I generate a random number between 1 and 2 as a radius for my circle. Then I plot my circle and saves it as a png. Also, several data points are generated both inside and outside the circle to make it noisy.
Then I will use Hough Transform to estimate the radius for the circle but, the number which it returns is more than 100. Although they are the same circles(I plotted to make sure).
I have tried to use polyfit function to map these two numbers but, the estimated radius seems to be smaller than the real one in some examples. After mapping It returns the same number. For example, a random radius is 1.2 and Hough estimates it 110 however it seems that it should be near 1.(Because I plot it and it is clear that is it smaller). Also, for Hough Transform I am using this code https://www.mathworks.com/matlabcentral/fileexchange/35223-circle-detection-using-hough-transforms
I tried the predefined Matlab function (imfindcircles) but it returns null for all my circles.
r1 = 1 + abs((1)*randn(1,1)); %radius of inner circle
r1 = abs(r1);
r2 = (1.2)*r1; %radius of outercircle
k = 500; %number of random numbers
% circle coordinate points
t = 0:2*pi/59:2*pi;
xv = cos(t)';
yv = sin(t)';
%%%%%%%%random points
xq = 2*randn(k,1)-1;
yq = 2*randn(k,1)-1; %distribution 1
xq1 = 2*(rand(k,1))-1; %distribution 2
yq1 = 2*(rand(k,1))-1;
in = inpolygon(xq1,yq1,r1*xv,r1*yv); %number of points inside the geometry
in1= inpolygon(xq,yq,r2*xv,r2*yv); %points inside outer circle
in2 = xor(in,in1); %points between circle
in3= inpolygon(xq1,yq1,r2*xv,r2*yv); %points inside outer circle
fig = figure(22);hold on;
% Random points
plot(xq(in2),yq(in2),'bo','MarkerFaceColor','r','MarkerSize',5,'Marker','o','MarkerEdgeColor','none');
axis equal;
plot(xq1(~in2),yq1(~in2),'bo','MarkerFaceColor','r','MarkerSize',5,'Marker','o','MarkerEdgeColor','none');
axis equal;
img= getframe(fig);
figure;
imshow(img.cdata)
hold on;
[r,c,rad] = circlefinder(img.cdata);
[m I] = max(rad);
%ploting the bigest estimated circle
angle = 2*pi*randn(k,1);
haffpointX = rad(I).*cos(angle)+ c(I);
haffpointY = rad(I).*sin(angle)+ r(I);
scatter(haffpointX,haffpointY,'g');
axis equal
I expect that for different random circles with noisy data Hough Transform estimates its circle with the number between the range of 1 and 2 so, I can use its results.
Thank you in advance
Related
I'm trying to calculate the mean value of the pixels inside a circle. In the future this needs to be extended to 3D, but for now a 2D sollution would already help me out.
As can be seen in the image, some pixels are entirely inside the circle, but some are only partly inside the circle. The ones partly in the circle also need to contribute only partly to the mean value. The pixels are square. This will simplify the mathematics I hope.
I can calculate the distance from the pixelcorners to the central point, from this you can find the pixels enterly inside and enterly outside. The rest needs correction. But how to find this correction.
[edit] thanks to Heath Raftery the problem is solved! [/edit]
the integral of a circle with radius r
As an example: I want to know the average pixelvalue of pixels in this circle. I know it is 0.3425, since 34.25% of the circle has a value of 1 and the rest is 0.
