Asked by ROMIL
on 19 Feb 2014

Please provide me the matlab code to identify shapes on this image and classify them as square, rectangle, circle and triangle.

Answer by Matt Kindig
on 21 Feb 2014

Edited by Matt Kindig
on 21 Feb 2014

Accepted Answer

Another approach is to calculate the best-fit bounding rectangle of each object, such that the bounding rectangle can be oriented at an arbitrary angle. I took the following approach:

1) Identify the boundary (i.e. perimeter) of each object, using bwboundaries()

2) Calculate the smallest rectangular bounding box that contains this perimeter. To do this, I used the minboundrect function available at http://www.mathworks.com/matlabcentral/fileexchange/34767-a-suite-of-minimal-bounding-objects/content/MinBoundSuite/minboundrect.m.

3) I calculated the width, height, and area of each bounding rectangle. The aspect ratio (ratio between the width and height) can be used to determine whether it is a square (aspect ratio ~= 1.0) or rectangle.

4) For a rectangle or square, the filled area of the object (from regionprops()) should be almost the same as the area of its bounding rectangle, whereas for a triangle it should be substantially less.

5) For the circularity condition, I used the ratio between perimeter and area like Image Analyst suggested.

Enjoy!

im = imread('http://www.mathworks.com/matlabcentral/answers/uploaded_files/8372/abc.jpg');

%convert to 2D black and white with colors inverted

BW = im(:,:,1) < 10;

%get outlines of each object

[B,L,N] = bwboundaries(BW);

%get stats

stats= regionprops(L, 'Centroid', 'Area', 'Perimeter');

Centroid = cat(1, stats.Centroid);

Perimeter = cat(1,stats.Perimeter);

Area = cat(1,stats.Area);

CircleMetric = (Perimeter.^2)./(4*pi*Area); %circularity metric

SquareMetric = NaN(N,1);

TriangleMetric = NaN(N,1);

%for each boundary, fit to bounding box, and calculate some parameters

for k=1:N,

boundary = B{k};

[rx,ry,boxArea] = minboundrect( boundary(:,2), boundary(:,1)); %x and y are flipped in images

%get width and height of bounding box

width = sqrt( sum( (rx(2)-rx(1)).^2 + (ry(2)-ry(1)).^2));

height = sqrt( sum( (rx(2)-rx(3)).^2+ (ry(2)-ry(3)).^2));

aspectRatio = width/height;

if aspectRatio > 1,

aspectRatio = height/width; %make aspect ratio less than unity

end

SquareMetric(k) = aspectRatio; %aspect ratio of box sides

TriangleMetric(k) = Area(k)/boxArea; %filled area vs box area

end

%define some thresholds for each metric

%do in order of circle, triangle, square, rectangle to avoid assigning the

%same shape to multiple objects

isCircle = (CircleMetric < 1.1);

isTriangle = ~isCircle & (TriangleMetric < 0.6);

isSquare = ~isCircle & ~isTriangle & (SquareMetric > 0.9);

isRectangle= ~isCircle & ~isTriangle & ~isSquare; %rectangle isn't any of these

%assign shape to each object

whichShape = cell(N,1);

whichShape(isCircle) = {'Circle'};

whichShape(isTriangle) = {'Triangle'};

whichShape(isSquare) = {'Square'};

whichShape(isRectangle)= {'Rectangle'};

%now label with results

RGB = label2rgb(L);

imshow(RGB); hold on;

Combined = [CircleMetric, SquareMetric, TriangleMetric];

for k=1:N,

%display metric values and which shape next to object

Txt = sprintf('C=%0.3f S=%0.3f T=%0.3f', Combined(k,:));

text( Centroid(k,1)-20, Centroid(k,2), Txt);

text( Centroid(k,1)-20, Centroid(k,2)+20, whichShape{k});

end

Priya Raj
on 20 Mar 2018

thank you sir!!

Priya Raj
on 20 Mar 2018

Mark Steven Billones
on 21 Mar 2018

sir can't run in matlab 2015, whats the problem?

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Answer by Image Analyst
on 19 Feb 2014

Edited by Image Analyst
on 19 Mar 2016

Look at the perimeter squared to area ratio. Use regionprops as shown in my Image Segmentation Tutorial: http://www.mathworks.com/matlabcentral/fileexchange/?term=authorid%3A31862

[EDIT]

See attached demo.

