使用Hough变换椭圆检测

问题描述

使用Hough变换，我怎么可以检测并得到（X0，Y0）坐标和A和B在二维空间中的椭圆形？

的

这是ellipse01.bmp：

  I = imread（'ellipse01.bmp'）;
[M N] =尺寸（Ⅰ）;
C = 0;
对于i = 1：M
    对于j = 1：N
        如果我（I，J）== 1
        C = C + 1;
        P（C，1）= I;
        P（C，2）= j的;
        结束
    结束
结束
边=转置（对）;
Size_Ellipse =大小（边）;
B = 1为：ceil（Size_Ellipse（1）/ 2）;
度Acc =零（长度（B），1）;
A1 = 0; A2 = 0; B1 = 0; B2 = 0;
Ellipse_Minor = []; Ellipse_Major = []; Ellipse_X0 = []; Ellipse_Y0 = [];
Global_Threshold = CEIL（Size_Ellipse（2）/ 6）;％用于长轴比较
Local_Threshold = CEIL（Size_Ellipse（1）/ 25）;％用于短轴比较
[Y，X] =找到（边）;
极限= numel（Y）;
脱粒= 150;
帕拉= [];

对于Count_01 = 1：（限-1）
  对于Count_02 =（Count_01 + 1）：限制
    如果（（Count_02＆GT;限制）||（Count_01＆GT;限制））
      继续
    结束
    A1 = Y（Count_01）; B1 = X（Count_01）;
    A2 = Y（Count_02）; B2 = X（Count_02）;
    Dist_01 =（SQRT（（A1-A2）^ 2 +（B1-B2）^ 2））;
    如果（Dist_01＆GT; Global_Threshold）
      Center_X0 =（B1 + B2）/ 2; Center_Y0 =（A1 + A2）/ 2;
      主要= Dist_01 / 2.0;α= ATAN（（A2-A1）/（B2-B1））;
      如果（阿尔法== 0）
        对于Count_03 = 1：极限
          如果（（Count_03〜= Count_01）||（Count_03〜= Count_02））
            A3 = Y（Count_03）; B3 = X（Count_03）;
            Dist_02 =（开方（（A3  -  Center_Y0）^ 2 +（B3  -  Center_X0）^ 2））;
            如果（Dist_02＆GT; Local_Threshold）
              Cos_Tau =（（主要）^ 2 +（Dist_02）^ 2  - （A3-A2）^ 2  - （B3-B2）^ 2）/（2 *主要* Dist_02）;
              Sin_Tau = 1  - （Cos_Tau）^ 2;
              Minor_Temp =（（主要* Dist_02 * Sin_Tau）^ 2）/（主要^ 2  - （（Dist_02 * Cos_Tau）^ 2））;
              如果（（Minor_Temp→1）及及（Minor_Temp＆LT; B（完）））
                ACC（圆（Minor_Temp））= ACC（圆（Minor_Temp））+ 1;
              结束
            结束
          结束
        结束
      结束
      小=找到（ACC == MAX（ACC（:)））;
      如果（ACC（次要）GT; Thresh的）
        Ellipse_Minor（完+ 1）=小（1）; Ellipse_Major（完+ 1）=少校;
        Ellipse_X0（完+ 1）= Center_X0; Ellipse_Y0（完+ 1）= Center_Y0;
        对数= 1：numel（X）
          Para_X =（（X（计数）-Ellipse_X0（完））^ 2）/（Ellipse_Major（完）^ 2）;
          Para_Y =（（Y（计数）-Ellipse_Y0（完））^ 2）/（Ellipse_Minor（完）^ 2）;
          如果（（（Para_X + Para_Y）GT; =  -  2）及及（（Para_X + Para_Y）其中; = 2））
            边（X（计数），Y（计数））= 0;
          结束
        结束
      结束
      ACC =零（大小（ACC））;
    结束
  结束
结束
 

解决方案

如果你用圆圈粗糙的变换给出RHO = X * COS（THETA）+ Y * SIN（THETA）对于椭圆既然是

您可以转换方程 RHO = A * X * COS（THETA）+ B * Y * SIN（THETA）虽然我不知道你是否使用标准Hough变换，为椭圆状的转换，你可以操纵的第一个给定的函数。

using Hough Transform, how can I detect and get coordinates of (x0,y0) and "a" and "b" of an ellipse in 2D space?