Function to check what part of a pixel is in the circle:
function [ a ] = incirc( x,y,r )
%only handles the top right quadrant of a circle
if x<0||y<0,error('only positive x,y');end
%integral of sqrt(r^2-x^2) dx
F = #(x,r) (1/2)*(x*sqrt(r^2-x^2)+r^2*atan(x/sqrt(r^2-x^2)));
%find corner locations
x=[x-0.5,x+0.5];
y=[y-0.5,y+0.5];
d = sqrt(x.^2+y.^2); %distance to closed and furthest corner
if max(d)<r,a=1;return;end %inside circle
if min(d)>r,a=0;return;end %outside circle
%intersections with edges (r^2 = x^2+y^2)
inters = [sqrt(r^2-y(1)^2),sqrt(r^2-y(2)^2),sqrt(r^2-x(1)^2),sqrt(r^2-x(2)^2)]; %x(1) x(2) y(1) y(2)
%remove imaginary and out of range intersections
inters(imag(inters)~=0)=NaN;
inters(inters<1E-5)=NaN; %to find values that are zero
inters([~((x(1)<inters(1:2))&(inters(1:2)<x(2))),~((y(1)<inters(3:4))&(inters(3:4)<y(2)))])=NaN;
idx = find(~isnan(inters));
if numel(idx)~=2,error('need two intersections of circle with pixel');end
%check area of pixel inside circumference
if all(idx==[1,2]) %2 intersections on y-edge
a=(F(y(2),r)-F(y(1),r)) - x(1); %area
elseif all(idx==[3,4]) %2 intersections on x-edge
a=(F(x(2),r)-F(x(1),r)) - y(1); %area
elseif all(idx==[1,3]) %one intersection on y-edge one on x-edge (left&bottom)
a=(F(inters(1),r)-F(x(1),r))- (y(1)*(inters(1)-x(1)));
elseif all(idx==[2,4]) %one intersection on y-edge one on x-edge (top&right)
a=(inters(2)-x(1))+(F(x(2),r)-F(inters(2),r))-(y(1)*(x(2)-inters(2)));
else
error('geometry')
end
a=real(a);
if a<0||a>1
error('computational error');
end
end
Script to test the function
M = ones(100); %data
M(1:50,:)=0;
pos=[50.2,50];
r = 2;
%calculate what the result should be
h=50-pos(2)+0.5;
A=pi*r^2;
wedge = acos(h/r)/pi;
triangle = h*sqrt(r^2-h^2);
res=(A*wedge-triangle)/A
S=0;N=0;
for i = 1:size(M,1)
for j = 1:size(M,2)
x=abs(j-pos(1));
y=abs(i-pos(2));
n=incirc( x,y,r );
M_(i,j)=n;
S = S+M(i,j)*n;
N = N+n;
end
end
result = S/N
result = 0.3425
You can see the algorithm finds the part of the pixel in the circle.
The question is missing a question, but I'll assume that it's not how to calculate whether pixels are fully inside or outside the circle. That's a relatively simple task. That is, a pixel is fully inside if the furtherest corner of the pixel to the centre is less than a radius away from the centre, and a pixel is fully outside if the closest corner of the pixel to the centre is more than a radius away from the centre.
The question of what proportion of pixels on the circumference fall within the circumference is much trickier. There are two fundamental solutions:
Exact and hard.
Approximate and a bit easier.
In both cases, note the horizontal and vertical symmetry means only the top right quadrant need be considered.
Then, for (1), translate the circle centre to the origin (0, 0) and treat the circumference as the function y(x) = sqrt(r^2 - x^2). Then, the area of an overlapping pixel within the circle is the integral:
integral(y(x) - y0, from x0 to x1, with respect to x)
where y0 is the bottom coordinate of the pixel, x0 is the left coordinate and x1 is the right coordinate.
This integral can be solved exactly with a trigonometric identity and a trigonometric substitution.
For (2), just generate a set of random points within the pixel and count how many of them fall within the circumference. As the set gets larger, the proportion of points that fall within the circumference to the count of all point approaches the proportion of the pixel within the circumference.
You can use inpolygon, to get the indices which lie inside the circle, once you have those indices you can get your pixels and do what you want.
M = rand(100); %data
[nx,ny] = size(M) ;
[X,Y] = meshgrid(1:ny,1:nx) ;
pos=[20,20];
r = 5;
phi=linspace(0,2*pi,100);
imagesc(M);
axis image
hold on
plot(pos(1),pos(2),'rx')
xc = pos(1)+r*sin(phi) ;
yc = pos(2)+r*cos(phi) ;
plot(xc,yc,'-r');
% hold off
%% get indices which are inside the circle
idx = inpolygon(X(:),Y(:),xc,yc) ;
xi = X(idx) ; yi = Y(idx) ;
plot(xi,yi,'.r')
mypixels = M(idx) ;
You can also use rangesearch to get the points lying within the given radius of the circle. As below:
M = rand(100); %data
[nx,ny] = size(M) ;
[X,Y] = meshgrid(1:ny,1:nx) ;
pos=[20,20];
r = 5;
phi=linspace(0,2*pi,100);
imagesc(M);
axis image
hold on
plot(pos(1),pos(2),'rx')
xc = pos(1)+r*sin(phi) ;
yc = pos(2)+r*cos(phi) ;
plot(xc,yc,'-r');
% hold off
%% Use nearest neighbour search
idx = rangesearch([X(:),Y(:)],pos,r) ;
xi = X(idx{1}) ; yi = Y(idx{1}) ;
plot(xi,yi,'.r')
mypixels = M(idx{1}) ;
I have a paired matrix with pixels equal 0 or 255 and when it's 255 it makes disk shape.
