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How to detect the shape in matlab?

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I can't understand the technique how to analyse the shape. So any please help me to understand this concept.
Code is below
function W = Classify(ImageRead)
RGB = imread('test.bmp');
figure,
imshow(RGB),
title('Original Image');
GRAY = rgb2gray(RGB);
figure,
imshow(GRAY),
title('Gray Image');
threshold = graythresh(GRAY);
BW = im2bw(GRAY, threshold);
figure,
imshow(BW),
title('Binary Image');
BW = ~ BW;
figure,
imshow(BW),
title('Inverted Binary Image');
[B,L] = bwboundaries(BW, 'noholes');
STATS = regionprops(L, 'all'); % we need 'BoundingBox' and 'Extent'
% Step 7: Classify Shapes according to properties
% Square = 3 = (1 + 2) = (X=Y + Extent = 1)
% Rectangular = 2 = (0 + 2) = (only Extent = 1)
% Circle = 1 = (1 + 0) = (X=Y , Extent < 1)
% UNKNOWN = 0
figure,
imshow(RGB),
title('Results');
hold on
for i = 1 : length(STATS)
W(i) = uint8(abs(STATS(i).BoundingBox(3)-STATS(i).BoundingBox(4)) < 0.1);
W(i) = W(i) + 2 * uint8((STATS(i).Extent - 1) == 0 );
centroid = STATS(i).Centroid;
switch W(i)
case 1
plot(centroid(1),centroid(2),'wO');
case 2
plot(centroid(1),centroid(2),'wX');
case 3
plot(centroid(1),centroid(2),'wS');
end
end
return

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Accepted Answer

Image Analyst
Image Analyst on 27 Dec 2013
What are the kinds of shapes you have there?
  1. polygons (everything is a polygon)
  2. quadrilaterals, polygons, and ellipsoids
  3. quadrilaterals, rectangles, polygons, and ellipsoids
  4. quadrilaterals, rectangles, polygons, circles, and ellipsoids
  5. quadrilaterals, rectangles, squares, polygons, circles, and ellipsoids
You might look at the solidity, area, and perimeter. And the circularity = perimeter.^2 ./ (4*pi*area).
You may also find this useful to determine how many sides a polygon has: http://matlab.wikia.com/wiki/FAQ#How_do_I_find_.22kinks.22_in_a_curve.3F

  3 Comments

saravanakumar D
saravanakumar D on 27 Dec 2013
i want to know why programmer using boundingbox in this programme. what is meaning of BoundingBox(3)and BoundingBox(4)
hold on for i = 1 : length(STATS) W(i) = uint8(abs(STATS(i).BoundingBox(3)-STATS(i).BoundingBox(4)) < 0.1); W(i) = W(i) + 2 * uint8((STATS(i).Extent - 1) == 0 ); centroid = STATS(i).Centroid; switch W(i) case 1 plot(centroid(1),centroid(2),'wO'); case 2 plot(centroid(1),centroid(2),'wX'); case 3 plot(centroid(1),centroid(2),'wS'); end
Marc
Marc on 27 Dec 2013
Boundingbox is a property from regionprops. It's being stored in the structure STATS.
doc regionprops
I assume you have the image processing toolbox?
Image Analyst
Image Analyst on 27 Dec 2013
bounding box won't help unless one of the shape classes bounding box size is known in advance and specified as a characteristic of that shape. In other words "if the width of the bounding box is this and the height of the bounding box is that, then the shape must be this (circle or whatever)."

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More Answers (3)

Shawn Fernandes
Shawn Fernandes on 21 Mar 2018
Hi All,
Bounding box gives the smallest possible rectangle / cuboid that fits the given shape, and would support n dimensions. [x_cordinate,y_cordinate,z_cordinate,....nth_cordinate,x_width,y_width,z_width.....nth_width] in this 2 D image, we have bounding box defined for each shapes as [x_cordinate,y_cordinate,x_width,y_width]
Extent gives the ratio of area of the bounding box to area of the region. For squares and rectangles, as the bounding box matches the shape, extent = 1. For circles and ellipses, the ratio of area of region to bounding box is always a constant = pi/4, [ (pi * a * b) / (2*x * 2 * y) is extent of circular region, for circle, a = b = x =y, for ellipse, a=x and b =y ]
So
(1)for Circles, we have x_width = y_width,extent = pi/4
(2)for squares, we have x_width = y_width,extent =1,
(3)for rectangles, we have x_width != y_width,extent =1
(4)For ellipse, we have we have x_width != y_width,extent = pi/4
Reference:-
https://www.mathworks.com/help/images/ref/regionprops.html
the below code has been tested and it works
for i = 1 : length(STATS)
centroid = STATS(i).Centroid;
if((STATS(i).BoundingBox(3)~=STATS(i).BoundingBox(4)) && (STATS(i).Extent<1))
plot(centroid(1),centroid(2),'w+');
text(centroid(1),centroid(2),num2str(i),'Color','y');
end
if((STATS(i).BoundingBox(3)==STATS(i).BoundingBox(4)) && (STATS(i).Extent==1))
plot(centroid(1),centroid(2),'wS');
text(centroid(1),centroid(2),num2str(i),'Color','y');
end
if((STATS(i).BoundingBox(3)==STATS(i).BoundingBox(4)) && (STATS(i).Extent > 0.76 && STATS(i).Extent < .795))
plot(centroid(1),centroid(2),'wO');
text(centroid(1),centroid(2),num2str(i),'Color','y');
end
if((STATS(i).BoundingBox(3)~=STATS(i).BoundingBox(4)) && (STATS(i).Extent==1))
plot(centroid(1),centroid(2),'wX');
text(centroid(1),centroid(2),num2str(i),'Color','y');
end
if((STATS(i).BoundingBox(3)~=STATS(i).BoundingBox(4)) && (STATS(i).Extent > 0.76 && STATS(i).Extent < .795))
plot(centroid(1),centroid(2),'w*');
text(centroid(1),centroid(2),num2str(i),'Color','y');
end
end
Hope this helps..

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Show 2 older comments
Ravi Singh
Ravi Singh on 1 Apr 2019
hii amar could you share your method to detect similar objects and finding the count of it ?? would be a great help. I need to detect multiple objects in image and based on color or shape and count the similar objects..

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sss
sss on 26 Dec 2016
Edited: Image Analyst on 26 Dec 2016
what is the meaning of this for loop? -----
for i = 1 : length(STATS)
W(i) = uint8(abs(STATS(i).BoundingBox(3)-STATS(i).BoundingBox(4)) < 0.1);
W(i) = W(i) + 2 * uint8((STATS(i).Extent - 1) == 0 );
centroid = STATS(i).Centroid;
switch W(i)
case 1
plot(centroid(1),centroid(2),'wO');
case 2
plot(centroid(1),centroid(2),'wX');
case 3
plot(centroid(1),centroid(2),'wS');
end

  1 Comment

Image Analyst
Image Analyst on 26 Dec 2016
It plots w0, wx, or xS at the centroid of blobs in a binary image. If the blob is roughly square it puts a wS at the centroid. If it's a rectangle it will put up wX. Otherwise it will put up w0 for arbitrarily-shaped blobs that fit in a bounding box that is roughly square. I don't see anything being put up for arbitrarily-shaped blobs that have a rectangular bounding box.

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RUTARAKA GIDEON
RUTARAKA GIDEON on 22 Jan 2018
i have a question,what does this function really do?

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