detect certain shapes in binary

hi,in my image has 2 shapes ;triangular and circular(in binary). how can I eliminate the circular shape so that it just detect triangular shape.many thanks!

Answer 1

[Walter Roberson]

Use regionprops() to calculate the Eccentricity. Eccentricity should be nearly 0 for a circle.

[Tulips]

how to use the eccentricity.can it be used for diamond or square shape?

[Walter Roberson]

BW = bwlabel(YourImage); rp = regionprops(BW, 'Eccentricity');

Then look at rp(1).Eccentricity, rp(2).Eccentricity and so on. The circles will be close to 0, squares would be around 1/sqrt(2). diamonds might also be around the same as squares. Triangles would generally be somewhere between 0 and 1/sqrt(2) -- experiment and find out.

If you have identified a blob as being "square or diamond", and want to know which, use regionprops to check the Orientation of the object.

[Tulips]

do have any tutorial on this.what's the meaning of eccentricity?it takes which coordinates/pixels?still dont get it..btw,thanks!

[Tulips]

dear Walter Roberson, do u have any tutorial on this?how to eliminate the unwanted shape by using eccentricity.hope u can help!tqvm!

[Image Analyst]

You can try my image segmentation tutorial: http://www.mathworks.com/matlabcentral/fileexchange/?term=authorid%3A31862

Another metric you can look at is the "circularity" which is perimeter^2 / (4*pi*area). It's 1 for a circle and gets bigger the more tortuous the perimeter shape is.

[Ryan]

regionprops documentation:

'Eccentricity' — Scalar that specifies the eccentricity of the ellipse that has the same second-moments as the region. The eccentricity is the ratio of the distance between the foci of the ellipse and its major axis length. The value is between 0 and 1. (0 and 1 are degenerate cases; an ellipse whose eccentricity is 0 is actually a circle, while an ellipse whose eccentricity is 1 is a line segment.) This property is supported only for 2-D input label matrices.

This figure demonstrates the major (longer) and minor (shorter) axes:

[Tulips]

dear Walter Roberson, can I know whis is the difference between rp(1).Eccentricity, rp(2).Eccentricity, u mentioned abt this in the earlier comment.why it shud have rp(1) and rp(2)?

[Image Analyst]

Those (the 1 and 2) refer to different blob numbers. For example, blob #1 is the circle or disk, and blob #2 is the triangle, and blob #3 is some other blob, etc. Each blob will have its own eccentricity value, and that (rp(n).Eccentricity) is how you refer to them.

[Tulips]

then how to set the perimeter value for several shapes such as diamond/triangle/square/circle/octagon..?is that from the enccentricity value?how to declare the value.how do I know the formula/calculation on this..Pls advice.thanks, sir..

[Image Analyst]

Perimeter is one of the measurements that can be returned by regionprops(), just like Eccentricity, Area, and so on. So you just do (untested)

measurements = regionprops(labeledImage, 'Perimeter', 'Area', 'Eccentricity');
allPerimeters = [measurements.Perimeter];
allAreas = [measurements.Area];
circularities = allPerimeters .^ 2 ./ (4 * pi * allAreas);

[Tulips]

thanks Mr Analysis for the detail answer. Can i know the circularities are the value to find the circular shape, isn't it?then, what about the diamond/triangle/octagon shape.should I used the circularities too to find these shapes values..thanks in advance!-_-

[Image Analyst]

For a perfect circle the circularity is 1 and it goes up from there for less and less round shapes. I don't know the exact numbers - you'll just have to examine them to see. The octagon will be slightly higher, a hexagon higher still, and a triangle will have an even higher circularity. Something that has a very high perimeter to area ration, like a splat shape or asterisk shape, will have a much higher circularity value. To be most precise you might have to construct a feature vector of the circularity, eccentricity, solidity, etc. and come up with some kind of model that uses all of those to predict the blobs closest standard shape (circle, triangle, octagon, or whatever).

[Tulips]

sir, from the result I get. the eccentricity of square is 0.7 and eccentricity of circle is 0.22. Can I know what is the matlab code to remove the square shape based on the these eccentricity values?hope u can advice.thanks!

[Image Analyst]

See my image segmentation tutorial. http://www.mathworks.com/matlabcentral/fileexchange/?term=authorid%3A31862 It's done in there, and very well documented about how you use ismember() to filter out objects based on certain criteria.

[Tulips]

sir,which file should I refer to?there are a lot..