I would like to labelize the differents disks and get in a tab for each label : their radius and their central point. How can I do for this last two points?
Below an exemple with a small matrix
Mat=zeros(12,12);
Mat(2,6:7)=255; Mat(3,5)=255; Mat(3,8)=255; Mat(4,5)=255
Mat(4,8)=255; Mat(5,6:7)=255;
Mat(10,10)=255; Mat(11,9)=255; Mat(12,10)=255; Mat(11,11)=255;
CC=bwconncomp(Mat,8);
MatL=labelmatrix(CC);
figure, imagesc(Mat)
you can use regionprops to compute centroid and area, and then use area to compute approximate radius:
% generate matrix
Mat=zeros(12,12);
Mat(2,6:7)=255; Mat(3,5)=255; Mat(3,8)=255; Mat(4,5)=255;
Mat(4,8)=255; Mat(5,6:7)=255;
Mat(10,10)=255; Mat(11,9)=255; Mat(12,10)=255; Mat(11,11)=255;
% convert to binary
MatBin = Mat > 0;
% fill circles
MatFull = imfill(MatBin,4,'holes');
% get centroids and areas
props = regionprops(MatFull,{'Area','Centroid'});
Area = [props(:).Area];
Centroid = reshape([props(:).Centroid],[],2)';
% compute radius
Radius = sqrt(Area ./ pi);
% plotting
imshow(MatFull,[],'InitialMagnification','fit')
hold on
for ii = 1:numel(Radius)
text(Centroid(ii,1),Centroid(ii,2),['r = ' num2str(Radius(ii))],...
'VerticalAlignment','middle','HorizontalAlignment',...
'center','FontSize',12,'Color','b');
end
hold off
I have a set of matrixes with a number of blobs (this changes from matrix to matrix), and I'm looking for a way to find the center of mass of the blobs. To do so I use regionprops with the Centroid option.
B = bwlabel(A);
stat = regionprops(B,'Centroid');
number_centers = numel(stat);
coord_centers = zeros(number_centers,2);
xy_cylinder = zeros(number_centers,2);
for x = 1: number_centers
coord_centers(x,1) = stat(x).Centroid(1); % Angle
coord_centers(x,2) = stat(x).Centroid(2); % Radius
end
If I plot the matrix with the centroids superimposed (code below), I noticed that they slightly offset (they are on the corner of the pixel where the centroid is). Do you think I'm right? And do you know why is that?
axis equal;
pcolor(A), colorbar, hold on;
xlabel('Angle'); ylabel('Radius');
for x = 1: numel(stat)
plot(stat(x).Centroid(1),stat(x).Centroid(2),'r+');
end
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:
BW = poly2mask(x, y, m, n) computes a
binary region of interest (ROI) mask,
BW, from an ROI polygon, represented
by the vectors x and y. The size of BW
is m-by-n.
poly2mask sets pixels in BW
that are inside the polygon (X,Y) to 1
and sets pixels outside the polygon to
0.
Problem:
Given such a binary mask BW of a convex quadrilateral, what would be the most efficient way to determine the four corners?
E.g.,
Best Solution so far:
Use edge to find the bounding lines, the Hough transform to find the 4 lines in the edge image and then find the intersection points of those 4 lines or use a corner detector on the edge image. Seems complicated, and I can't help feeling there's a simpler solution out there.
Btw, convhull doesn't always return 4 points (maybe someone can suggest qhull options to prevent that) : it returns a few points along the edges as well.
EDIT:
Amro's answer seems quite elegant and efficient. But there could be multiple "corners" at each real corner since the peaks aren't unique. I could cluster them based on θ and average the "corners" around a real corner but the main problem is the use of order(1:10).
Is 10 enough to account for all the corners or will this exclude a "corner" at a real corner?
This is somewhat similar to what #AndyL suggested. However I'm using the boundary signature in polar coordinates instead of the tangent.
Note that I start by extracting the edges, getting the boundary, then converting it to signature. Finally we find the points on the boundary that are furthest from the centroid, those points constitute the corners found. (Alternatively we can also detect peaks in the signature for corners).
