Related
This is the problem I have: I have an image as shown below. I want to detect the circular region which I have marked with a red line for display here (that particular bright ring).
Initially, this is what I do for now: (MATLAB)
binaryImage = imdilate(binaryImage,strel('disk',5));
binaryImage = imfill(binaryImage, 'holes'); % Fill holes.
binaryImage = bwareaopen(binaryImage, 20000); % Remove small blobs.
binaryImage = imerode(binaryImage,strel('disk',300));
out = binaryImage;
img_display = immultiply(binaryImage,rgb2gray(J1));
figure, imshow(img_display);
The output seems to be cut on one of the parts of the object (for a different image as input, not the one displayed above). I want an output in such a way that it is symmetric (its not always a perfect circle, when it is rotated).
I want to strictly avoid im2bw since as soon as I binarize, I lose a lot of information about the shape.
This is what I was thinking of:
I can detect the outer most circular (almost circular) contour of the image (shown in yellow). From this, I can find out the centroid and maybe find a circle which has a radius of 50% (to locate the region shown in red). But this won't be exactly symmetric since the object is slightly tilted. How can I tackle this issue?
I have attached another image where object is slightly tilted here
I'd try messing around with the 'log' filter. The region you want is essentially low values of the 2nd order derivative (i.e. where the slope is decreasing), and you can detect these regions by using a log filter and finding negative values. Here's a very basic outline of what you can do, and then tweak it to your needs.
img = im2double(rgb2gray(imread('wheel.png')));
img = imresize(img, 0.25, 'bicubic');
filt_img = imfilter(img, fspecial('log',31,5));
bin_img = filt_img < 0;
subplot(2,2,1);
imshow(filt_img,[]);
% Get regionprops
rp = regionprops(bin_img,'EulerNumber','Eccentricity','Area','PixelIdxList','PixelList');
rp = rp([rp.EulerNumber] == 0 & [rp.Eccentricity] < 0.5 & [rp.Area] > 2000);
bin_img(:) = false;
bin_img(vertcat(rp.PixelIdxList)) = true;
subplot(2,2,2);
imshow(bin_img,[]);
bin_img(:) = false;
bin_img(rp(1).PixelIdxList) = true;
bin_img = imfill(bin_img,'holes');
img_new = img;
img_new(~bin_img) = 0;
subplot(2,2,3);
imshow(img_new,[]);
bin_img(:) = false;
bin_img(rp(2).PixelIdxList) = true;
bin_img = imfill(bin_img,'holes');
img_new = img;
img_new(~bin_img) = 0;
subplot(2,2,4);
imshow(img_new,[]);
Output:
I have an image and a subimage which is cropped out of the original image.
Here's the code I have written so far:
val1 = imread(img);
val2 = imread(img_w);
gray1 = rgb2gray(val1);%grayscaling both images
gray2 = rgb2gray(val2);
matchingval = normxcorr2(gray1,gray2);%normalized cross correlation
[max_c,imax]=max(abs(matchingval(:)));
After this I am stuck. I have no idea how to change the whole image grayscale except for the sub image which should be in color.
How do I do this?
Thank you.
If you know what the coordinates are for your image, you can always just use the rgb2gray on just the section of interest.
For instance, I tried this on an image just now:
im(500:1045,500:1200,1)=rgb2gray(im(500:1045,500:1200,1:3));
im(500:1045,500:1200,2)=rgb2gray(im(500:1045,500:1200,1:3));
im(500:1045,500:1200,3)=rgb2gray(im(500:1045,500:1200,1:3));
Where I took the rows (500 to 1045), columns (500 to 1200), and the rgb depth (1 to 3) of the image and applied the rgb2gray function to just that. I did it three times as the output of rgb2gray is a 2d matrix and a color image is a 3d matrix, so I needed to change it layer by layer.
This worked for me, making only part of the image gray but leaving the rest in color.
The issue you might have though is this, a color image is 3 dimensions while a gray scale need only be 2 dimensions. Combining them means that the gray scale must be in a 3d matrix.
Depending on what you want to do, this technique may or may not help.
Judging from your code, you are reading the image and the subimage in MATLAB. What you need to know are the coordinates of where you extracted the subimage. Once you do that, simply take your original colour image, convert that to grayscale, then duplicate this image in the third dimension three times. You need to do this so that you can place colour pixels in this image.
For RGB images, grayscale images have the RGB components to all be the same. Duplicating this image in the third dimension three times creates the RGB version of the grayscale image. Once you do that, simply use the row and column coordinates of where you extracted the subimage and place that into the equivalent RGB grayscale image.
