This is a more specific problem with complex settings. I have calculated the distance of all 1-pixels to its nearest 0-pixels of raw image and transformed to an image with local maxima shown as below:
I used the following code to extract the local maxima from this transformed matrix:
loc_max = imregionalmax(loc,4);
figure;imshow(loc_max)
The loc_max gives out disconnected dots of those local maxima points. Tried a bunch of ways to connect them, but not working as it was supposed to do yet.
Here is one of my ideas:
[yy xx]=find(loc_max ==1);
d = [];
dmin = [];
idx = [];
for i = 1: size(yy,1)
for j = 1:size(xx,1)
if i ~= j
d(j) = sqrt(sum(bsxfun(#minus,[yy(j) xx(j)],[yy(i) xx(i)]).^2,2));
%calculate the distance between current 1-pixel in loc_max and all other local maxima pixels in loc_max.
end
end
[dmin(i),idx(i)] = min(d(:));%find the minimum distance between current 1-pixel to others
end
I tried to find the nearest 1-pixels in loc_max to the current 1-pixel in loc_max, and then connect them. But this isn't the solution yet. Because it won't connect to its next 1-pixel if the previous pixel is the nearest pixel of the current one.
In addition, I want to preserve the pixel information of those 0-pixels along the connected line between two disconnected 1-pixels. I want to be able to reconstruct the whole image out of this simplified skeleton later.
Any help would be greatly appreciated!
I have tried erode and dilate (like Ander Biguri advised).
I tried setting the angles of the kernels to be correct.
Check the following:
%Fix the input (remove JPEG artifacts and remove while margins)
%(This is not part of the solution - just a little cleanup).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
I = imread('https://i.stack.imgur.com/4L3fP.jpg'); %Read image from imgur.
I = im2bw(I);
[y0, x0] = find(I == 0);
[y1, x1] = find(I == 0, 1, 'last');
I = I(y0:y1, x0:x1);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
se0 = strel('disk', 3);
J = imdilate(I, se0); %Dilate - make line fat
se1 = strel('line', 30, 150); %150 degrees.
se2 = imdilate(se1.getnhood , se0); %Create 150 degrees fat line
J = imclose(J, se2); %Close - (dilate and erode)
se1 = strel('line', 30, 140); %140 degrees.
se2 = imdilate(se1.getnhood , se0);
J = imclose(J, se2);
se1 = strel('line', 80, 60); %60 degrees.
se2 = imdilate(se1.getnhood , se0); %Create 60 degrees fat line
J = imclose(J, se2);
se4 = strel('disk', 2);
J = imerode(J, se4); %Erode - make lines little thinner.
figure;imshow(J);
Result:
Do you think it's good enough?
Ander Biguri gets the credit for the following solution:
J = bwmorph(J,'skel',Inf);
Related
I'm trying to write a MATLAB script that does the following:
Given: pixel coordinates(x,y) for a .jpg image
Goal: Check, within a 5 pixel radius of given coordinates, if there is a pixel of a certain value.
For example, let's say I'm given the coordinates (100,100), then I want to check the neighborhood of (100,100) within my image for any pixels that are black (0,0,0). So perhaps, pixel (103, 100) and (104,100) might have the value (0,0,0).
Current code:
x_coord = uint32(coord(:,1));
y_coord = uint32(coord(:,2));
count = 0;
for i = 1:length(x_coord)
%(img(x,y) returns pixel value at that (x,y)
%Note 0 = black. Indicating that, at that position, the image is just
% black
if img(x_coord(i),y_coord(i)) == 0
count = count + 1;
end
end
It currently only checks at an exact location. Not in a local neighborhood. How to could I extend this?
EDIT: Also note, as long as there as at least one pixel in the neighborhood with the value, I increment count. I'm not trying to enumerate how many pixels in the neighborhood have that value, just trying to find evidence of at least one pixel that has that value.
