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Non overlapping randomly located circles
(4 answers)
Closed 5 years ago.
I have written a MATLAB code to be able to visualize some Circles. Please have a look at my below code and the attached figure as the output.
clc;
clear;
close all;
% X and Y of each Center
Xloc = [1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5];
Yloc = [1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4,5,5,5,5,5];
% Radius of each circle
radius = unifrnd(0,1,[1 numel(Xloc)]);
% Random colours
allColours = rand(numel(Xloc),3);
% Transform the data into position = [left bottom width height]
pos = [Xloc(:)-radius(:) Yloc(:)-radius(:) 2*radius(:)*[1 1]];
% Create and format the axes
H = axes;
hold on;
axis equal;
box on;
set(H,'XTickLabel',[],'YTickLabel',[]);
% Create the circles
for idx = 1:numel(Xloc);
rectangle(...
'Position',pos(idx,:),...
'Curvature',[1 1],...
'FaceColor',allColours(idx,:),...
'EdgeColor','none');
end
The output figure is (Circles' radius is generated randomly, so if you execute the code, you will face with a new output):
As you can see in the figure, there is overlap between circles. I was wondering how can I separate centers from each other to do not overlap each others, and at the same time they keep the original distance (or similar distance) from each other in [Xloc Yloc]
You can try to formulate an optimisation problem:
You have to constrain the minimal distance between the centers based on the desired radii.
You have to minimise the deviations from its desired center location.
Related
I want to convert an image from Cartesian to Polar and to use it for opengl texture.
So I used matlab referring to the two articles below.
Link 1
Link 2
My code is exactly same with Link 2's anwser
% load image
img = imread('my_image.png');
% convert pixel coordinates from cartesian to polar
[h,w,~] = size(img);
[X,Y] = meshgrid((1:w)-floor(w/2), (1:h)-floor(h/2));
[theta,rho] = cart2pol(X, Y);
Z = zeros(size(theta));
% show pixel locations (subsample to get less dense points)
XX = X(1:8:end,1:4:end);
YY = Y(1:8:end,1:4:end);
tt = theta(1:8:end,1:4:end);
rr = rho(1:8:end,1:4:end);
subplot(121), scatter(XX(:),YY(:),3,'filled'), axis ij image
subplot(122), scatter(tt(:),rr(:),3,'filled'), axis ij square tight
% show images
figure
subplot(121), imshow(img), axis on
subplot(122), warp(theta, rho, Z, img), view(2), axis square
The result was exactly what I wanted, and I was very satisfied except for one thing. It's the area (red circled area) in the picture just below. Considering that the opposite side (blue circled area) is not, I think this part should also be filled. Because of this part is empty, so there is a problem when using it as a texture.
And I wonder how I can fill this part. Thank you.
(little difference from Link 2's answer code like degree<->radian and axis values. but i think it is not important.)
Those issues you show in your question happen because your algorithm is wrong.
What you did (push):
throw a grid on the source image
transform those points
try to plot these colored points and let MATLAB do some magic to make it look like a dense picture
Do it the other way around (pull):
throw a grid on the output
transform that backwards
sample the input at those points
The distinction is called "push" (into output) vs "pull" (from input). Only Pull gives proper results.
Very little MATLAB code is necessary. You just need pol2cart and interp2, and a meshgrid.
With interp2 you get to choose the interpolation (linear, cubic, ...). Nearest-neighbor interpolation leaves visible artefacts.
im = im2single(imread("PQFax.jpg"));
% center of polar map, manually picked
cx = 10 + 409/2;
cy = 7 + 413/2;
% output parameters
radius = 212;
dRho = 1;
dTheta = 2*pi / (2*pi * radius);
Thetas = pi/2 - (0:dTheta:2*pi);
Rhos = (0:dRho:radius);
% polar mesh
[Theta, Rho] = meshgrid(Thetas, Rhos);
% transform...
