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
Related
Im trying to create a project in MATLAB where i project a heat map matrix on to a cylinder in MATLAB but i'm bumping in to all kinds of issues.
Im using MATLAB version R2019b(Latest version).
Now my first issue id like to adress is is the modells being absolutely tiny even when i zoom in to the max.
Is there any way to make these models larger or to display them in a separate window ??
Also my second question is the regarding the scaling. As you can see the scale on the Z-axis of the cylinder goes to 512. any way to get a larger more detailed image where i can see each point on that scale?
clear vars
filename =
['/Users/gomer/Documents/Datavisualisering/Projekt/data/day/filenames.txt'];
%This line simply gives us a table of all filenames in this file.
T = readtable(filename);
tsize = size(T);
%extracts the content of the table as a categorical array.
%%
%
% {rownumber,variablenumber}. {100,1} = row 100, variable 1.
%converts the categorical array into a string array.
%joins the string array across the column.
%string(T{100:105,1}); implies from row 100 to row 105 and use variable 1.
%This line simply adds the name of the file at row 100 to the path of the
%file. Hnece we get a full filepath.
filename = strcat('/Users/gomer/Documents/Datavisualisering/Projekt/data/day/', string(T{100,1}));
map100 = getHeatMap(filename);
%%
%
filename = strcat('/Users/gomer/Documents/Datavisualisering/Projekt/data/day/', string(T{1000,1}));
map1000 = getHeatMap(filename);
%creates a image
k=imshow(map100);
%creates a colormap.
%gca returns the current axes (or standalone visualization) in the current figure.
%hence the command just works top down and affects last image displayed.
colormap(gca, 'jet');
k=imshow(map1000);
colormap(gca, "jet");
function heat = getHeatMap(filename)
%Returns a struct with info about the file.
s = dir(filename);
%Opens a file for reading, hence the 'r'.
fin=fopen(filename,'r');
%A = fread(fileID,sizeA,precision)
%reads file data into an array, A, with dimensions, sizeA, and positions the file pointer after the last value read.
%fread populates A in column order.
I=fread(fin,s.bytes,'uint8=>uint8');
%The uint16 function converts a Input array, specified as a scalar, vector, matrix, or multidimensional array.
%Converts the values in to type uint16 and creates a Matrix with the values.
%Looks more like we are just calculating the size of values to be
%deducted from matrix size (no idea why).
w = uint16(I(1))+256*uint16(I(2));
h = uint16(I(3))+256*uint16(I(4));
skip = s.bytes - w*h + 1;
IN = I(skip:1:s.bytes);
%reshape(A,sz) reshapes A using the size vector, sz, to define size(B). For example, reshape(A,[2,3]) reshapes A into a 2-by-3 matrix.
%sz must contain at least 2 elements, and prod(sz) must be the same as numel(A).
%single(X) converts the values in X to single precision.
Z=single(reshape(IN,w,h));
%Interpolate the values.
%telling the system which points to interpolate in between.
Z=griddedInterpolant(Z');
y_range = linspace(1.0,single(h),360);
x_range = linspace(1.0,single(w),512);
%Used to apecify the query points (points in between which we want
%interpolation) as vectors (this is to get a more continues image).
%specifies the query points as grid vectors.
%Use this syntax to conserve memory when you want to query a large grid of points.
heat = uint8(Z({y_range, x_range}));
end
Thank you
Resizing Figure Window
Copying your script to a .m and running it instead of a live .mlx file will automatically create all the figures in a separate window by default. To adjust the size of the figures the following the position property can be modified. To adjust the scale functions xticks(), yticks() and zticks() can be used. These three scale functions take in an array representing all the line markers along the respective axis/scale.
Test Plot Script:
X_Position = 10;
Y_Position = 10;
Width = 1000;
Height = 500;
%Configuring the figure settings%
figure('Position', [X_Position Y_Position Width Height])
%Test plot (replace with your plot)%
Data = [1 2 3 4 5 6];
plot(Data);
Plotting Heatmaps on a Cylindrical Surfaces
Method 1: Using warp() Function
Haven't delved into which orientation the data should be set as but this is an option to use the warp() function to wrap the heatmap around a cylinder. They're most likely other 3D plotting options if specific points are of interest. The points of the cylinder to be plotted are generated using the cylinder() function which returns the xyz-coordinates of the cylinder's wireframe. The fourth input argument into the warp() function serves and a colormap in this case it is proportional to the heatmap values.