EDIT:
Even though I am unable to identify an error with the code, I am not able to get the exact results I want. Here is the code I am using.
val = 0; %pixel value to check
N = 50; % neighbourhood radius
%2D grid of coordinates surrounding center coordinate
[R, C] = ndgrid(1 : size(img, 1), 1 : size(img, 2));
for kk = 1 : size(coord, 1)
r = coord(kk, 1); c = coord(kk, 2); % Get pixel locations
% mask of valid locations within the neighbourhood (avoid boundary problems)
mask = (R - r).^2 + (C - c).^2 <= N*N;
pix = img(mask); % Get the valid pixels
valid = any(pix(:) ~= val);
% Add either 0 or 1 depending if we have found any matching pixels
if(valid == 1)
img = insertMarker(img, [r c], 'x', 'color', 'red', 'size', 10);
imwrite(img, images(i).name,'tiff');
end
count = count + valid;
end
An easier way to do this would be to use indexing to grab a neighbourhood, then to check to see if any of the pixels in the neighbourhood have the value that you're looking for, use any on a flattened version of this neighbourhood. The trick with grabbing the right neighbourhood is to first generate a 2D grid of coordinates that span the entire dimensions of your image, then simply use the equation of a circle with the centre of it being each coordinate you are looking at and determine those locations that satisfy the following equation:
(x - a)^2 + (y - b)^2 <= N^2
N is the radius of the observation window, (a, b) is a coordinate of interest while (x, y) is a coordinate in the image. Use meshgrid to generate the coordinates.
You would use the above equation to create a logical mask, index into your image to pull the locations that are valid within the mask and check how many pixels match the one you want. Another added benefit with the above approach is that you are not subject to any out of bounds errors. Because you are pre-generating the list of all valid coordinates in your image, generating the mask will confine you within the boundaries of the image so you never have to check for out of boundaries conditions.... even when you specify coordinates to search that are out of bounds.
Specifically, assuming your image is stored in img, you would do:
count = 0; % Remembers total count of pixels matching a value
val = 0; % Value to match
N = 50; % Radius of neighbourhood
% Generate 2D grid of coordinates
[x, y] = meshgrid(1 : size(img, 2), 1 : size(img, 1));
% For each coordinate to check...
for kk = 1 : size(coord, 1)
a = coord(kk, 1); b = coord(kk, 2); % Get the pixel locations
mask = (x - a).^2 + (y - b).^2 <= N*N; % Get a mask of valid locations
% within the neighbourhood
pix = img(mask); % Get the valid pixels
count = count + any(pix(:) == val); % Add either 0 or 1 depending if
% we have found any matching pixels
end
The proposed solution:
fc = repmat(-5:5,11,1);
I = (fc.^2+fc'.^2)<=25;
fc_x = fc(I);
fc_y = fc'; fc_y = fc_y(I);
for i = 1:length(x_coord)
x_toCheck = fc_x + x_coord(i);
y_toCheck = fc_y + y_coord(i);
I = x_toCheck>0 & x_toCheck<=yourImageWidth;
I = I.*(y_toCheck>0 & y_toCheck<=yourImageHeight);
x_toCheck = x_toCheck(logical(I));
y_toCheck = y_toCheck(logical(I));
count = sum(img(x_toCheck(:),y_toCheck(:)) == 0);
end
If your img function can only check one pixel at a time, just add a for loop:
for i = 1:length(x_coord)
x_toCheck = fc_x + x_coord(i);
y_toCheck = fc_y + y_coord(i);
I = x_toCheck>0 & x_toCheck<=yourImageWidth;
I = I.*(y_toCheck>0 & y_toCheck<=yourImageHeight);
x_toCheck = x_toCheck(logical(I));
y_toCheck = y_toCheck(logical(I));
for j = 1:length(x_toCheck)
count = count + (img(x_toCheck(j),y_toCheck(j)) == 0);
end
end
Step-by-step:
You first need to get all the coordinates within 5 pixels range of the given coordinate.
We start by building a square of 11 pixels in length/width.
fc = repmat(-5:5,11,1);
fc_x = fc;
fc_y = fc';
plot(fc_x,fc_y,'.');
We now need to build a filter to get rid of those points outside the 5-pixel radius.