[Walter Roberson]

http://www.mathworks.com/matlabcentral/fileexchange/25157-image-segmentation-tutorial-blobsdemo/content/BlobsDemo.m

Look for the line that says

% Now I'll demonstrate how to select certain blobs based using the ismember function.

Answer 2

[Image Analyst]

I prefer the perimeter squared to area ratio - the "circularity". For things like an asterisk shape, the eccentricity would be similar to similar to a circle bit the "circularity" of an asterisk would be much higher than that of a circle.

circularity = perimeter^2 / (4 * pi * area);

This is a very commonly used metric in shape analysis. So common that I don't know why it's not built into regionprops(). See my shape recognition demo:

% Demo to find certain shapes in an image based on their shape.
clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
imtool close all;  % Close all imtool figures.
clear;  % Erase all existing variables.
workspace;  % Make sure the workspace panel is showing.
fontSize = 20;

% Read in a standard MATLAB gray scale demo image.
folder = fullfile(matlabroot, '\toolbox\images\imdemos');
baseFileName = 'pillsetc.png';
% Get the full filename, with path prepended.
fullFileName = fullfile(folder, baseFileName);
% Check if file exists.
if ~exist(fullFileName, 'file')
  % File doesn't exist -- didn't find it there.  Check the search path for it.
  fullFileName = baseFileName; % No path this time.
  if ~exist(fullFileName, 'file')
    % Still didn't find it.  Alert user.
    errorMessage = sprintf('Error: %s does not exist in the search path folders.', fullFileName);
    uiwait(warndlg(errorMessage));
    return;
  end
end

% Read in image into an array.
rgbImage = imread(fullFileName); 
[rows, columns, numberOfColorBands] = size(rgbImage); 
% Display it.
subplot(2, 2, 1);
imshow(rgbImage, []);
title('Input Image', 'FontSize', fontSize);
% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0 0 1 1]);
% Give a name to the title bar.
set(gcf,'name','Shape Recognition Demo','numbertitle','off') 

% If it's monochrome (indexed), convert it to color. 
if numberOfColorBands > 1
  grayImage = rgbImage(:,:,2);
else
  % It's already a gray scale image.
  grayImage = rgbImage;
end

% Make a triangle on it.
triangleXCoordinates = [360 420 480];
triangleYCoordinates = [350 252 350];
traiangleBinaryImage = poly2mask(triangleXCoordinates, triangleYCoordinates, rows, columns);
% Burn it into the gray scale image.
grayImage(traiangleBinaryImage) = 255;

% Display it.
subplot(2, 2, 2);
imshow(grayImage, []);
title('Grayscale Image', 'FontSize', fontSize);

% Binarize the image.
binaryImage = grayImage > 120;
% Display it.
subplot(2, 2, 3);
imshow(binaryImage, []);
title('Initial (Noisy) Binary Image', 'FontSize', fontSize);

% Remove small objects.
binaryImage = bwareaopen(binaryImage, 300);
% Display it.
subplot(2, 2, 4);
imshow(binaryImage, []);
title('Cleaned Binary Image', 'FontSize', fontSize);

[labeledImage numberOfObjects] = bwlabel(binaryImage);
blobMeasurements = regionprops(labeledImage,...
  'Perimeter', 'Area', 'FilledArea', 'Solidity', 'Centroid'); 

% Get the outermost boundaries of the objects, just for fun.
filledImage = imfill(binaryImage, 'holes');
boundaries = bwboundaries(filledImage);

% Collect some of the measurements into individual arrays.
perimeters = [blobMeasurements.Perimeter];
areas = [blobMeasurements.Area];
filledAreas = [blobMeasurements.FilledArea];
solidities = [blobMeasurements.Solidity];
% Calculate circularities:
circularities = perimeters .^2 ./ (4 * pi * filledAreas);
% Print to command window.
fprintf('#, Perimeter,        Area, Filled Area, Solidity, Circularity\n');
for blobNumber = 1 : numberOfObjects
  fprintf('%d, %9.3f, %11.3f, %11.3f, %8.3f, %11.3f\n', ...
    blobNumber, perimeters(blobNumber), areas(blobNumber), ...
    filledAreas(blobNumber), solidities(blobNumber), circularities(blobNumber));
end

% Say what shape they are.
% IMPORTANT NOTE: depending on the aspect ratio of the rectangle or triangle
% their circularity can go from some minimum number up to a huge number.
for blobNumber = 1 : numberOfObjects
  % Outline the object so the user can see it.
  thisBoundary = boundaries{blobNumber};
  subplot(2, 2, 2); % Switch to upper right image.
  hold on;
  % Display prior boundaries in blue
  for k = 1 : blobNumber-1
    thisBoundary = boundaries{k};
    plot(thisBoundary(:,2), thisBoundary(:,1), 'b', 'LineWidth', 3);
  end
  % Display this bounary in red.
  thisBoundary = boundaries{blobNumber};
  plot(thisBoundary(:,2), thisBoundary(:,1), 'r', 'LineWidth', 3);
  subplot(2, 2, 4); % Switch to lower right image.
  