Venkatesh A
on 12 Dec 2015

then mozhi
on 5 Jan 2017

Image Analyst
on 5 Jan 2017

I just copied and pasted my demo from above and it works fine (as usual). Virtually every time someone says my demo didn't work it's because they modified it somehow. For example a common problem is the demo was expecting a gray scale image and the user changed it to feed it a color image. In your case since it was saying that numSidesCircularity is undefined, you may have deleted this line:

[binaryImage, numSidesCircularity] = CreateDemoImage();

and replaced it with an imread() of your own image. The only way I can tell is if you upload the version you are using along with your full error message (ALL the red text).

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Answer by phaneendra ch
on 22 Nov 2015

Answer by Venkatesh A
on 12 Dec 2015

Image Analyst
on 12 Dec 2015

Venkatesh A, you forgot to give the link to your question where you attached your code and image.

If you do that, we can go to your question and tell you how to use imfill(binaryImage, 'holes') to fill holes in your image.

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Answer by Venkatesh A
on 14 Dec 2015

Sorry sir. here is the image

Jan
on 18 Feb 2016

No, there is no image.

B.k Sumedha
on 19 Mar 2016

There is no image attached.

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Answer by AKHIL RAJAGOPAL
on 23 Apr 2016

I am getting an error at isTriangle = ~isCircle & (TriangleMetric < 0.6); as: Error using & Matrix dimensions must agree.

Please help me solve this problem.

kaz
on 25 Apr 2016

kaz
on 25 Apr 2016

thank you so much

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Answer by noor jahan m
on 8 Dec 2016

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Answer by Rahul Chauhan
on 23 Oct 2017

Rahul Chauhan
on 24 Oct 2017

Image Analyst
on 24 Oct 2017

Again, it's minboundrect(), not minboundract().

Supply your image and code in a new question (not as an Answer here in ROMIL's discussion thread - he probably doesn't care anymore since he posted this three and a half years ago.)

Rahul Chauhan
on 25 Oct 2017

again sir i corrected the function but the code is not working here i am attaching the code and image file know help me.... here is the code :(1)file attached as test.m

(2)minboundrect.m

here is the image: abc.jpg

ty in advance sir for helping me.....

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Answer by Pavel Vilbik
on 11 Dec 2017

Image Analyst
on 11 Dec 2017

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Answer by Michelle de Bock
on 28 Dec 2018

Hi Sir,

Is it also possible to classify the direction of the triangle. e.g. left pointing triangle or right pointing triangle? Also classifying the thrid/fourth object in the picture? This is not really a rectangle, but how to separate this from a real rectangle?

Kind regards,

Michelle

Image Analyst
on 28 Dec 2018

Yes, I'm sure you could. Just modify my attached shape recognition demo. Once you have the blob, find its bounding box and centroid with regionprops. Then if the centroid is to the left of the centerline of the bounding box, it's pointing to the left. If it's below, it's pointing up.

For the other object, you'll also have to look for how many vertices it has and then perhaps scale a template to its size and see if enough pixels match to be considered that object. You could also do the template matching method with the triangles if you want. No, I don't have code for that but, being smart engineer, I'm sure you will find it easy to do.

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## 5 Comments

## Ashish Uthama (view profile)

Direct link to this comment:https://it.mathworks.com/matlabcentral/answers/116793-how-to-classify-shapes-of-this-image-as-square-rectangle-triangle-and-circle#comment_197193

## ROMIL (view profile)

Direct link to this comment:https://it.mathworks.com/matlabcentral/answers/116793-how-to-classify-shapes-of-this-image-as-square-rectangle-triangle-and-circle#comment_197261

## phaneendra ch (view profile)

Direct link to this comment:https://it.mathworks.com/matlabcentral/answers/116793-how-to-classify-shapes-of-this-image-as-square-rectangle-triangle-and-circle#comment_329013

## phaneendra ch (view profile)

Direct link to this comment:https://it.mathworks.com/matlabcentral/answers/116793-how-to-classify-shapes-of-this-image-as-square-rectangle-triangle-and-circle#comment_329014

## Image Analyst (view profile)

Direct link to this comment:https://it.mathworks.com/matlabcentral/answers/116793-how-to-classify-shapes-of-this-image-as-square-rectangle-triangle-and-circle#comment_329118

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