I = (fc.^2+fc'.^2)<=25;
Apply the filter, so we can get a circle of 5-pixel radius.
fc_x = fc_x(I);
fc_y = fc_y(I);
Next translate the centre of the circle to the given coordinate:
x_toCheck = fc_x + x_coord(i);
y_toCheck = fc_y + y_coord(i);
You need to check whether part of the circle is outside the range of your image:
I = x_toCheck>0 & x_toCheck<=yourImageWidth;
I = I.*(y_toCheck>0 & y_toCheck<=yourImageHeight);
x_toCheck = x_toCheck(logical(I));
y_toCheck = y_toCheck(logical(I));
Finally count the pixels:
count = sum(img(x_toCheck,y_toCheck) == 0);
I was trying to implement the IBVS algorithm (the one explained in the Introduction here) in MATLAB myself, but I am facing the following problem : The algorithm seems to work only for the cases that the camera does not have to change its orientation in respect to the world frame.For example, if I just try to make one vertex of the initial (almost) square go closer to its opposite vertex, the algorithm does not work, as can be seen in the following image
The red x are the desired projections, the blue circles are the initial ones and the green ones are the ones I get from my algorithm.
Also the errors are not exponentially dereasing as they should.
What am I doing wrong? I am attaching my MATLAB code which is fully runable. If anyone could take a look, I would be really grateful. I took out the code that was performing the plotting. I hope it is more readable now. Visual servoing has to be performed with at least 4 target points, because else the problem has no unique solution. If you are willing to help, I would suggest you take a look at the calc_Rotation_matrix() function to check that the rotation matrix is properly calculated, then verify that the line ds = vc; in euler_ode is correct. The camera orientation is expressed in Euler angles according to this convention. Finally, one could check if the interaction matrix L is properly calculated.
function VisualServo()
global A3D B3D C3D D3D A B C D Ad Bd Cd Dd
%coordinates of the 4 points wrt camera frame
A3D = [-0.2633;0.27547;0.8956];
B3D = [0.2863;-0.2749;0.8937];
C3D = [-0.2637;-0.2746;0.8977];
D3D = [0.2866;0.2751;0.8916];
%initial projections (computed here only to show their relation with the desired ones)
A=A3D(1:2)/A3D(3);
B=B3D(1:2)/B3D(3);
C=C3D(1:2)/C3D(3);
D=D3D(1:2)/D3D(3);
%initial camera position and orientation
%orientation is expressed in Euler angles (X-Y-Z around the inertial frame
%of reference)
cam=[0;0;0;0;0;0];
%desired projections
Ad=A+[0.1;0];
Bd=B;
Cd=C+[0.1;0];
Dd=D;
t0 = 0;
tf = 50;
s0 = cam;
%time step
dt=0.01;
t = euler_ode(t0, tf, dt, s0);
end
function ts = euler_ode(t0,tf,dt,s0)
global A3D B3D C3D D3D Ad Bd Cd Dd
s = s0;
ts=[];
for t=t0:dt:tf
ts(end+1)=t;
cam = s;
% rotation matrix R_WCS_CCS
R = calc_Rotation_matrix(cam(4),cam(5),cam(6));
r = cam(1:3);
% 3D coordinates of the 4 points wrt the NEW camera frame
A3D_cam = R'*(A3D-r);
B3D_cam = R'*(B3D-r);
C3D_cam = R'*(C3D-r);
D3D_cam = R'*(D3D-r);
% NEW projections
A=A3D_cam(1:2)/A3D_cam(3);
B=B3D_cam(1:2)/B3D_cam(3);
C=C3D_cam(1:2)/C3D_cam(3);
D=D3D_cam(1:2)/D3D_cam(3);
% computing the L matrices
L1 = L_matrix(A(1),A(2),A3D_cam(3));
L2 = L_matrix(B(1),B(2),B3D_cam(3));
L3 = L_matrix(C(1),C(2),C3D_cam(3));
L4 = L_matrix(D(1),D(2),D3D_cam(3));
L = [L1;L2;L3;L4];
%updating the projection errors
e = [A-Ad;B-Bd;C-Cd;D-Dd];
%compute camera velocity
vc = -0.5*pinv(L)*e;
%change of the camera position and orientation
ds = vc;
%update camera position and orientation
s = s + ds*dt;
end
ts(end+1)=tf+dt;
end