  % Determine the shape.
  if circularities(blobNumber) < 1.2
    message = sprintf('The circularity of object #%d is %.3f,\nso the object is a circle',...
      blobNumber, circularities(blobNumber));
    shape = 'circle';
  elseif circularities(blobNumber) < 1.6
    message = sprintf('The circularity of object #%d is %.3f,\nso the object is a square',...
      blobNumber, circularities(blobNumber));
    shape = 'square';
  elseif circularities(blobNumber) > 1.6 && circularities(blobNumber) < 1.8
    message = sprintf('The circularity of object #%d is %.3f,\nso the object is an isocoles triangle',...
      blobNumber, circularities(blobNumber));
    shape = 'triangle';
  else
    message = sprintf('The circularity of object #%d is %.3f,\nso the object is something else.',...
      blobNumber, circularities(blobNumber));
    shape = 'something else';
  end
  % Display in overlay above the object
  overlayMessage = sprintf('Object #%d = %s\ncirc = %.2f, s = %.2f', ...
    blobNumber, shape, circularities(blobNumber), solidities(blobNumber));
  text(blobMeasurements(blobNumber).Centroid(1), blobMeasurements(blobNumber).Centroid(2), ...
    overlayMessage, 'Color', 'r');
  button = questdlg(message, 'Continue', 'Continue', 'Cancel', 'Continue');
  if strcmp(button, 'Cancel')
    break;
  end
end

[Tulips]

thanks for spending your time helping me out..!appreciated it!;)

[Tulips]

hi mr expert, can I know what is the function of this poly2mask(triangleXCoordinates, triangleYCoordinates, rows, columns);??? the poly2mask ?also the purpose of solidity..thanks !

[Image Analyst]

poly2mask takes some x,y coordinates and turns them into an image with the coordinates defining the polygon. Solidity is the fraction that a shape takes up of its convex hull. For triangles, squares, and circles with perfect sides, this should be 1. If the sides were wavy/wiggly then it could be less than 1.

[Tulips]

what is the purpose of convex hull?what is the meaning behind it?

[Image Analyst]

Think of the convex hull as if you wrapped a rubber band around your object. If an object had no concavities, like a perfect square or circle, then the solidity = 1 because there are no points on the perimeter the rubber band does not touch. Now if the blob looked like a boomerang, then there would be a huge concavity where the rubber band would not touch the perimeter, so it's solidity would be much less than 1.

[Tulips]

can I know from the code given, how can I remove the rectangle such as in the photo http://www.flickr.com/photos/80166922@N06/7681815436/in/photostream/ I just want the inverted triangle shape only.many thanks.

[Image Analyst]

Detect the rectangle - it will have the smallest area. Then use ismember to remove it. Have you run my BlobsDemo? It goes over all this.

[nawaf]

@Image Analyst, Could you please explain your note: "IMPORTANT NOTE: depending on the aspect ratio of the rectangle or triangle their circularity can go from some minimum number up to a huge number". What is the best way to tackle this problem?

[Image Analyst]

As the shape get smaller, the shape is less defined. For example, if the shape is

0 1
1 1

That's a pretty small triangle. What is the area? is it 3? Or is it the square root of 2? What is the perimeter? Is it 3? Or is it 2+sqrt(2)? The circularity for small blocky/quantized shapes like this is affected greatly by the answers to those questions, and to the size of them. In addition, the perimeter might depend on what orientation the shape is at, like 0, 20 degrees, 40 degrees, 75 degrees, or whatever. Just try it and see.

You will find it useful to run my shape recongition demo, attached.

[nawaf]

Thank you! this is very helpful, appreciated!

[Laveena Kewlani]

Image Analyst, Is there any way to know the circularity of other important shape, hexagon, star etc?

Answer 3

[murk hassan memon]

Hi Image Analyst, Currently i am working on microscopic blood images in order to detect infected parasites of malaria and for this implementation first i need much data set . i have gone through two different image resources (ASH,CDC) but did not find much data set there. so its my request to you that i you have data set of microscopic blood images(normal,abnormal)then kindly help me out or suggest me another image resources to get from there. waiting for your response