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:
Related
I want to detect the points shown in the image below:
I have done this so far:
[X,map] = rgb2ind(img,0.0);
img = ind2gray(X,map); % Convert indexed to grayscale
level = graythresh(img); % Compute an appropriate threshold
img_bw = im2bw(img,level);% Convert grayscale to binary
mask = zeros(size(img_bw));
mask(2:end-2,2:end-2) = 1;
img_bw(mask<1) = 1;
%invert image
img_inv =1-img_bw;
% find blobs
img_blobs = bwmorph(img_inv,'majority',10);
% figure, imshow(img_blobs);
[rows, columns] = size(img_blobs);
for col = 1 : columns
thisColumn = img_blobs(:, col);
topRow = find(thisColumn, 1, 'first');
bottomRow = find(thisColumn, 1, 'last');
img_blobs(topRow : bottomRow, col) = true;
end
inverted = imcomplement(img_blobs);
ed = edge(inverted,'canny');
figure, imshow(ed),title('inverted');
Now how to proceed to get the coordinates of the desired position?
The top point is obviously the white pixel with the highest ordinate, which is easily obtained.
The bottom point is not so well defined. What you can do is
follow the peak edges until you reach a local minimum, on the left and on the right. That gives you a line segment, which you can intersect with the vertical through the top point.
if you know a the peak width, try every pixel on the vertical through the top point, downward, and stop until it has no left nor right neighbors at a distance equal to the peak with.
as above, but stop when the distance between the left and right neighbors exceeds a threshold.
In this particular case, you could consider using houghlines in matlab. Setting the required Theta and MinLength parameter values, you should be able to get the two vertical lines parallel to your peak. You can use the end points of the vertical lines, to get the point at the bottom.
Here is a sample code.
[H,theta,rho] = hough(bw,'Theta',5:1:30);%This is the angle range
P = houghpeaks(H,500,'NHoodSize',[11 11]);
lines = houghlines(bw,theta,rho,P,'FillGap',10,'MinLength',300);
Here is a complete description of how houghlines actually works.
i am working on a research about the swimming of fishes using analysis of videos, then i need to be carefully with the images (obtained from video frames) with emphasis in the tail.
The images are in High-Resolution and the software that i customize works with binary images, because is easy to use maths operations on this.
For obten this binary images i use 2 methods:
1)Convert the image to gray, invert the colors,later to bw and finally to binary with a treshold that give me images like this, with almost nothing of noise. The images sometimes loss a bit of area and doesn't is very exactly with the tail(now i need more acurracy for determinate the amplitude of tail moves)
image 1
2)i use this code, for cut the border that increase the threshold, this give me a good image of the edge, but i dont know like joint these point and smooth the image, or fitting binary images, the app fitting of matlab 2012Rb doesn't give me a good graph and i don't have access to the toolboxs of matlab.
s4 = imread('arecorte.bmp');
A=[90 90 1110 550]
s5=imcrop(s4,A)
E = edge(s5,'canny',0.59);
image2
My question is that
how i can fit the binary image or joint the points and smooth without disturb the tail?
Or how i can use the edge of the image 2 to increase the acurracy of the image 1?
i will upload a image in the comments that give me the idea of the method 2), because i can't post more links, please remember that i am working with iterations and i can't work frame by frame.
Note: If i ask this is because i am in a dead point and i don't have the resources to pay to someone for do this, until this moment i was able to write the code but in this final problem i can't alone.