[Xq,Yq] = pol2cart(Theta, Rho);
% translate to sit on the circle's center
Xq = Xq + cx;
Yq = Yq + cy;
% sample image at those points
Ro = interp2(im(:,:,1), Xq,Yq, "cubic");
Go = interp2(im(:,:,2), Xq,Yq, "cubic");
Bo = interp2(im(:,:,3), Xq,Yq, "cubic");
Vo = cat(3, Ro, Go, Bo);
Vo = imrotate(Vo, 180);
imshow(Vo)
The other way around (get a "torus" from a "ribbon") is quite similar. Throw a meshgrid on the torus space, subtract center, transform from cartesian to polar, and use those to sample from the "ribbon" image into the "torus" image.
I'm more familiar with OpenCV than with MATLAB. Perhaps MATLAB has something like OpenCV's warpPolar(), or a generic remap(). In any case, the operation is trivial to do entirely "by hand" but there are enough supporting functions to take the heavy lifting off your hands (interp2, pol2cart, meshgrid).
1.- The white arcs tell that the used translation pol-cart introduces significant errors.
2.- Reversing the following script solves your question.
It's a script that goes from cart-pol without introducing errors or ignoring input data, which is what happens when such wide white arcs show up upon translation apparently correct.
clear all;clc;close all
clc,cla;
format long;
A=imread('shaffen dass.jpg');
[sz1 sz2 sz3]=size(A);
szx=sz2;szy=sz1;
A1=A(:,:,1);A2=A(:,:,2);A3=A(:,:,3); % working with binary maps or grey scale images this wouldn't be necessary
figure(1);imshow(A);
hold all;
Cx=floor(szx/2);Cy=floor(szy/2);
plot(Cx,Cy,'co'); % because observe image centre not centered
Rmin=80;Rmax=400; % radius search range for imfindcircles
[centers, radii]=imfindcircles(A,[Rmin Rmax],... % outer circle
'ObjectPolarity','dark','Sensitivity',0.9);
h=viscircles(centers,radii);
hold all; % inner circle
[centers2, radii2]=imfindcircles(A,[Rmin Rmax],...
'ObjectPolarity','bright');
h=viscircles(centers2,radii2);
% L=floor(.5*(radii+radii2)); % this is NOT the length X that should have the resulting XY morphed graph
L=floor(2*pi*radii); % expected length of the morphed graph
cx=floor(.5*(centers(1)+centers2(1))); % coordinates of averaged circle centres
cy=floor(.5*(centers(2)+centers2(2)));
plot(cx,cy,'r*'); % check avg centre circle is not aligned to figure centre
plot([cx 1],[cy 1],'r-.');
t=[45:360/L:404+1-360/L]; % if step=1 then we only get 360 points but need an amount of L points
% if angle step 1/L over minute waiting for for loop to finish
R=radii+5;x=R*sind(t)+cx;y=R*cosd(t)+cy; % build outer perimeter
hL1=plot(x,y,'m'); % axis equal;grid on;
% hold all;
% plot(hL1.XData,hL1.YData,'ro');
x_ref=hL1.XData;y_ref=hL1.YData;
% Sx=zeros(ceil(R),1);Sy=zeros(ceil(R),1);
Sx={};Sy={};
for k=1:1:numel(hL1.XData)
Lx=floor(linspace(x_ref(k),cx,ceil(R)));
Ly=floor(linspace(y_ref(k),cy,ceil(R)));
% plot(Lx,Ly,'go'); % check
% plot([cx x(k)],[cy y(k)],'r');
% L1=unique([Lx;Ly]','rows');
Sx=[Sx Lx'];Sy=[Sy Ly'];
end
sx=cell2mat(Sx);sy=cell2mat(Sy);
[s1 s2]=size(sx);
B1=uint8(zeros(s1,s2));
B2=uint8(zeros(s1,s2));
B3=uint8(zeros(s1,s2));
for n=1:1:s2
for k=1:1:s1
B1(k,n)=A1(sx(k,n),sy(k,n));
B2(k,n)=A2(sx(k,n),sy(k,n));
B3(k,n)=A3(sx(k,n),sy(k,n));
end
end
C=uint8(zeros(s1,s2,3));
C(:,:,1)=B1;
C(:,:,2)=B2;
C(:,:,3)=B3;
figure(2);imshow(C);
the resulting
3.- let me know if you'd like some assistance writing pol-cart from this script.
