How to classify shapes of this image as square, rectangle, triangle and circle?

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ROMIL
ROMIL 2014 年 2 月 19 日
コメント済み: Image Analyst 2021 年 10 月 6 日
Please provide me the matlab code to identify shapes on this image and classify them as square, rectangle, circle and triangle.
  5 件のコメント
phaneendra ch
phaneendra ch 2015 年 12 月 11 日
Ur code is working very excellent.
Image Analyst
Image Analyst 2015 年 12 月 12 日
Not sure whose code you're talking about, but glad you finally got it working. If you have any questions in the future, post your image and code in a new question.

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採用された回答

Matt Kindig
Matt Kindig 2014 年 2 月 21 日
編集済み: Matt Kindig 2014 年 2 月 21 日
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
  12 件のコメント
Beenish Ishtiaq
Beenish Ishtiaq 2021 年 8 月 5 日
error in minboundrect

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その他の回答 (10 件)

Image Analyst
Image Analyst 2014 年 2 月 19 日
編集済み: Image Analyst 2016 年 3 月 19 日
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.
  21 件のコメント
aliya
aliya 2021 年 10 月 6 日
hye sir, i use your shape_recognition_demo.m. code. but i want to use my own image, i already use your method which replace the
% Now create a demo image.
[binaryImage, numSidesCircularity] = CreateDemoImage();
with imread() but i got an error 'unrecognized function or variable 'binaryImage' , how to solve this? thankyou in advanced!
Image Analyst
Image Analyst 2021 年 10 月 6 日
@aliya I just downloaded it and it works fine. You modified it somehow but didn't attach your code so I can't fix it.
I'm attaching the lastest version I have of the demo (not sure if it's changed over the last 7 years, but probably).

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phaneendra ch
phaneendra ch 2015 年 11 月 22 日
While executing the above code I am getting an error in the code I.e.,(undefined function or variable 'minbounderect'). As I am new to MATLAB plz solve this prblm.tq in advance sir
  2 件のコメント
Image Analyst
Image Analyst 2015 年 11 月 22 日
I'm attaching John D'Errico's function, minboundrect().
kaz
kaz 2016 年 4 月 25 日
i am so noob with matlab. sir how to add this function to the codes at the top. which i am getting the same error minboundrect. ty

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Venkatesh A
Venkatesh A 2015 年 12 月 12 日
Mr.matt kindig, I am working with your "various shapes detecting" code which is in this page. The code u have written is working very well for the above black and white image(the image which you chosen to write code). But I am using some other images( 'shapes.jpg' which is now I am submitting) which might have some holes in the inside of the image and I am getting error( error is the matrix cannot be generated). How to fill the small holes in that image. Can I draw subplots for your code for various shapes that are present in the image?
  1 件のコメント
Image Analyst
Image Analyst 2015 年 12 月 12 日
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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Venkatesh A
Venkatesh A 2015 年 12 月 14 日
Sorry sir. here is the image
  2 件のコメント
Jan
Jan 2016 年 2 月 18 日
No, there is no image.
B.k Sumedha
B.k Sumedha 2016 年 3 月 19 日
There is no image attached.

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AKHIL RAJAGOPAL
AKHIL RAJAGOPAL 2016 年 4 月 23 日
I am getting an error at isTriangle = ~isCircle & (TriangleMetric < 0.6); as: Error using & Matrix dimensions must agree.
Please help me solve this problem.
  10 件のコメント
Image Analyst
Image Analyst 2016 年 4 月 25 日
See attached spatial calibration demo.
kaz
kaz 2016 年 4 月 25 日
thank you so much

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noor jahan m
noor jahan m 2016 年 12 月 8 日
i am a beginner in matlab. I tried the code for rectangle detection. I cud also find the function minboundrect . but i got error in convex hull of this function. Can u please help further? thank u

Rahul Chauhan
Rahul Chauhan 2017 年 10 月 23 日
Ty very much sir for the code but I'm getting error as "undefined function or variable 'minboundract'" So plz help me sir getting this error correct asap.... Again Ty in advance sir
  4 件のコメント
Image Analyst
Image Analyst 2017 年 10 月 24 日
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
Rahul Chauhan 2017 年 10 月 25 日
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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Pavel Vilbik
Pavel Vilbik 2017 年 12 月 11 日
How to find the qr code( this plastic card) in this photo, as it has a begining job, I need a coordinate and axis, and that I would be marked with a ractangle.
  1 件のコメント
Image Analyst
Image Analyst 2017 年 12 月 11 日
You should start your own question on this. In that question I'll tell you how to find the blobs, perhaps based on color saturation, and then to take the histogram of each blob looking for a fairly bimodal histogram. The standard deviation of the histogram of the all black and all white objects will be much less than a checkerboard object. If you still can't figure it out, I might post code over in that new question you're going to post.

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Michelle de Bock
Michelle de Bock 2018 年 12 月 28 日
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
  1 件のコメント
Image Analyst
Image Analyst 2018 年 12 月 28 日
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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Ahaana Khurana
Ahaana Khurana 2020 年 2 月 4 日
ROMIL can you pls provide the code for SHAPE DETECTION on ahaanakhurana@gmail.com.
  3 件のコメント
Dina Abd El-twab
Dina Abd El-twab 2020 年 2 月 24 日
編集済み: Dina Abd El-twab 2020 年 2 月 24 日
pp=alexnet;
ppl=pp.Layers;
pp=pp.Layers(1:19);
ppp=[pp
fullyConnectedLayer(2)
softmaxLayer()
classificationLayer()]
options = trainingOptions('sgdm',...
'InitialLearnRate',1e-3,...
'MaxEpochs',10,...
'CheckpointPath',tempdir);
train1 = trainFasterRCNNObjectDetector(gTruth,ppp,options, ...
'NegativeOverlapRange',[0 0.1], ...
'PositiveOverlapRange',[0.5 1], ...
'SmallestImageDimension',300);
a = imread('US0018_0131.png');
a = imresize(a,[227 227]);
[bbox,score,label] = detect(train1,a);
detect= insertShape(a,'rectangle',bbox);
figure
imshow(detect)
@Image Analyst
Image Analyst
Dina Abd El-twab
Dina Abd El-twab 2020 年 2 月 24 日
I applied this code to draw a rectangle on the region of interest that i want after taining using faster RCNN .I want to convert the drawn rectangle to be circle in the next step , could you help me please ?
@Image Analyst
Image Analyst

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