I think you should use connected component labling and discard the small labels and than extract the labels boundary to get the pixels of each part
the code:
clear all
% Read image
I = imread('fish.jpg');
% You don't need to do it you haef allready a bw image
Ibw = rgb2gray(I);
Ibw(Ibw < 100) = 0;
% Find size of image
[row,col] = size(Ibw);
% Find connceted components
CC = bwconncomp(Ibw,8);
% Find area of the compoennts
stats = regionprops(CC,'Area','PixelIdxList');
areas = [stats.Area];
% Sort the areas
[val,index] = sort(areas,'descend');
% Take the two largest comonents ids and create filterd image
IbwFilterd = zeros(row,col);
IbwFilterd(stats(index(1,1)).PixelIdxList) = 1;
IbwFilterd(stats(index(1,2)).PixelIdxList) = 1;
imshow(IbwFilterd);
% Find the pixels of the border of the main component and tail
boundries = bwboundaries(IbwFilterd);
yCorrdainteOfMainFishBody = boundries{1}(:,1);
xCorrdainteOfMainFishBody = boundries{1}(:,2);
linearCorrdMainFishBody = sub2ind([row,col],yCorrdainteOfMainFishBody,xCorrdainteOfMainFishBody);
yCorrdainteOfTailFishBody = boundries{2}(:,1);
xCorrdainteOfTailFishBody = boundries{2}(:,2);
linearCorrdTailFishBody = sub2ind([row,col],yCorrdainteOfTailFishBody,xCorrdainteOfTailFishBody);
% For visoulaztion put color for the boundries
IFinal = zeros(row,col,3);
IFinalChannel = zeros(row,col);
IFinal(:,:,1) = IFinalChannel;
IFinalChannel(linearCorrdMainFishBody) = 255;
IFinal(:,:,2) = IFinalChannel;
IFinalChannel = zeros(row,col);
IFinalChannel(linearCorrdTailFishBody) = 125;
IFinal(:,:,3) = IFinalChannel;
imshow(IFinal);
The final image:
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 want to stretch an elliptical object in an image until it forms a circle. My program currently inputs an image with an elliptical object (eg. coin at an angle), thresholds and binarizes it, isolates the region of interest using edge-detect/bwboundaries(), and performs regionprops() to calculate major/minor axis lengths.
Essentially, I want to use the 'MajorAxisLength' as the diameter and stretch the object on the minor axis to form a circle. Any suggestions on how I should approach this would be greatly appreciated. I have appended some code for your perusal (unfortunately I don't have enough reputation to upload an image, the binarized image looks like a white ellipse on a black background).
EDIT: I'd also like to apply this technique to the gray-scale version of the image, to examine what the stretch looks like.
code snippet:
rgbImage = imread(fullFileName);
redChannel = rgbImage(:, :, 1);
binaryImage = redChannel < 90;
labeledImage = bwlabel(binaryImage);
area_measurements = regionprops(labeledImage,'Area');
allAreas = [area_measurements.Area];
biggestBlobIndex = find(allAreas == max(allAreas));
keeperBlobsImage = ismember(labeledImage, biggestBlobIndex);
measurements = regionprops(keeperBlobsImage,'Area','MajorAxisLength','MinorAxisLength')
You know the diameter of the circle and you know the center is the location where the major and minor axes intersect. Thus, just compute the radius r from the diameter, and for every pixel in your image, check to see if that pixel's Euclidean distance from the cirlce's center is less than r. If so, color the pixel white. Otherwise, leave it alone.
[M,N] = size(redChannel);
new_image = zeros(M,N);
for ii=1:M
for jj=1:N
if( sqrt((jj-center_x)^2 + (ii-center_y)^2) <= radius )
new_image(ii,jj) = 1.0;
end
end
end
This can probably be optimzed by using the meshgrid function combined with logical indices to avoid the loops.
I finally managed to figure out the transform required thanks to a lot of help on the matlab forums. I thought I'd post it here, in case anyone else needed it.
stats = regionprops(keeperBlobsImage, 'MajorAxisLength','MinorAxisLength','Centroid','Orientation');
alpha = pi/180 * stats(1).Orientation;
Q = [cos(alpha), -sin(alpha); sin(alpha), cos(alpha)];
x0 = stats(1).Centroid.';
a = stats(1).MajorAxisLength;
b = stats(1).MinorAxisLength;
S = diag([1, a/b]);
C = Q*S*Q';
d = (eye(2) - C)*x0;
tform = maketform('affine', [C d; 0 0 1]');
Im2 = imtransform(redChannel, tform);
subplot(2, 3, 5);
imshow(Im2);
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.