load('HeatMapMatrix.mat');
%Setting up the figure%
clf;
close all;
figure('Position', [10 10 1000 500])
%Creating the cylinder%
Cylinder_Radius = 360;
Cylinder_Height = 512;
[X,Y,Z] = cylinder(Cylinder_Radius,Cylinder_Height-1);
%Warping the heatmap along the cylinder%
subplot(1,2,1); warp(X,Y,Cylinder_Height.*Z,map100');
colormap(gca,'jet');
subplot(1,2,2); warp(X,Y,Cylinder_Height.*Z,map100);
colormap(gca,'jet');
Method 2: Plotting All Points and Using surf() Function:
In this case, the coordinates for a cylinder are generated by first finding the coordinates of the circumference of the cylinder. This is done by using the sin() and cos() relationships:
X-Points = Radius × cos(𝜃)
Y-Points = Radius × sin(𝜃)
This results in the xy-coordinates of the cylinder's circle. These xy-coordinates need to be duplicated using repmat() to be later used for the varying heights. The process can be described best with a diagram as follows:
Four matrices above are created to plot each Heat Data point corresponding to an xyz-coordinate. The x and y coordinates are repeated in every row of matrices X-Points and Y_Points since those are constant for the repeating circles. The columns of the matrix Z-Points are also duplicates of each other since the heights should be constant for each row corresponding to each circle.
load('HeatMapMatrix.mat');
Radius = 20;
Number_Of_Data_Points = 360;
Theta = linspace(0,2*pi,Number_Of_Data_Points);
%The xy values according to radius and number of points%
X_Points = Radius*cos(Theta);
Y_Points = Radius*sin(Theta);
map100 = rot90(map100);
Sample_Range = 255 - 0;
Temperature_Range = 450 - 50;
Multiplier = Temperature_Range/Sample_Range;
map100 = map100.*Multiplier + 50;
Height = 512;
X_Points = repmat(X_Points,Height,1);
Y_Points = repmat(Y_Points,Height,1);
Z_Points = (1:512)';
Z_Points = repmat(Z_Points,1,Number_Of_Data_Points);
clf;
close;
figure('Position', [10 10 800 500])
Offset = 200;
subplot(1,3,1:2); Surface = surf(Y_Points,X_Points,Z_Points,'Cdata',map100);
title("3D Heatmap Plot");
zlabel("Height");
shading interp
colorbar
% direction = [0 1 0];
% rotate(Surface,direction,90)
Maximum_Value = 450;
Minimum_Value = 50;
caxis([Minimum_Value Maximum_Value]);
subplot(1,3,3); imshow(rot90(rot90(map100)));
colormap(gca, 'default');
title("Flat Heatmap Plot");
caxis([Minimum_Value Maximum_Value]);
colorbar;
Ran using MATLAB R2019b
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
How to count circle objects in a bright image using MATLAB?
The input image is:
imfindcircles function can't find any circle in this image.
Based on well known image processing techniques, you can write your own processing tool:
img = imread('Mlj6r.jpg'); % read the image
imgGray = rgb2gray(img); % convert to grayscale
sigma = 1;
imgGray = imgaussfilt(imgGray, sigma); % filter the image (we will take derivatives, which are sensitive to noise)
imshow(imgGray) % show the image
[gx, gy] = gradient(double(imgGray)); % take the first derivative
[gxx, gxy] = gradient(gx); % take the second derivatives
[gxy, gyy] = gradient(gy); % take the second derivatives
k = 0.04; %0.04-0.15 (see wikipedia)
blob = (gxx.*gyy - gxy.*gxy - k*(gxx + gyy).^2); % Harris corner detector (high second derivatives in two perpendicular directions)
blob = blob .* (gxx < 0 & gyy < 0); % select the top of the corner (i.e. positive second derivative)
figure
imshow(blob) % show the blobs
blobThresshold = 1;
circles = imregionalmax(blob) & blob > blobThresshold; % find local maxima and apply a thresshold
figure
imshow(imgGray) % show the original image
hold on
[X, Y] = find(circles); % find the position of the circles
plot(Y, X, 'w.'); % plot the circle positions on top of the original figure
nCircles = length(X)
This code counts 2710 circles, which is probably a slight (but not so bad) overestimation.
The following figure shows the original image with the circle positions indicated as white dots. Some wrong detections are made at the border of the object. You can try to make some adjustments to the constants sigma, k and blobThresshold to obtain better results. In particular, higher k may be beneficial. See wikipedia, for more information about the Harris corner detector.