The following is a complete implementation:
I = imread('oxyjj.png');
if ndims(I)==3
I = rgb2gray(I);
end
subplot(221), imshow(I), title('org')
%%# Process Image
%# edge detection
BW = edge(I, 'sobel');
subplot(222), imshow(BW), title('edge')
%# dilation-erosion
se = strel('disk', 2);
BW = imdilate(BW,se);
BW = imerode(BW,se);
subplot(223), imshow(BW), title('dilation-erosion')
%# fill holes
BW = imfill(BW, 'holes');
subplot(224), imshow(BW), title('fill')
%# get boundary
B = bwboundaries(BW, 8, 'noholes');
B = B{1};
%%# boudary signature
%# convert boundary from cartesian to ploar coordinates
objB = bsxfun(#minus, B, mean(B));
[theta, rho] = cart2pol(objB(:,2), objB(:,1));
%# find corners
%#corners = find( diff(diff(rho)>0) < 0 ); %# find peaks
[~,order] = sort(rho, 'descend');
corners = order(1:10);
%# plot boundary signature + corners
figure, plot(theta, rho, '.'), hold on
plot(theta(corners), rho(corners), 'ro'), hold off
xlim([-pi pi]), title('Boundary Signature'), xlabel('\theta'), ylabel('\rho')
%# plot image + corners
figure, imshow(BW), hold on
plot(B(corners,2), B(corners,1), 's', 'MarkerSize',10, 'MarkerFaceColor','r')
hold off, title('Corners')
EDIT:
In response to Jacob's comment, I should explain that I first tried to find the peaks in the signature using first/second derivatives, but ended up taking the furthest N-points. 10 was just an ad-hoc value, and would be difficult to generalize (I tried taking 4 same as number of corners, but it didn't cover all of them). I think the idea of clustering them to remove duplicates is worth looking into.
As far as I see it, the problem with the 1st approach was that if you plot rho without taking θ into account, you will get a different shape (not the same peaks), since the speed by which we trace the boundary is different and depends on the curvature. If we could figure out how to normalize that effect, we can get more accurate results using derivatives.
If you have the Image Processing Toolbox, there is a function called cornermetric which can implement a Harris corner detector or Shi and Tomasi's minimum eigenvalue method. This function has been present since version 6.2 of the Image Processing Toolbox (MATLAB version R2008b).
Using this function, I came up with a slightly different approach from the other answers. The solution below is based on the idea that a circular area centered at each "true" corner point will overlap the polygon by a smaller amount than a circular area centered over an erroneous corner point that is actually on the edge. This solution can also handle cases where multiple points are detected at the same corner...
The first step is to load the data:
rawImage = imread('oxyjj.png');
rawImage = rgb2gray(rawImage(7:473, 9:688, :)); % Remove the gray border
subplot(2, 2, 1);
imshow(rawImage);
title('Raw image');
Next, compute the corner metric using cornermetric. Note that I am masking the corner metric by the original polygon, so that we are looking for corner points that are inside the polygon (i.e. trying to find the corner pixels of the polygon). imregionalmax is then used to find the local maxima. Since you can have clusters of greater than 1 pixel with the same corner metric, I then add noise to the maxima and recompute so that I only get 1 pixel in each maximal region. Each maximal region is then labeled using bwlabel:
cornerImage = cornermetric(rawImage).*(rawImage > 0);
maxImage = imregionalmax(cornerImage);
noise = rand(nnz(maxImage), 1);
cornerImage(maxImage) = cornerImage(maxImage)+noise;
maxImage = imregionalmax(cornerImage);
labeledImage = bwlabel(maxImage);
The labeled regions are then dilated (using imdilate) with a disk-shaped structuring element (created using strel):
diskSize = 5;
dilatedImage = imdilate(labeledImage, strel('disk', diskSize));
subplot(2, 2, 2);
imshow(dilatedImage);
title('Dilated corner points');
Now that the labeled corner regions have been dilated, they will partially overlap the original polygon. Regions on an edge of the polygon will have about 50% overlap, while regions that are on a corner will have about 25% overlap. The function regionprops can be used to find the areas of overlap for each labeled region, and the 4 regions that have the least amount of overlap can thus be considered as the true corners:
maskImage = dilatedImage.*(rawImage > 0); % Overlap with the polygon
stats = regionprops(maskImage, 'Area'); % Compute the areas
[sortedValues, index] = sort([stats.Area]); % Sort in ascending order
cornerLabels = index(1:4); % The 4 smallest region labels
maskImage = ismember(maskImage, cornerLabels); % Mask of the 4 smallest regions
subplot(2, 2, 3);
imshow(maskImage);
title('Regions of minimal overlap');
And we can now get the pixel coordinates of the corners using find and ismember:
[r, c] = find(ismember(labeledImage, cornerLabels));
subplot(2, 2, 4);
imshow(rawImage);
hold on;
plot(c, r, 'r+', 'MarkerSize', 16, 'LineWidth', 2);
title('Corner points');
And here's a test with a diamond shaped region:
I like to solve this problem by working with a boundary, because it reduces this from a 2D problem to a 1D problem.
