在MATLAB中检测图像内的圆形 [英] Detect a circular shape inside image in MATLAB

问题描述

检测图像中这些圆形的最快方法是什么?

半径始终在(80-100mm)之间.背景始终为白色.圆圈将始终居中.

我尝试了.

以下是较大算法中的相关代码

 ％在下方应用霍夫[累计，马戏团，西拉德] = ...CircularHough_Grd(gR，[89 93]，...17.4、13、1)；％的执行时间为0.72秒让我们看看我们得到了什么imshow(gR);坚持，稍等;plot(circen(:，1)，circen(:，2)，'r +');对于ii = 1:size(circen，1)矩形('Position'，[circen(ii，1)-cirrad(ii)，circen(ii，2)-cirrad(ii)，2 * cirrad(ii)，2 * cirrad(ii)]，...'曲率'，[1,1]，'edgecolor'，'b'，'linewidth'，1.5)；结尾暂缓 

有意义的圆圈是中间的圆圈.

解决方案

这是我的建议:
1.转换为灰色图像，增强与白色的区别"

  gimg = min(img，[]，3); 

2.去除白色区域的阈值

  BW = im2bw(gimg，.4); 

3.获取图像区域的面积和质心性

  st = regionprops(〜BW，'Area'，'Centroid'，'PixelIdxList'); 

4.只选择足够大的区域

  sel = [st.Area]>编号(BW)* 0.025;％至少为图像尺寸的2.5％st = st(sel); 

5.计算区域到图像中心的距离

  cntr = 0.5 * [size(BW，2)size(BW，1)];％X-Y坐标和NOT Row/Cold = sqrt(sum(bsxfun(@ minus，vertcat(st.Centroid)，cntr).^ 2，2)); 

6.选择最靠近中心的区域

  [mn idx] = min(d); 

7.创建一个面具

  res = false(size(BW));res(st(idx).PixelIdxList)= true; 

您还可以考虑使用其他区域属性(例如'Eccentricity')来更好地拒绝非圆形区域.

Whats the fastest way to detect these circular shapes in a image?

The radius is always between(80-100mm). Background is always white. And the circle will always be in center.

I have tried Hough Transform but I couldn't really get it to work. I am new to this, and I get a feeling like Hough Transform is a overkill for this. Kindly suggest me the right approach to do this.

---

UPDATE

Here is what I got after applying hough transform.

I have used the algorithm mentioned here.

Following is the relevant code from the bigger algorithm

% applying Hough Below
[accum, circen, cirrad] = ...
    CircularHough_Grd(gR, [89 93],...
    17.4, 13, 1);   % this executes in 0.72 sec

% Lets see what we got
imshow(gR);
hold on;
plot(circen(:,1), circen(:,2), 'r+');
for ii = 1 : size(circen, 1)
    rectangle('Position',[circen(ii,1) - cirrad(ii), circen(ii,2) - cirrad(ii), 2*cirrad(ii), 2*cirrad(ii)],...
        'Curvature', [1,1], 'edgecolor', 'b', 'linewidth', 1.5);
end
hold off;

The meaningful circle is the one in the middle.

解决方案

Here's what I propose:
1. convert to gray image, enhancing the "difference from white"

gimg = min( img, [], 3 );

2. Threshold to remove white area

BW = im2bw( gimg, .4 ); 

3. Get area and centroid porperties of image regions

st = regionprops( ~BW, 'Area', 'Centroid', 'PixelIdxList' );

4. select only large enough regions

sel = [st.Area] > numel(BW)*0.025; % at least 2.5% of image size
st = st(sel);

5. compute region distance to center of image

cntr = .5 * [size(BW,2) size(BW,1)]; % X-Y coordinates and NOT Row/Col
d = sqrt( sum( bsxfun(@minus,vertcat( st.Centroid ), cntr ).^2, 2 ) );

6. pick the region closest to center

[mn idx] = min(d);

7. Create a mask

res = false(size(BW)); 
res( st(idx).PixelIdxList ) = true;

You might also consider using other region properties (e.g., 'Eccentricity') to better reject regions that are not circular.

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