Regards
John BG
I have a series of the experimental images. In order to process it, at first, I need to determine the area of interest. This area is between two concentric circles. These circles can be a little different on each image (e.g. center can be shifted by a small distance).
In order to find these circles, I converted my image to binary image. Now it looks like this:
I am interested in the biggest circle and the one after the biggest (See note on the picture). Does anyone know a fast algorithm that can find it? Exhausted search seems to be a bad option. I know, that at each image the locations and radii of these circles can change only by a little. Here is a link to .mat file with the picture I attached link.
Thanks,
You can achieve this task even without using inbuilt function imfindcircles. I follow the following method. Get all the non-zero pixels using find. Now you have all the points, I want to classify them into different circles, for this I will use histogram. I can fix the number of bins (which is number of circles) here. Once I got all the points separated, I have scattered points along a circle..with these points, I will fit a circle along those points. You may check the below code:
load Sample.mat ;
I = Immm ;
[y,x] = find(I) ;
%% Get Bounding box
x0 = min(x) ; x1 = max(x) ;
y0 = min(y) ; y1 = max(y) ;
% length and breadth
L = abs(x1-x0) ;
B = abs(y1-y0) ;
% center bounding box
C = [x0+B/2 y0+L/2] ;
%% Get distances of the points from center of bounding box
data = repmat(C,[length(x),1])-[x,y] ;
dist = sqrt(data(:,1).^2+data(:,2).^2);
%% Classify the points into circles
nbins = 4 ; % number of circles you want
[N,edges,bin] = histcounts(dist,nbins) ;
% plot the classified circle points for check
figure(1)
imshow(I)
hold on
for i = 1:nbins
plot((x(bin==i)),y(bin==i),'.','color',rand(1,3)) ;
end
%% Circle's radii and center
Circ = cell(nbins,1) ;
for i = 1:nbins
[xc1,yc1,R] = circfit(x(bin==i),y(bin==i)) ;
Circ{i} = [xc1 yc1 R] ;
end
figure(2)
imshow(I)
hold on
th = linspace(0,2*pi) ;
for i = 1:nbins
xc = Circ{i}(1)+Circ{i}(3)*cos(th) ;
yc = Circ{i}(2)+Circ{i}(3)*sin(th) ;
plot(xc,yc,'color',rand(1,3),'linewidth',3) ;
end
Circ is cell, which centers and Radii of circles. You can download the function circfit from this link: http://matlab.wikia.com/wiki/FAQ#How_can_I_fit_a_circle_to_a_set_of_XY_data.3F
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I have one vector of radii and second vector of hundreds of [X,Y] coordinates. For each possible coordinate-radius pair I have count all black pixels within a circle (whose center is placed in the coordinate) on an input binary image.
What is the fastest way to do it? My only idea is to iterate through every pixel of an image, check the circle equation and then the pixel color but it doesn't seem much optimized for several hundred such operations.
Matlab is great for working with images thanks to the matrix syntax. It does also work with indices so most time you can avoid "iterating through pixels" (although sometimes you'll still have to).
Instead of checking all the pixels within each circle, and having to detect how many pixels were counted twice, another approach is to create a mask, the same size of you image. Blank this mask for each of your circles (so overlapping pixels are only 'blanked' once), then apply the mask on your original picture and count the remaining illuminated pixels.
For an example, I have to take some sample data, the image:
load trees
BW = im2bw(X,map,0.4);
imshow(BW)
And 20 random point/circle coordinates (you can change the number of points and the min/max radii easily):
%// still building sample data
s = size(BW) ;
npt = 20 ; Rmin=5 ; Rmax=20 ; %// problem constants
x = randi([1 s(2)] ,npt,1); %// random X coordinates
y = randi([1 s(1)] ,npt,1); %// Y
r = randi([Rmin Rmax],npt,1); %// radius size between 5 to 20 pixels.