As such, given your colour image that is stored in img and your subimage stored in imgsub, and specifying the rows and columns of where you extracted the subimage in row1,col1 and row2,col2 - with row1,col1 being the top left corner of the subimage and row2,col2 is the bottom right corner, do this:
img_gray = rgb2gray(img);
img_gray = cat(3, img_gray, img_gray, img_gray);
img_gray(row1:row2, col1:col2,:) = imgsub;
To demonstrate this, let's try this with an image in MATLAB. We'll use the onion.png image that's part of the image processing toolbox in MATLAB. Therefore:
img = imread('onion.png');
Let's also define our row1,col1,row2,col2:
row1 = 50;
row2 = 90;
col1 = 80;
col2 = 150;
Let's get the subimage:
imgsub = img(row1:row2,col1:col2,:);
Running the above code, this is the image we get:
I took the same example as rayryeng's answer and tried to solve by HSV conversion.
The basic idea is to set the second layer i.e saturation layer to 0 (so that they are grayscale). then rewrite the layer with the original saturation layer only for the sub image area, so that, they alone have the saturation values.
Code:
img = imread('onion.png');
img = rgb2hsv(img);
sPlane = zeros(size(img(:,:,1)));
sPlane(50:90,80:150) = img(50:90,80:150,2);
img(:,:,2) = sPlane;
img = hsv2rgb(img);
imshow(img);
Output: (Same as rayryeng's output)
Related Answer with more details here
I have some code which takes a fish eye images and converts it to a rectangular image in each RGB channels. I am having trouble with the fact the the output image is square instead of rectangular. (this means that the image is distorted, compressed horizontally.) I have tried changing the output matrix to a more suitable format, without success. Besides this i have also discovered that for the code to work the input image must be square like 500x500. Any idea how to solve this issue? This is the code:
The code is inspired by Prakash Manandhar "Polar To/From Rectangular Transform of Images" file exchange on mathworks.
EDIT. Code now works.
function imP = FISHCOLOR2(imR)
rMin=0.1;
rMax=1;
[Mr, Nr, Dr] = size(imR); % size of rectangular image
xRc = (Mr+1)/2; % co-ordinates of the center of the image
yRc = (Nr+1)/2;
sx = (Mr-1)/2; % scale factors
sy = (Nr-1)/2;
reduced_dim = min(size(imR,1),size(imR,2));
imR = imresize(imR,[reduced_dim reduced_dim]);
M=size(imR,1);N=size(imR,2);
dr = (rMax - rMin)/(M-1);
dth = 2*pi/N;
r=rMin:dr:rMin+(M-1)*dr;
th=(0:dth:(N-1)*dth)';
[r,th]=meshgrid(r,th);
x=r.*cos(th);
y=r.*sin(th);
xR = x*sx + xRc;
yR = y*sy + yRc;
for k=1:Dr % colors
imP(:,:,k) = interp2(imR(:,:,k), xR, yR); % add k channel
end
imP = imresize(imP,[size(imP,1), size(imP,2)/3]);
imP = imrotate(imP,270);
SOLVED
Input image <- Image link
Output image <- Image link
PART A
To remove the requirement of a square input image, you may resize the input image into a square one with this -
%%// Resize the input image to make it square
reduced_dim = min(size(imR,1),size(imR,2));
imR = imresize(imR,[reduced_dim reduced_dim]);
Few points I would like to raise here though about this image-resizing to make it a square image. This was a quick and dirty approach and distorts the image for a non-square image, which you may not want if the image is not too "squarish". In many of those non-squarish images, you would find blackish borders across the boundaries of the image. If you can remove that using some sort of image processing algorithm or just manual photoshoping, then it would be ideal. After that even if the image is not square, imresize could be considered a safe option.
PART B
Now, after doing the main processing of flattening out the fisheye image,
at the end of your code, it seemed like the image has to be rotated
90 degrees clockwise or counter-clockwise depending on if the fisheye
image have objects inwardly or outwardly respectively.
%%// Rotating image
imP = imrotate(imP,-90); %%// When projected inwardly
imP = imrotate(imP,-90); %%// When projected outwardly
Note that the flattened image must have the height equal to the half of the
size of the input square image, that is the radius of the image.
Thus, the final output image must have number of rows as - size(imP,2)/2
Since you are flattening out a fisheye image, I assumed that the width
of the flattened image must be 2*PI times the height of it. So, I tried this -
imP = imresize(imP,[size(imP,2)/2 pi*size(imP,2)]);
But the results looked too flattened out. So, the next logical experimental
value looked like PI times the height, i.e. -
imP = imresize(imP,[size(imP,2)/2 pi*size(imP,2)/2]);
Results in this case looked good.
I'm not very experienced in the finer points of image processing in MATLAB, but depending on the exact operation of the imP fill mechanism, you may get what you're looking for by doing the following. Change:
M = size(imR, 1);
N = size(imR, 2);
To:
verticalScaleFactor = 0.5;
M = size(imR, 1) * verticalScaleFactor;
N = size(imR, 2);
If my hunch is right, you should be able to tune that scale factor to get the image just right. It may, however, break your code. Let me know if it doesn't work, and edit your post to flesh out exactly what each section of code does. Then we should be able to give it another shot. Good luck!
