detect Circles and squares on the image using regionprops

Question

Ahmed Emad 2019-12-6

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/495131-detect-circles-and-squares-on-the-image-using-regionprops

评论： Image Analyst 2019-12-6

Now , i have image called "test.png" and my question is how to detect circles with & without holes and display each of them in single figure and here is my code but desn't work with me , so need help

I = imread('test.png');
imshow(I);
I = rgb2gray(I);
I = im2bw(I,0.01);
[L ,num] = bwlabel(I);
stats1 = regionprops (L,'EulerNumber' , 'Area' , 'Perimeter' , 'BoundingBox');
sum = 0;
for R=1:num
% this way desn't work but maybe it just need a little changes     
%     x = uint8 (stats1(R).BoundingBox(1));
%     y = uint8 (stats1(R).BoundingBox(2));
%     if (I(x,y) == 0)
%         sum = sum + 1;
%         bb = stats1(R).BoundingBox;
%         rectangle('position',bb,'edgecolor','r','linewidth',1.3);
%     end
        circularity = (stats1(R).Perimeter .^ 2) ./ (4 * pi * stats1(R).Area);
        if (circularity >= 1.1)
            sum = sum + 1;
            bb = stats1(R).BoundingBox;
            rectangle('position',bb,'edgecolor','r','linewidth',1.3);
        end
end
disp(sum);

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Image Analyst 2019-12-6

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Marcel Kreuzberg 2019-12-6

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/495131-detect-circles-and-squares-on-the-image-using-regionprops#answer_405141

try this

I = imread('test.png');
imshow(I);
I = rgb2gray(I);
I = im2bw(I,0.01);
[L ,num] = bwlabel(I);
stats1 = regionprops (L,'EulerNumber' , 'Area' ,'ConvexArea', 'Perimeter' , 'BoundingBox');
sum1 = 0;
sum2 = 0;
for R=1:num
    circularity = (stats1(R).Perimeter .^ 2) ./ (4 * pi * stats1(R).ConvexArea);
    if (circularity < 1) %Circles
        if stats1(R).EulerNumber == 1
            sum1 = sum1 + 1;
            bb = stats1(R).BoundingBox;
            rectangle('position',bb,'edgecolor','r','linewidth',1.3);
        end
        if stats1(R).EulerNumber < 1
            sum2 = sum2 + 1;
            bb = stats1(R).BoundingBox;
            rectangle('position',bb,'edgecolor','y','linewidth',1.3);
        end
        
    end
end
disp(sum1); %circle without holes
disp
(sum2); %circle with holes

2 个评论
显示无隐藏无

Ahmed Emad 2019-12-6

Thank you very much Marcel Kreuzberg but could you please tell me the difference between 'Area' and 'ConvexArea' ? and explain me the circularity equation because i still misunderstand it. Thank you again!

Image Analyst 2019-12-6

Convex area does not include holes or "bays" in the outer perimeter. Basically think like if you put a rubber band around your shape. The area inside the rubber band is the convex hull area. The regular area would be the actual area of the white pixels, i.e., it does not include holes.

Circularity is 1 for a circle and higher for non-circular things. If the blob is a perfect circle the perimeter squared equals 4 * pi * the area. The more non-circular and tortuous the shape becomes, the perimeter gets larger faster than the area does and so the circularity metric will get larger. The smallest it can theoretically be is 1 for a perfect circle, though I've seen some as low as 0.9 presumably due to quantization error because we have an image compoaed of square blocks (pixels).

For another demo, see my attached shape recognition demos and see some of the links on the left of this page.

请先登录，再进行评论。