function R = calc_Rotation_matrix(theta_x, theta_y, theta_z)
Rx = [1 0 0; 0 cos(theta_x) -sin(theta_x); 0 sin(theta_x) cos(theta_x)];
Ry = [cos(theta_y) 0 sin(theta_y); 0 1 0; -sin(theta_y) 0 cos(theta_y)];
Rz = [cos(theta_z) -sin(theta_z) 0; sin(theta_z) cos(theta_z) 0; 0 0 1];
R = Rx*Ry*Rz;
end
function L = L_matrix(x,y,z)
L = [-1/z,0,x/z,x*y,-(1+x^2),y;
0,-1/z,y/z,1+y^2,-x*y,-x];
end
Cases that work:
Ad=2*A;
Bd=2*B;
Cd=2*C;
Dd=2*D;
Ad=A+1;
Bd=B+1;
Cd=C+1;
Dd=D+1;
Ad=2*A+1;
Bd=2*B+1;
Cd=2*C+1;
Dd=2*D+1;
Cases that do NOT work:
Rotation by 90 degrees and zoom out (zoom out alone works, but I am doing it here for better visualization)
Ad=2*D;
Bd=2*C;
Cd=2*A;
Dd=2*B;
Your problem comes from the way you move the camera from the resulting visual servoing velocity. Rather than
cam = cam + vc*dt;
you should compute the new camera position using the exponential map
cam = cam*expm(vc*dt)
I've an image over which I would like to compute a local histogram within a circular neighborhood. The size of the neighborhood is given by a radius. Although the code below does the job, it's computationally expensive. I run the profiler and the way I'm accessing to the pixels within the circular neighborhoods is already expensive.
Is there any sort of improvement/optimization based maybe on vectorization? Or for instance, storing the neighborhoods as columns?
I found a similar question in this post and the proposed solution is quite in the spirit of the code below, however the solution is still not appropriate to my case. Any ideas are really welcomed :-) Imagine for the moment, the image is binary, but the method should also ideally work with gray-level images :-)
[rows,cols] = size(img);
hist_img = zeros(rows, cols, 2);
[XX, YY] = meshgrid(1:cols, 1:rows);
for rr=1:rows
for cc=1:cols
distance = sqrt( (YY-rr).^2 + (XX-cc).^2 );
mask_radii = (distance <= radius);
bwresponses = img(mask_radii);
[nelems, ~] = histc(double(bwresponses),0:255);
% do some processing over the histogram
...
end
end
EDIT 1 Given the received feedback, I tried to update the solution. However, it's not yet correct
radius = sqrt(2.0);
disk = diskfilter(radius);
fun = #(x) histc( x(disk>0), min(x(:)):max(x(:)) );
output = im2col(im, size(disk), fun);
function disk = diskfilter(radius)
height = 2*ceil(radius)+1;
width = 2*ceil(radius)+1;
[XX,YY] = meshgrid(1:width,1:height);
dist = sqrt((XX-ceil(width/2)).^2+(YY-ceil(height/2)).^2);
circfilter = (dist <= radius);
end
Following on the technique I described in my answer to a similar question you could try to do the following:
compute the index offsets from a particular voxel that get you to all the neighbors within a radius
Determine which voxels have all neighbors at least radius away from the edge
Compute the neighbors for all these voxels
Generate your histograms for each neighborhood
It is not hard to vectorize this, but note that
It will be slow when the neighborhood is large
It involves generating an intermediate matrix that is NxM (N = voxels in image, M = voxels in neighborhood) which could get very large
Here is the code:
% generate histograms for neighborhood within radius r
A = rand(200,200,200);
radius = 2.5;
tic
sz=size(A);
[xx yy zz] = meshgrid(1:sz(2), 1:sz(1), 1:sz(3));
center = round(sz/2);
centerPoints = find((xx - center(1)).^2 + (yy - center(2)).^2 + (zz - center(3)).^2 < radius.^2);
centerIndex = sub2ind(sz, center(1), center(2), center(3));
% limit to just the points that are "far enough on the inside":
inside = find(xx > radius+1 & xx < sz(2) - radius & ...
yy > radius + 1 & yy < sz(1) - radius & ...