This is the code which works.
function imP = FISHCOLOR2(imR)
rMin=0.1;
rMax=1;
[Mr, Nr, Dr] = size(imR); % size of rectangular image
xRc = (Mr+1)/2; % co-ordinates of the center of the image
yRc = (Nr+1)/2;
sx = (Mr-1)/2; % scale factors
sy = (Nr-1)/2;
reduced_dim = min(size(imR,1),size(imR,2));
imR = imresize(imR,[reduced_dim reduced_dim]);
M=size(imR,1);N=size(imR,2);
dr = (rMax - rMin)/(M-1);
dth = 2*pi/N;
r=rMin:dr:rMin+(M-1)*dr;
th=(0:dth:(N-1)*dth)';
[r,th]=meshgrid(r,th);
x=r.*cos(th);
y=r.*sin(th);
xR = x*sx + xRc;
yR = y*sy + yRc;
for k=1:Dr % colors
imP(:,:,k) = interp2(imR(:,:,k), xR, yR); % add k channel
end
imP = imresize(imP,[size(imP,1), size(imP,2)/3]);
imP1 = imrotate(imP1,270);
I have a binary image lu and when I rotate the image the size of the image lu changes but i need to preserve the size of the image :
m=2048;
n=3072;
ODcenter =1.0e+03 *[2.0345 0.9985]
OD=ODcenter ;
X=zeros(m,n); %% m,n is size of image
t = 0:.1:2*pi;
ODradius = norm(ODcenter(2) - ODcenter(1)) / 2;
xm2 = round(2*ODradius*cos(t)+OD(1));
ym2 = round(2*ODradius*sin(t)+OD(2));
imCircleAlphaData2 = roipoly(X,xm2,ym2);
figure; imshow(imCircleAlphaData2);
lu=imCircleAlphaData2;
mask1 = true(size(lu)); %# Create a matrix of true values the same size
mask1(ODcenter(2):end,:) = false; %# Set the lower half to false
lu(~mask1) = 0; %# Set all elements in lu corresponding to mask 1==0
mask2 = true(size(lu));
mask2(:,ODcenter(1):end) = false; %# Set the right of the upper half to false
lu(~mask2) = 0; %# Set all elements in lu corresponding mask 2==0
figure;
imshow(lu); % shows left upper
lurot= imrotate(lu,45);
figure,imshow(lurot)
Size of lurot and lu is different . How can I preserve the size of image even if some part of image will be cropped after rotation
Basically, you have two options with Matlab imrotate:
Use crop which will make the output image the same size as the input image, cropping the rotated image to make it fit
Use loose which will make the output image large enough to contain the entire original rotated image. Generally, this will make the output image larger than the input image.
lurot= imrotate(lu,45,'nearest','crop');
As you see, I have shapes and their white boundaries. I want to fill the shapes in white color.
The input is:
I would like to get this output:
Can anybody help me please with this code? it doesn't change the black ellipses to white.
Thanks alot :]]
I = imread('untitled4.bmp');
Ibw = im2bw(I);
CC = bwconncomp(Ibw); %Ibw is my binary image
stats = regionprops(CC,'pixellist');
% pass all over the stats
for i=1:length(stats),
size = length(stats(i).PixelList);
% check only the relevant stats (the black ellipses)
if size >150 && size < 600
% fill the black pixel by white
x = round(mean(stats(i).PixelList(:,2)));
y = round(mean(stats(i).PixelList(:,1)));
Ibw = imfill(Ibw, [x, y]);
end;
end;
imshow(Ibw);
Your code can be improved and simplified as follows. First, negating Ibw and using BWCONNCOMP to find 4-connected components will give you indices for each black region. Second, sorting the connected regions by the number of pixels in them and choosing all but the largest two will give you indices for all the smaller circular regions. Finally, the linear indices of these smaller regions can be collected and used to fill in the regions with white. Here's the code (quite a bit shorter and not requiring any loops):
I = imread('untitled4.bmp');
Ibw = im2bw(I);
CC = bwconncomp(~Ibw, 4);
[~, sortIndex] = sort(cellfun('prodofsize', CC.PixelIdxList));
Ifilled = Ibw;
Ifilled(vertcat(CC.PixelIdxList{sortIndex(1:end-2)})) = true;
imshow(Ifilled);
And here's the resulting image:
If your images are all black&white, and you have the image processing toolkit, then this looks like what you need:
http://www.mathworks.co.uk/help/toolbox/images/ref/imfill.html
Something like:
imfill(image, [startX, startY])
where startX, startY is a pixel in the area that you want to fill.