Then we build your custom mask :
%// create empty mask with enough overlap for the circles on the border of the image
mask = false( s+2*Rmax ) ;
%// prepare grid for a sub-mask of pixels, as wide as the maximum circle
xmask = -Rmax:Rmax ;
[xg,yg] = ndgrid(xmask,xmask) ;
rg = sqrt( (xg.^2+yg.^2) ) ; %// radius of each pixel in the subgrid
for ip=1:npt
mrow = xmask+Rmax+y(ip) ; %// calc coordinates of subgrid on original mask
mcol = xmask+Rmax+x(ip) ; %// calc coordinates of subgrid on original mask
cmask = rg <= r(ip) ; %// calculate submask for this radius
mask(mrow,mcol) = mask(mrow,mcol) | cmask ; %// apply the sub-mask at the x,y coordinates
end
%// crop back the mask to image original size (=remove border overlap)
mask = mask(Rmax+1:end-Rmax,Rmax+1:end-Rmax) ;
imshow(mask)
Then you just apply the mask and count :
%% // Now apply on original image
BWm = ~BW & mask ; %// note the ~ (not) operator because you want the "black" pixels
nb_black_pixels = sum(sum(BWm)) ;
imshow(BWm)
nb_black_pixels =
5283
Here is one implementation:
Advantages:
No loops, meshgrid/ndgrid. Instead used faster bsxfun and pdist2
Dots are counted only once, even when the circles overlap.
Variable radius used (radius of all circle is not same)
Code:
%// creating a binary image with little black dots
A = randi(600,256);
imbw = A ~= 1;
%// Your binary image with black dots
imshow(imbw);
%// getting the index of black dots
[dotY, dotX] = find(~imbw);
nCoords = 10; %// required number of circles
%// generating its random coordinates as it is unknown here
Coords = randi(size(A,1),nCoords,2);
%// calculating the distance from each coordinate with every black dots
out = pdist2(Coords,[dotX, dotY]).'; %//'
%// Getting only the black dots within the radius
%// using 'any' avoids calculating same dot twice
radius = randi([10,25],1,size(Coords,1));
pixelMask = any(bsxfun(#lt, out, radius),2);
nPixels = sum(pixelMask);
%// visualizing the results by plotting
hold on
scatter(dotX(pixelMask),dotY(pixelMask));
viscircles([Coords(:,1),Coords(:,2)],radius.'); %//'
hold off
Output:
>> nPixels
nPixels =
19
I am relatively new to Matlab and I want to generate a hexagonal grid that represents a cellular network where each hexagon has a specific behaviour.
My question is how does one go from a square grid to hex grid (transfer matrix)?. This is what I have so far.
[X,Y] = meshgrid(0:60);
figure(1), plot(X,Y,'b')
hold on, plot(Y,X,'b')
axis square
A few years ago I wrote some code to do just that:
%// Define input data:
close all
clear all
M_max = 14; %// number of cells in vertical direction
N_max = 10; %// number of cells in horizontal direction
trans = 1; %// hexagon orientation (0 or 1)
%// Do the plotting:
hold on
for m = -M_max:M_max
for n = -N_max:N_max
center = [.5 sqrt(3)/2] + m*[0 -sqrt(3)] + n*[3/2 sqrt(3)/2];
if ~trans
plot([center(1)-1 center(1)],[center(2) center(2)])
plot([center(1) center(1)+1/2],[center(2) center(2)+sqrt(3)/2])
plot([center(1) center(1)+1/2],[center(2) center(2)-sqrt(3)/2])
else %// exchange the two arguments to `plot`
plot([center(2) center(2)],[center(1)-1 center(1)])
plot([center(2) center(2)+sqrt(3)/2],[center(1) center(1)+1/2])
plot([center(2) center(2)-sqrt(3)/2],[center(1) center(1)+1/2])
end %if
end %for
end %for
plot([-15 15],[0 0],'-.') %// adjust length manually
plot([-15 15],[-15*sqrt(3) 15*sqrt(3)],'-.') %// adjust length manually
axis image
set(gca,'xtick',[])
set(gca,'ytick',[])
axis([-10 10 -13.3 13.3]) %// adjust axis size manually
set(gca,'Visible','off') %// handy for printing the image
For example, this is the image generated with the above data:
This is the processed image and I can't increase the bwareaopen() as it won't work for my other image.
Anyway I'm trying to find the shortest points in the centre points of the barcode, to get the straight line across the centre points in the barcode.
Example:
After doing a centroid command, the points in the barcode are near to each other. Therefore, I just wanted to get the shortest points(which is the barcode) and draw a straight line across.