zz > radius + 1 & zz < sz(3) - radius);
offsets = centerPoints - centerIndex;
allPoints = 1:prod(sz);
insidePoints = allPoints(inside);
indices = bsxfun(#plus, offsets, insidePoints);
hh = histc(A(indices), 0:0.1:1); % <<<< modify to give you the histogram you want
toc
A 2D version of the same code (which might be all you need, and is considerably faster):
% generate histograms for neighborhood within radius r
A = rand(200,200);
radius = 2.5;
tic
sz=size(A);
[xx yy] = meshgrid(1:sz(2), 1:sz(1));
center = round(sz/2);
centerPoints = find((xx - center(1)).^2 + (yy - center(2)).^2 < radius.^2);
centerIndex = sub2ind(sz, center(1), center(2));
% limit to just the points that are "far enough on the inside":
inside = find(xx > radius+1 & xx < sz(2) - radius & ...
yy > radius + 1 & yy < sz(1) - radius);
offsets = centerPoints - centerIndex;
allPoints = 1:prod(sz);
insidePoints = allPoints(inside);
indices = bsxfun(#plus, offsets, insidePoints);
hh = histc(A(indices), 0:0.1:1); % <<<< modify to give you the histogram you want
toc
You're right, I don't think that colfilt can be used as you're not applying a filter. You'll have to check the correctness, but here's my attempt using im2col and your diskfilter function (I did remove the conversion to double so it now output logicals):
function circhist
% Example data
im = randi(256,20)-1;
% Ranges - I do this globally for the whole image rather than for each neighborhood
mini = min(im(:));
maxi = max(im(:));
edges = linspace(mini,maxi,20);
% Disk filter
radius = sqrt(2.0);
disk = diskfilter(radius); % Returns logical matrix
% Pad array with -1
im_pad = padarray(im, (size(disk)-1)/2, -1);
% Convert sliding neighborhoods to columns
B = im2col(im_pad, size(disk), 'sliding');
% Get elements from each column that correspond to disk (logical indexing)
C = B(disk(:), :);
% Apply histogram across columns to count number of elements
out = histc(C, edges)
% Display output
figure
imagesc(out)
h = colorbar;
ylabel(h,'Counts');
xlabel('Neighborhood #')
ylabel('Bins')
axis xy
function disk = diskfilter(radius)
height = 2*ceil(radius)+1;
width = 2*ceil(radius)+1;
[XX,YY] = meshgrid(1:width,1:height);
dist = sqrt((XX-ceil(width/2)).^2+(YY-ceil(height/2)).^2);
disk = (dist <= radius);
If you want to set your ranges (edges) based on each neighborhood then you'll need to make sure that the vector is always the same length if you want to build a big matrix (and then the rows of that matrix won't correspond to each other).
You should note that the shape of the disk returned by fspecial is not as circular as what you were using. It's meant to be used a smoothing/averaging filter so the edges are fuzzy (anti-aliased). Thus when you use ~=0 it will grab more pixels. It'd stick with your own function, which is faster anyways.
You could try processing with an opposite logic (as briefly explained in the comment)
hist = zeros(W+2*R, H+2*R, Q);
for i = 1:R+1;
for j = 1:R+1;
if ((i-R-1)^2+(j-R-1)^2 < R*R)
for q = 0:1:Q-1;
hist(i:i+W-1,j:j+H-1,q+1) += (image == q);
end
end
end
end
I am new to Matlab and to Image Processing as well. I am working on separating background and foreground in images like this
I have hundreds of images like this, found here. By trial and error I found out a threshold (in RGB space): the red layer is always less than 150 and the green and blue layers are greater than 150 where the background is.
so if my RGB image is I and my r,g and b layers are
redMatrix = I(:,:,1);
greenMatrix = I(:,:,2);
blueMatrix = I(:,:,3);
by finding coordinates where in red, green and blue the values are greater or less than 150 I can get the coordinates of the background like
[r1 c1] = find(redMatrix < 150);
[r2 c2] = find(greenMatrix > 150);
[r3 c3] = find(blueMatrix > 150);
now I get coordinates of thousands of pixels in r1,c1,r2,c2,r3 and c3.
My questions:
How to find common values, like the coordinates of the pixels where red is less than 150 and green and blue are greater than 150?
I have to iterate every coordinate of r1 and c1 and check if they occur in r2 c2 and r3 c3 to check it is a common point. but that would be very expensive.
Can this be achieved without a loop ?
If somehow I came up with common points like [commonR commonC] and commonR and commonC are both of order 5000 X 1, so to access this background pixel of Image I, I have to access first commonR then commonC and then access image I like
I(commonR(i,1),commonC(i,1))
that is expensive too. So again my question is can this be done without loop.