Use bwtraceboundary() from the image processing toolkit to extract a list of points on the boundary. Then convert the boundary into a series of tangent vectors (there are a number of ways to do this, one way would be to subrtact the
ith point along the boundary from the i+deltath point.) Once you have a list of vectors, take the dot product of adjacent vectors. The four points with the smallest dot products are your corners!
If you want your algorithm to work on polygons with an abritrary number of vertices, then simply search for dot products that are a certain number of standard deviations below the median dot product.
I decided to use a Harris corner detector (here's a more formal description) to obtain the corners. This can be implemented as follows:
%% Constants
Window = 3;
Sigma = 2;
K = 0.05;
nCorners = 4;
%% Derivative masks
dx = [-1 0 1; -1 0 1; -1 0 1];
dy = dx'; %SO code color fix '
%% Find the image gradient
% Mask is the binary image of the quadrilateral
Ix = conv2(double(Mask),dx,'same');
Iy = conv2(double(Mask),dy,'same');
%% Use a gaussian windowing function and compute the rest
Gaussian = fspecial('gaussian',Window,Sigma);
Ix2 = conv2(Ix.^2, Gaussian, 'same');
Iy2 = conv2(Iy.^2, Gaussian, 'same');
Ixy = conv2(Ix.*Iy, Gaussian, 'same');
%% Find the corners
CornerStrength = (Ix2.*Iy2 - Ixy.^2) - K*(Ix2 + Iy2).^2;
[val ind] = sort(CornerStrength(:),'descend');
[Ci Cj] = ind2sub(size(CornerStrength),ind(1:nCorners));
%% Display
imshow(Mask,[]);
hold on;
plot(Cj,Ci,'r*');
Here, the problem with multiple corners thanks to Gaussian windowing function which smooths the intensity change. Below, is a zoomed version of a corner with the hot colormap.
Here's an example using Ruby and HornetsEye. Basically the program creates a histogram of the quantised Sobel gradient orientation to find dominant orientations. If four dominant orientations are found, lines are fitted and the intersections between neighbouring lines are assumed to be the corners of the projected rectangle.
#!/usr/bin/env ruby
require 'hornetseye'
include Hornetseye
Q = 36
img = MultiArray.load_ubyte 'http://imgur.com/oxyjj.png'
dx, dy = 8, 6
box = [ dx ... 688, dy ... 473 ]
crop = img[ *box ]
crop.show
s0, s1 = crop.sobel( 0 ), crop.sobel( 1 )
mag = Math.sqrt s0 ** 2 + s1 ** 2
mag.normalise.show
arg = Math.atan2 s1, s0
msk = mag >= 500
arg_q = ( ( arg.mask( msk ) / Math::PI + 1 ) * Q / 2 ).to_int % Q
hist = arg_q.hist_weighted Q, mag.mask( msk )
segments = ( hist >= hist.max / 4 ).components
lines = arg_q.map segments
lines.unmask( msk ).normalise.show
if segments.max == 4
pos = MultiArray.scomplex *crop.shape
pos.real = MultiArray.int( *crop.shape ).indgen! % crop.shape[0]
pos.imag = MultiArray.int( *crop.shape ).indgen! / crop.shape[0]
weights = lines.hist( 5 ).major 1.0
centre = lines.hist_weighted( 5, pos.mask( msk ) ) / weights
vector = pos.mask( msk ) - lines.map( centre )
orientation = lines.hist_weighted( 5, vector ** 2 ) ** 0.5
corner = Sequence[ *( 0 ... 4 ).collect do |i|
i1, i2 = i + 1, ( i + 1 ) % 4 + 1
l1, a1, l2, a2 = centre[i1], orientation[i1], centre[i2], orientation[i2]
( l1 * a1.conj * a2 - l2 * a1 * a2.conj -
l1.conj * a1 * a2 + l2.conj * a1 * a2 ) /
( a1.conj * a2 - a1 * a2.conj )
end ]
result = MultiArray.ubytergb( *img.shape ).fill! 128
result[ *box ] = crop
corner.to_a.each do |c|
result[ c.real.to_i + dx - 1 .. c.real.to_i + dx + 1,
c.imag.to_i + dy - 1 .. c.imag.to_i + dy + 1 ] = RGB 255, 0, 0
end
result.show
end