This is ellipse01.bmp:

I = imread('ellipse01.bmp');
[m n] = size(I);
c=0;
for i=1:m
    for j=1:n
        if I(i,j)==1
        c=c+1;
        p(c,1)=i;
        p(c,2)=j;
        end
    end
end
Edges=transpose(p);
Size_Ellipse = size(Edges);
B = 1:ceil(Size_Ellipse(1)/2);
Acc = zeros(length(B),1);
a1=0;a2=0;b1=0;b2=0;
Ellipse_Minor=[];Ellipse_Major=[];Ellipse_X0 = [];Ellipse_Y0 = [];
Global_Threshold = ceil(Size_Ellipse(2)/6);%Used for Major Axis Comparison
Local_Threshold = ceil(Size_Ellipse(1)/25);%Used for Minor Axis Comparison
[Y,X]=find(Edges);
Limit=numel(Y);
Thresh = 150;
Para=[];

for Count_01 =1:(Limit-1)
  for Count_02 =(Count_01+1):Limit
    if ((Count_02>Limit) || (Count_01>Limit))
      continue
    end
    a1=Y(Count_01);b1=X(Count_01);
    a2=Y(Count_02);b2=X(Count_02);
    Dist_01 = (sqrt((a1-a2)^2+(b1-b2)^2));
    if (Dist_01 >Global_Threshold)
      Center_X0 = (b1+b2)/2;Center_Y0 = (a1+a2)/2;
      Major = Dist_01/2.0;Alpha = atan((a2-a1)/(b2-b1));
      if(Alpha == 0)
        for Count_03 = 1:Limit
          if( (Count_03 ~= Count_01) || (Count_03 ~= Count_02))
            a3=Y(Count_03);b3=X(Count_03);
            Dist_02 = (sqrt((a3 - Center_Y0)^2+(b3 - Center_X0)^2));
            if(Dist_02 > Local_Threshold)
              Cos_Tau = ((Major)^2 + (Dist_02)^2 - (a3-a2)^2 - (b3-b2)^2)/(2*Major*Dist_02);
              Sin_Tau = 1 - (Cos_Tau)^2;
              Minor_Temp = ((Major*Dist_02*Sin_Tau)^2)/(Major^2 - ((Dist_02*Cos_Tau)^2));
              if((Minor_Temp>1) && (Minor_Temp<B(end)))
                Acc(round(Minor_Temp)) = Acc(round(Minor_Temp))+1;
              end
            end
          end
        end
      end
      Minor = find(Acc == max(Acc(:)));
      if(Acc(Minor)>Thresh)
        Ellipse_Minor(end+1)=Minor(1);Ellipse_Major(end+1)=Major;
        Ellipse_X0(end+1) = Center_X0;Ellipse_Y0(end+1) = Center_Y0;
        for Count = 1:numel(X)
          Para_X = ((X(Count)-Ellipse_X0(end))^2)/(Ellipse_Major(end)^2);
          Para_Y = ((Y(Count)-Ellipse_Y0(end))^2)/(Ellipse_Minor(end)^2);
          if (((Para_X + Para_Y)>=-2)&&((Para_X + Para_Y)<=2))
            Edges(X(Count),Y(Count))=0;
          end
        end
      end
      Acc = zeros(size(Acc));
    end
  end
end

解决方案

If you use circle for rough transform is given as rho = x*cos(theta) + y*sin(theta)For ellipse since it is

You could transform the equation as rho = a*x*cos(theta) + b*y*sin(theta)Although I am not sure if you use standard Hough Transform, for ellipse-like transforms, you could manipulate the first given function.

这篇关于使用Hough变换椭圆检测的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！