Any help would be appreciated.
I got solution with #Science_Fiction answer's
Just elaborating his/her answer
I used
mask = I(:,:,1) < 150 & I(:,:,2) > 150 & I(:,:,3) > 150;
No loop is needed. You could do it like this:
I = imread('image.jpg');
redMatrix = I(:,:,1);
greenMatrix = I(:,:,2);
blueMatrix = I(:,:,3);
J(:,:,1) = redMatrix < 150;
J(:,:,2) = greenMatrix > 150;
J(:,:,3) = blueMatrix > 150;
J = 255 * uint8(J);
imshow(J);
A greyscale image would also suffice to separate the background.
K = ((redMatrix < 150) + (greenMatrix > 150) + (blueMatrix > 150))/3;
imshow(K);
EDIT
I had another look, also using the other images you linked to.
Given the variance in background colors, I thought you would get better results deriving a threshold value from the image histogram instead of hardcoding it.
Occasionally, this algorithm is a little to rigorous, e.g. erasing part of the clothes together with the background. But I think over 90% of the images are separated pretty well, which is more robust than what you could hope to achieve with a fixed threshold.
close all;
path = 'C:\path\to\CUHK_training_cropped_photos\photos';
files = dir(path);
bins = 16;
for f = 3:numel(files)
fprintf('%i/%i\n', f, numel(files));
file = files(f);
if isempty(strfind(file.name, 'jpg'))
continue
end
I = imread([path filesep file.name]);
% Take the histogram of the blue channel
B = I(:,:,3);
h = imhist(B, bins);
h2 = h(bins/2:end);
% Find the most common bin in the *upper half*
% of the histogram
m = bins/2 + find(h2 == max(h2));
% Set the threshold value somewhat below
% the value corresponding to that bin
thr = m/bins - .25;
BW = im2bw(B, thr);
% Pad with ones to ensure background connectivity
BW = padarray(BW, [1 1], 1);
% Find connected regions in BW image
CC = bwconncomp(BW);
L = labelmatrix(CC);
% Crop back again
L = L(2:end-1,2:end-1);
% Set the largest region in the orignal image to white
for c = 1:3
channel = I(:,:,c);
channel(L==1) = 255;
I(:,:,c) = channel;
end
% Show the results with a pause every 16 images
subplot(4,4,mod(f-3,16)+1);
imshow(I);
title(sprintf('Img %i, thr %.3f', f, thr));
if mod(f-3,16)+1 == 16
pause
clf
end
end
pause
close all;
Results:
Your approach seems basic but decent. Since for this particular image the background is composed of mainly blue so you be crude and do:
mask = img(:,:,3) > 150;
This will set those pixels which evaluate to true for > 150 to 0 and false to 1. You will have a black and white image though.
imshow(mask);
To add colour back
mask3d(:,:,1) = mask;
mask3d(:,:,2) = mask;
mask3d(:,:,3) = mask;
img(mask3d) = 255;
imshow(img);
Should give you the colour image of face hopefully, with a pure white background. All this requires some trial and error.
using Hough Transform, how can I detect and get coordinates of (x0,y0) and "a" and "b" of an ellipse in 2D space?