All the points need not be join, best fit points will do.
Step 1
Step 2
Step 3
If you dont have the x,y elements Andrey uses, you can find them by segmenting the image and using a naive threshold value on the area to avoid including the number below the bar code.
I've hacked out a solution in MATLAB doing the following:
Loading the image and making it binary
Extracting all connected components using bwlabel().
Getting useful information about each of them via regionprops() [.centroid will be a good approximation to the middel point for the lines].
Thresholded out small regions (noise and numbers)
Extracted x,y coordinates
Used Andreys linear fit solution
Code:
set(0,'DefaultFigureWindowStyle','docked');
close all;clear all;clc;
Im = imread('29ekeap.jpg');
Im=rgb2gray(Im);
%%
%Make binary
temp = zeros(size(Im));
temp(Im > mean(Im(:)))=1;
Im = temp;
%Visualize
f1 = figure(1);
imagesc(Im);colormap(gray);
%Find connected components
LabelIm = bwlabel(Im);
RegionInfo = regionprops(LabelIm);
%Remove background region
RegionInfo(1) = [];
%Get average area of regions
AvgArea = mean([RegionInfo(1:end).Area]);
%Vector to keep track of likely "bar elements"
Bar = zeros(length(RegionInfo),1);
%Iterate over regions, plot centroids if area is big enough
for i=1:length(RegionInfo)
if RegionInfo(i).Area > AvgArea
hold on;
plot(RegionInfo(i).Centroid(1),RegionInfo(i).Centroid(2),'r*')
Bar(i) = 1;
end
end
%Extract x,y points for interpolation
X = [RegionInfo(Bar==1).Centroid];
X = reshape(X,2,length(X)/2);
x = X(1,:);
y = X(2,:);
%Plot line according to Andrey
p = polyfit(x,y,1);
xMin = min(x(:));
xMax = max(x(:));
xRange = xMin:0.01:xMax;
yRange = p(1).*xRange + p(2);
plot(xRange,yRange,'LineWidth',2,'Color',[0.9 0.2 0.2]);
The result is a pretty good fitted line. You should be able to extend it to the ends by using the 'p' polynomal and evaluate when you dont encounter any more '1's if needed.
Result:
If you already found the x,y of the centers, you should use polyfit function:
You will then find the polynomial coefficients of the best line. In order to draw a segment, you can take the minimal and maximal x
p = polyfit(x,y,1);
xMin = min(x(:));
xMax = max(x(:));
xRange = xMin:0.01:xMax;
yRange = p(1).*xRange + p(2);
plot(xRange,yRange);
If your ultimate goal is to generate a line perpendicular to the bars in the bar code and passing roughly through the centroids of the bars, then I have another option for you to consider...
A simple solution would be to perform a Hough transform to detect the primary orientation of lines in the bar code. Once you find the angle of the lines in the bar code, all you have to do is rotate that by 90 degrees to get the slope of a perpendicular line. The centroid of the entire bar code can then be used as an intercept for this line. Using the functions HOUGH and HOUGHPEAKS from the Image Processing Toolbox, here's the code starting with a cropped version of your image from step 1:
img = imread('bar_code.jpg'); %# Load the image
img = im2bw(img); %# Convert from RGB to BW
[H, theta, rho] = hough(img); %# Perform the Hough transform
peak = houghpeaks(H); %# Find the peak pt in the Hough transform
barAngle = theta(peak(2)); %# Find the angle of the bars
slope = -tan(pi*(barAngle + 90)/180); %# Compute the perpendicular line slope
[y, x] = find(img); %# Find the coordinates of all the white image points
xMean = mean(x); %# Find the x centroid of the bar code
yMean = mean(y); %# Find the y centroid of the bar code
xLine = 1:size(img,2); %# X points of perpendicular line
yLine = slope.*(xLine - xMean) + yMean; %# Y points of perpendicular line
imshow(img); %# Plot bar code image
hold on; %# Add to the plot
plot(xMean, yMean, 'r*'); %# Plot the bar code centroid
plot(xLine, yLine, 'r'); %# Plot the perpendicular line
And here's the resulting image: