so I am working for my disertation thesis and I have to detect the pupil from images using Hough Transform. So far I wrote a code that identifies 2 circles on my image, but right now I have to keep the black circle from the pupil.
When I run the code, it identifies me the pupil, but also a random circle on the cheek. My professor said that I should calculate the pixels mean and, considering the fact that the pupil is black, to keep the pixels from only that region. I don't know how to do this.
I will let my code here to have a look and if someone has an ideea on how should I write this and keep only the black pixels would be great. I also attached to this the final image to see what I obtained.
close all
clear all
path='C:\Users\Ioana PMEC\OneDrive\Ioana personal\Disertatie\test.jpg';
%Citire imagine initiala
xx = imread(path);
figure
imshow(xx)
title('Imagine initiala');% Binarizarea imaginii initiale
yy = rgb2gray(xx);
figure
imshow(yy);
title('Imagine binarizata');
e = edge(yy, 'canny');
imshow(e);
radii = 11:1:30;
h = circle_hough(e, radii, 'same', 'normalise');
peaks = circle_houghpeaks(h, radii, 'nhoodxy', 15, 'nhoodr', 21, 'npeaks', 2);
imshow(yy);
hold on;
for peak = peaks
[x, y]=circlepoints(peak(3));
plot(x+peak(1), y+peak(2), 'r-');
end
hold off
testimage
finalimage
I implemented something which should get the task done for you. The example is done with the image you provided.
Step 1: Read the file(s) and convert them to grayscale.
path = %user input;
RGB = imread(path);
lab = rgb2lab(RGB);
grayscale_image = rgb2gray(RGB);
Step 2: Do the Hough transformation with given parameters.
These, and also the sensitivity can be adapted according to your task. Hint: Play around in the Image Segmenter toolbox for fast parameter finding. Next, the inferred circles are converted to integer values, since these are required for indexing.
min_radius = 10;
max_radius = 50;
% Find circles
[centers,radii,~] = imfindcircles(RGB,[min_radius max_radius],'ObjectPolarity','dark','Sensitivity',0.95);
centers = uint16(centers);
radii = uint16(radii);
The annotated image appears as follows:
Step 3: Get the brightness values of the circles.
From the circle center and radius values, we infer their respective brightness. It is sufficient to only check for the x and y pixel values left/right and above/below the center. (-1 is just a safety margin to completely stay within the circles.)
brightness_checker = zeros(2, max(radii), 2);
for i=1:size(centers,1)
current_radii = radii(i)-1;
for j=1:current_radii
% X-center minus radius, step along x-axis
brightness_checker(i, j, 1) = grayscale_image((centers(i,2) - current_radii/2) + j,...
(centers(i,1) - radii(i)/2) + j);
% Y-center minus radius, step along y-axis
brightness_checker(i, j, 2) = grayscale_image((centers(i,2) - current_radii/2) + j,...
(centers(i,1) - current_radii/2) + j);
end
end
Step 4: Check which circle is a pupil.
The determined value of 30 could potentially be enhanced.
median_x = median(brightness_checker(:,:,1),2);
median_y = median(brightness_checker(:,:,2),2);
is_pupil = (median_x<30)&(median_y<30);
pupils_center = centers(is_pupil == true,:);
Step 5: Draw the pupils.
The marker can be changed. Refer to:
https://de.mathworks.com/help/matlab/ref/matlab.graphics.chart.primitive.line-properties.html
figure
imshow(grayscale_image);
hold on
plot(centers(:,1), centers(:,2), 'r+', 'MarkerSize', 20, 'LineWidth', 2);
hold on
plot(pupils_center(:,1), pupils_center(:,2), 'b+', 'MarkerSize', 20, 'LineWidth', 2);
This is the final 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 have created a synthetic image that consists of a circle at the centre of a box with the code below.
%# Create a logical image of a circle with image size specified as follows:
imageSizeY = 400;
imageSizeX = 300;
[ygv, xgv] = meshgrid(1:imageSizeY, 1:imageSizeX);
%# Next create a logical mask for the circle with specified radius and center
centerY = imageSizeY/2;
centerX = imageSizeX/2;
radius = 100;
Img = double( (ygv - centerY).^2 + (xgv - centerX).^2 <= radius.^2 );
%# change image labels from double to numeric
for ii = 1:numel(Img)
if Img(ii) == 0
Img(ii) = 2; %change label from 0 to 2
end
end
%# plot image
RI = imref2d(size(Img),[0 size(Img, 2)],[0 size(Img, 1)]);
figure, imshow(Img, RI, [], 'InitialMagnification','fit');
Now, i need to create a rectangular mask (with label == 3, and row/col dimensions: 1 by imageSizeX) across the image from top to bottom and at known angles with the edges of the circle (see attached figure). Also, how can i make the rectangle thicker than 1 by imageSizeX?. As another option, I would love to try having the rectangle stop at say column 350. Lastly, any ideas how I can improve on the resolution? I mean is it possible to keep the image size the same while increasing/decreasing the resolution.
I have no idea how to go about this. Please i need any help/advice/suggestions that i can get. Many thanks!.
You can use the cos function to find the x coordinate with the correct angle phi.
First notice that the angle between the radius that intersects the vertex of phi has angle with the x-axis given by:
and the x coordinate of that vertex is given by
so the mask simply needs to set that row to 3.
Example:
phi = 45; % Desired angle in degrees
width = 350; % Desired width in pixels
height = 50; % Desired height of bar in pixels
theta = pi-phi*pi/180; % The radius angle
x = centerX + round(radius*cos(theta)); % Find the nearest row
x0 = max(1, x-height); % Find where to start the bar
Img(x0:x,1:width)=3;
The resulting image looks like:
Note that the max function is used to deal with the case where the bar thickness would extend beyond the top of the image.
Regarding resolution, the image resolution is determined by the size of the matrix you create. In your example that is (400,300). If you want higher resolution simply increase those numbers. However, if you would like to link the resolution to a higher DPI (Dots per Inch) so there are more pixels in each physical inch you can use the "Export Setup" window in the figure File menu.
Shown here:
I'm extracting the outline of blob the following way:
bw = im2bw(image, threshold);
boundaries = bwboundaries(bw);
plot(boundaries(:, 2), boundaries(:, 1), 'k', 'LineWidth', 2);
what I would like to do now, is to scale boundaries so that I can plot a smaller version of the boundaries inside the original boundaries. Is there an easy way to do this?
Here's an example on what the result should look like: black is the original bounding box, and red is the same bounding box, just scaled (but with same center as black box).
EDIT:
I guess I can scale each point individually, but then I still have to recenter the coordinates. Is there a better way of doing this?
scale = 0.7
nbr_points = size(b, 1);
b_min = nan(nbr_points, 2);
for k = 1 : nbr_points
b_min(k, :) = ([scale 0; 0 scale] * b(k, 1:2)')';
end
Just creating a function which does this should be easy.
function scaledB = scaleBoundaries(B,scaleFactor)
% B is a cell array of boundaries. The output is a cell array
% containing the scaled boundaries, with the same center of mass
% as the input boundaries.
%%
for k = 1:length(B)
scaledB{k} = B{k} .* scaleFactor;
com = mean(B{k}); % Take the center of mass of each boundary
sCom = mean(scaledB{k}); % Take the center of mass of each scaled boundary
difference = com-sCom; % Difference between the centers of mass
% Recenter the scaled boundaries, by adding the difference in the
% centers of mass to the scaled boundaries:
scaledB{k}(:,1) = scaledB{k}(:,1) + difference(1);
scaledB{k}(:,2) = scaledB{k}(:,2) + difference(2);
end
end
Or did you want to avoid something unoptimized for speed purposes?
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);
I have two images. One is the original, and the another is rotated.
Now, I need to discover the angle that the image was rotated. Until now, I thought about discovering the centroids of each color (as every image I will use has squares with colors in it) and use it to discover how much the image was rotated, but I failed.
I'm using this to discover the centroids and the color in the higher square in the image:
i = rgb2gray(img);
bw = im2bw(i,0.01);
s = regionprops(bw,'Centroid');
centroids = cat(1, s.Centroid);
colors = impixel(img,centroids(1),centroids(2));
top = max(centroids);
topcolor = impixel(img,top(1),top(2));
You can detect the corners of one of the colored rectangles in both the image and the rotated version, and use these as control points to infer the transformation between the two images (like in image registration) using the CP2TFORM function. We can then compute the angle of rotation from the affine transformation matrix:
Here is an example code:
%# read first image (indexed color image)
[I1 map1] = imread('http://i.stack.imgur.com/LwuW3.png');
%# constructed rotated image
deg = -15;
I2 = imrotate(I1, deg, 'bilinear', 'crop');
%# find blue rectangle
BW1 = (I1==2);
BW2 = imrotate(BW1, deg, 'bilinear', 'crop');
%# detect corners in both
p1 = corner(BW1, 'QualityLevel',0.5);
p2 = corner(BW2, 'QualityLevel',0.5);
%# sort corners coordinates in a consistent way (counter-clockwise)
p1 = sortrows(p1,[2 1]);
p2 = sortrows(p2,[2 1]);
idx = convhull(p1(:,1), p1(:,2)); p1 = p1(idx(1:end-1),:);
idx = convhull(p2(:,1), p2(:,2)); p2 = p2(idx(1:end-1),:);
%# make sure we have the same number of corner points
sz = min(size(p1,1),size(p2,1));
p1 = p1(1:sz,:); p2 = p2(1:sz,:);
%# infer transformation from corner points
t = cp2tform(p2,p1,'nonreflective similarity'); %# 'affine'
%# rotate image to match the other
II2 = imtransform(I2, t, 'XData',[1 size(I1,2)], 'YData',[1 size(I1,1)]);
%# recover affine transformation params (translation, rotation, scale)
ss = t.tdata.Tinv(2,1);
sc = t.tdata.Tinv(1,1);
tx = t.tdata.Tinv(3,1);
ty = t.tdata.Tinv(3,2);
translation = [tx ty];
scale = sqrt(ss*ss + sc*sc);
rotation = atan2(ss,sc)*180/pi;
%# plot the results
subplot(311), imshow(I1,map1), title('I1')
hold on, plot(p1(:,1),p1(:,2),'go')
subplot(312), imshow(I2,map1), title('I2')
hold on, plot(p2(:,1),p2(:,2),'go')
subplot(313), imshow(II2,map1)
title(sprintf('recovered angle = %g',rotation))
If you can identify a color corresponding to only one component it is easier to:
Calculate the centroids for each image
Calculate the mean of the centroids (in x and y) for each image. This is the "center" of each image
Get the red component color centroid (in your example) for each image
Subtract the mean of the centroids for each image from the red component color centroid for each image
Calculate the ArcTan2 for each of the vectors calculated in 4), and subtract the angles. That is your result.
If you have more than one figure of each color, you need to calculate all possible combinations for the rotation and then select the one that is compatible with the other possible rotations.
I could post the code in Mathematica, if you think it is useful.
I would take a variant to the above mentioned approach:
% Crude binarization method to knock out background and retain foreground
% features. Note one looses the cube in the middle
im = im > 1
Then I would get the 2D autocorrelation:
acf = normxcorr2(im, im);
From this result, one can easily detect the peaks, and as rotation carries into the autocorrelation function (ACF) domain, one can ascertain the rotation by matching the peaks between the original ACF and the ACF from the rotated image, for example using the so-called Hungarian algorithm.