This is ellipse01.bmp:
I = imread('ellipse01.bmp');
[m n] = size(I);
c=0;
for i=1:m
for j=1:n
if I(i,j)==1
c=c+1;
p(c,1)=i;
p(c,2)=j;
end
end
end
Edges=transpose(p);
Size_Ellipse = size(Edges);
B = 1:ceil(Size_Ellipse(1)/2);
Acc = zeros(length(B),1);
a1=0;a2=0;b1=0;b2=0;
Ellipse_Minor=[];Ellipse_Major=[];Ellipse_X0 = [];Ellipse_Y0 = [];
Global_Threshold = ceil(Size_Ellipse(2)/6);%Used for Major Axis Comparison
Local_Threshold = ceil(Size_Ellipse(1)/25);%Used for Minor Axis Comparison
[Y,X]=find(Edges);
Limit=numel(Y);
Thresh = 150;
Para=[];
for Count_01 =1:(Limit-1)
for Count_02 =(Count_01+1):Limit
if ((Count_02>Limit) || (Count_01>Limit))
continue
end
a1=Y(Count_01);b1=X(Count_01);
a2=Y(Count_02);b2=X(Count_02);
Dist_01 = (sqrt((a1-a2)^2+(b1-b2)^2));
if (Dist_01 >Global_Threshold)
Center_X0 = (b1+b2)/2;Center_Y0 = (a1+a2)/2;
Major = Dist_01/2.0;Alpha = atan((a2-a1)/(b2-b1));
if(Alpha == 0)
for Count_03 = 1:Limit
if( (Count_03 ~= Count_01) || (Count_03 ~= Count_02))
a3=Y(Count_03);b3=X(Count_03);
Dist_02 = (sqrt((a3 - Center_Y0)^2+(b3 - Center_X0)^2));
if(Dist_02 > Local_Threshold)
Cos_Tau = ((Major)^2 + (Dist_02)^2 - (a3-a2)^2 - (b3-b2)^2)/(2*Major*Dist_02);
Sin_Tau = 1 - (Cos_Tau)^2;
Minor_Temp = ((Major*Dist_02*Sin_Tau)^2)/(Major^2 - ((Dist_02*Cos_Tau)^2));
if((Minor_Temp>1) && (Minor_Temp<B(end)))
Acc(round(Minor_Temp)) = Acc(round(Minor_Temp))+1;
end
end
end
end
end
Minor = find(Acc == max(Acc(:)));
if(Acc(Minor)>Thresh)
Ellipse_Minor(end+1)=Minor(1);Ellipse_Major(end+1)=Major;
Ellipse_X0(end+1) = Center_X0;Ellipse_Y0(end+1) = Center_Y0;
for Count = 1:numel(X)
Para_X = ((X(Count)-Ellipse_X0(end))^2)/(Ellipse_Major(end)^2);
Para_Y = ((Y(Count)-Ellipse_Y0(end))^2)/(Ellipse_Minor(end)^2);
if (((Para_X + Para_Y)>=-2)&&((Para_X + Para_Y)<=2))
Edges(X(Count),Y(Count))=0;
end
end
end
Acc = zeros(size(Acc));
end
end
end
Although this is an old question, perhaps what I found can help someone.
The main problem of using the normal Hough Transform to detect ellipses is the dimension of the accumulator, since we would need to vote for 5 variables (the equation is explained here):
There is a very nice algorithm where the accumulator can be a simple 1D array, for example, and that runs in . If you wanna see code, you can look at here (the image used to test was that posted above).
If you use circle for rough transform is given as rho = xcos(theta) + ysin(theta)
For ellipse since it is
You could transform the equation as
rho = axcos(theta) + bysin(theta)
Although I am not sure if you use standard Hough Transform, for ellipse-like transforms, you could manipulate the first given function.
If your ellipse is as provided, being a true ellipse and not a noisy sample of points;
the search for the two furthest points gives the ends of the major axis,
the search for the two nearest points gives the ends of the minor axis,
the intersection of these lines (you can check it's a right angle) occurs at the centre.
If you know the 'a' and 'b' of an ellipse then you can rescale the image by factor of a/b in one direction and look for circle. I am still thinking about what to do when a and b are unknown.
If you know that it is circle then use Hough transform for circles. Here is a sample code:
int accomulatorResolution = 1; // for each pixel
int minDistBetweenCircles = 10; // In pixels
int cannyThresh = 20;
int accomulatorThresh = 5*_accT+1;
int minCircleRadius = 0;
int maxCircleRadius = _maxR*10;
cvClearMemStorage(storage);
circles = cvHoughCircles( gryImage, storage,
CV_HOUGH_GRADIENT, accomulatorResolution,
minDistBetweenCircles,
cannyThresh , accomulatorThresh,
minCircleRadius,maxCircleRadius );
// Draw circles
for (int i = 0; i < circles->total; i++){
float* p = (float*)cvGetSeqElem(circles,i);
// Draw center
cvCircle(dstImage, cvPoint(cvRound(p[0]),cvRound(p[1])),
1, CV_RGB(0,255,0), -1, 8, 0 );
// Draw circle
cvCircle(dstImage, cvPoint(cvRound(p[0]),cvRound(p[1])),
cvRound(p[2]),CV_RGB(255,0,0), 1, 8, 0 );
}