I'm using the code on the MatLab website, "Color-Based Segmentation Using the Lab* Color Space":
http://www.mathworks.com/help/images/examples/color-based-segmentation-using-the-l-a-b-color-space.html
So I'm trying to select some areas myself instead of using the "load region_coordinates", using roipoly(fabric), but i get stuck. How do I save coordinates of the polygon I just drew? I'm actually following advice from lennon310 at Solution II, bottom of page:
A few questions about color segmentation with L*a*b*
I'm not sure when to save region_coordinates and do size(region_coordinates,1)
I made the following changes (Step 1):
1) Removed "load region_coordinates"
2) Added "region_coordinates = roipoly(fabric);"
Here's the code:
`
%% Step 1
fabric = imread(file);
figure(1); %Create figure window. "If h is not the handle and is not the Number property value of an existing figure, but is an integer, then figure(h) creates a figure object and assigns its Number property the value h."
imshow(fabric)
title('fabric')
%load regioncoordinates; % 6 marices(?) labelled val(:,:,1-6), 5x2 (row x column)
region_coordinates = roipoly(fabric);
nColors = 6;
sample_regions = false([size(fabric,1) size(fabric,2) nColors]); %Initializing an Image Dimension, 3x3 (:,:,:) to zero? Zeros() for arrays only I guess.
%Size one is column, size two is row?
for count = 1:nColors
sample_regions(:,:,count) = roipoly(fabric,region_coordinates(:,1,count),region_coordinates(:,2,count));
end
figure, imshow(sample_regions(:,:,2)),title('sample region for red');
%%Step 2
% Convert your fabric RGB image into an L*a*b* image using rgb2lab .
lab_fabric = rgb2lab(fabric);
%Calculate the mean a* and b* value for each area that you extracted with roipoly. These values serve as your color markers in a*b* space.
a = lab_fabric(:,:,2);
b = lab_fabric(:,:,3);
color_markers = zeros([nColors, 2]);%... I think this is initializing a 6x2 blank(0) array for colour storage. 6 for colours, 2 for a&b colourspace.
for count = 1:nColors
color_markers(count,1) = mean2(a(sample_regions(:,:,count))); %Label for repmat, Marker for
color_markers(count,2) = mean2(b(sample_regions(:,:,count)));
end
%For example, the average color of the red sample region in a*b* space is
fprintf('[%0.3f,%0.3f] \n',color_markers(2,1),color_markers(2,2));
%% Step 3: Classify Each Pixel Using the Nearest Neighbor Rule
%
color_labels = 0:nColors-1;
% Initialize matrices to be used in the nearest neighbor classification.
a = double(a);
b = double(b);
distance = zeros([size(a), nColors]);
%Perform classification, Elucidean Distance.
for count = 1:nColors
distance(:,:,count) = ( (a - color_markers(count,1)).^2 + (b - color_markers(count,2)).^2 ).^0.5;
end
[~, label] = min(distance,[],3);
label = color_labels(label);
clear distance;
%% Step 4: Display Results of Nearest Neighbor Classification
%
% The label matrix contains a color label for each pixel in the fabric image.
% Use the label matrix to separate objects in the original fabric image by color.
rgb_label = repmat(label,[1 1 3]);
segmented_images = zeros([size(fabric), nColors],'uint8');
for count = 1:nColors
color = fabric;
color(rgb_label ~= color_labels(count)) = 0;
segmented_images(:,:,:,count) = color;
end
%figure, imshow(segmented_images(:,:,:,1)), title('Background of Fabric');
%Looks different somehow.
figure, imshow(segmented_images(:,:,:,2)), title('red objects');
figure, imshow(segmented_images(:,:,:,3)), title('green objects');
figure, imshow(segmented_images(:,:,:,4)), title('purple objects');
figure, imshow(segmented_images(:,:,:,5)), title('magenta objects');
figure, imshow(segmented_images(:,:,:,6)), title('yellow objects');
`
You can retrieve the coordinates of the polygon using output arguments in the call to roipoly. You can then get a binary mask of the polygon, as well as vertices coordinates if you want.
Simple example demonstrating:
clear
clc
close all
A = imread('cameraman.tif');
figure;
imshow(A)
%// The vertices of the polygon are stored in xi and yi;
%// PolyMask is a binary image where pixels == 1 are white.
[polyMask, xi, yi] = roipoly(A);
This looks like this:
And if you wish to see the vertices with the binary mask:
%// display polymask
imshow(polyMask)
hold on
%// Highlight vertices in red
scatter(xi,yi,60,'r')
hold off
Which gives the following:
So to summarize:
1) The polygon's vertices are stored in xi and yi.
2) You can plot the binaryMask of the polygon using imshow(polyMask).
3) If you need the coordinates of the white pixels, you can use something like this:
[row_white,col_white] = find(polyMask == 1);
You're then good to go. Hope that helps!
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