins)
Magic doesn't exist so "you must choose, but choose wisely"
(Quote from Indiana Jones and the Last Crusade).
EDIT:
% Example with a 5x2 image, so an extra column will be added
% in order to use padarray.
im = [1 2; 3 4; 5 6; 7 8; 9 10];
nrows = size(a,1);
ncols = size(a,2);
d = abs(ncols-nrows); % difference between ncols and nrows:
if(mod(d,2) == 1) % if difference is an odd number
if (ncols > nrows) % we add a row at the end
im = [im; zeros(1, ncols)];
nrows = nrows + 1;
else % we add a col at the end
im = [im zeros(nrows, 1)];
ncols = ncols + 1;
end
end
if ncols > nrows
im = padarray(im, [(ncols-nrows)/2 0]);
else
im = padarray(im, [0 (nrows-ncols)/2]);
end
% Here im is a 5x5 matix, not perfectly centered
% because we added an odd number of columns: 3

Image Repetition from Binary to Cartesian

I'd like to take in an RGB image, find the points in the image that are white, and get the cartesian coordinates of those points in the image. I've gotten most of the way there, but when I try to plot the cartesian coordinates, I get a vertically tiled image (i.e. 5 overlapped copies of what I should see). Anyone know what could be causing this?
,
Code: (JPG comes in as 2448 x x3264 x 3 uint8)
I = imread('IMG_0245.JPG');
imshow(I); % display unaltered image
% Convert image to grayscale
I = rgb2gray(I);
% Convert image to binary (black/white)
I = im2bw(I, 0.9);
% Generate cartesian coordinates of image
imageSize = size(I);
[x, y] = meshgrid( 1:imageSize(1), 1:imageSize(2) );
PerspectiveImage = [x(:), y(:), I(:)];
% Get indices of white points only
whiteIndices = find(PerspectiveImage(:,3));
figure; plot( PerspectiveImage(whiteIndices, 1), PerspectiveImage(whiteIndices, 2),'.');
% Flip vertically to correct indexing vs. plotting issue
axis ij
Very simple. You're declaring your meshgrid wrong. It should be:
[x, y] = meshgrid( 1:imageSize(2), 1:imageSize(1) );
The first parameter denotes the horizontal extents of the 2D grid, and so you want to make this vary for as many columns as you have. Similarly, the second parameter denotes the vertical extents of the 2D grid, and so you want to make this for as many rows as you have.
I had to pre-process some of your image to get some good results because your original image had a large white border surrounding the image. I had to remove this border by removing all pure white pixels. I also read in the image directly from StackOverflow:
I = imread('http://s7.postimg.org/ovb53w4ff/Track_example.jpg');
mask = all(I == 255, 3);
I = bsxfun(#times, I, uint8(~mask));
This was the image I get after doing my pre-processing:
Once I do this and change your meshgrid call, I get this:

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