[求助代码问题出在哪？大人帮帮我！！谢谢 - MATLAB爱好者论坛-LabFans.com

langde · 2008-05-31, 11:33

:cry: 代码的功能是找到二值图象中灰度值=255的孤立点并且将其灰度值设为0
大概的思路是：
先扫描到一点然后以这个点作为起始点为基点扫描其周围的8个点当扫描到其周围8个点中有灰度值=255的点时再以这个新点为基点，重复上面的操作。。。。一直这样进行下去
当其周围8个点中没有灰度值=255的点，则认为该点是孤立点，将其灰度值设为0

代码：
%main.m
clc
clear all
bw1=[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0
0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0
0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 255 255 255 255 255 0 0 255 0 0 0 0 0 0 0 0 0 0 0
0 0 255 255 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 255 0 0 255 0 0 0 0 0 0 0 0 0 0 0
0 0 0 255 255 0 0 0 0 0 0 0 0 0 0 255 0 0 0 255 255 255 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0
0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0
0 0 255 255 0 0 0 0 0 0 255 0 0 0 0 0 0 255 255 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0
0 0 255 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 255 0 0 0 0
0 0 255 255 255 255 255 255 255 255 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 255 0 255 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 255 255 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 255 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 255 255 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 255 255 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0];
figure,imshow(bw1);
title('数据图象');
[m,n]=size(bw1);

%链码计算，长度小于某个值的边缘滤出
bianjie_length=0;
first_point=zeros(1,2); %起始点
temp1_im=zeros(m,n); %临时存放检测到的点
end_im=zeros(m,n); %最终检测到的点

biaoji=zeros(1,2,2*(m+n)); %用于避免循环的出现
for i=2:20
for j=2:20
if bw1(i,j)==255
first_point=[i,j];
temp1_im(first_point(1,1),first_point(1,2))=255;
bianjie_length=1;
biaoji(:,:,1)=first_point(:,

;

%以i,j为起始点开始计算边缘长度
while ((first_point(1,1))~=1&&(first_point(1,2)~=1))
first_point=fun_find(bw1,biaoji,first_point(1,1),first_point(1,2),bianjie_length,1,1);
temp1_im(first_point(1,1),first_point(1,2))=255;
temp_bianjie_length=bianjie_length;
bianjie_length=temp_bianjie_length+1;
biaoji(:,:,bianjie_length)=first_point(:,

;
figure,imshow(temp1_im);
title('temp1_image');
disp(first_point);
end
disp(bianjie_length);

if bianjie_length>=2 % 人为设定的阈值
for p=1:m % 存储符合要求的边界
for q=1:n
if temp1_im(p,q)==255
end_im(p,q)=temp1_im(p,q);
end
end
end
bianjie_length=0;
else % 不满足条件，就将temp1_im清零，进入下一轮查找
bianjie_length=0;
end
end
end
end

figure,imshow(end_im);
title('检测结果');

下面的是我自己编写的子函数

%fun_find
function re_point=fun_find(bw,bji,a,b,len,c,d)
%//////////////////////////////////////
% bw为图象数据
% m,n为图象数据的维数
% bji为以搜索点记录
% i,j为搜索的起始点坐标
% len为已扫描的点数
%//////////////////////////////////////
re_point=[c,d];

for k=a+1:-1:a-1
for l=b+1:-1:b-1
if ((bw(k,l)==255)&&(bw(k,l)~=bw(a,b)))
for bj=1:len
if (bji(:,:,bj)~=[k,l])
re_point(:,

=[k,l];
break;
end
end
end
end
end

end

%i-1 7 6 5
% i 8 O 4
% i+1 1 2 3
% j-1 j j+1
% 对应每一点，依次搜寻其左下方点、下方点，右下方点、右边点,即为逆时针方向查找

附件里也有代码可以运行但是不能得到预期的结果很郁闷哪位高人能帮我看看改后烦请发到我的油箱吧 [email protected] 谢谢了

pslpsl · 2008-08-20, 18:39

不知道这段代码的最终目的是什么？

程序调好了运行结果中3个孤立点被消除了

但如果只是为了消除孤立点完全没必要写得这么复杂的

代码:

clc
clear all
bw1=[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0
0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0
0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 255 255 255 255 255 0 0 255 0 0 0 0 0 0 0 0 0 0 0
0 0 255 255 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 255 0 0 255 0 0 0 0 0 0 0 0 0 0 0
0 0 0 255 255 0 0 0 0 0 0 0 0 0 0 255 0 0 0 255 255 255 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0
0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0
0 0 255 255 0 0 0 0 0 0 255 0 0 0 0 0 0 255 255 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0
0 0 255 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 255 0 0 0 0
0 0 255 255 255 255 255 255 255 255 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 255 0 255 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 255 255 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 255 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 255 255 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 255 255 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0];
figure,imshow(bw1);
title('数据图象');
[m,n]=size(bw1);
%  EDIT BY　PSL　2008　08 20
bianjie_length=0;                
first_point=zeros(1,2);                 %起始点
temp1_im=zeros(m,n);                    %临时存放检测到的点
end_im=zeros(m,n);                      %最终检测到的点
biaoji=zeros(1,2,2*(m+n));             %用于避免循环的出现
figure;
title('temp1_image');
for i=1:m
    for j=1:n
        if bw1(i,j)==255
            first_point=[i,j];
            bianjie_length=bianjie_length+1;
            biaoji(:,:,bianjie_length)=first_point(:,:);
            %以i,j为起始点开始计算边缘长度
            disp(first_point);
            while  ((first_point(1,1))~=0&(first_point(1,2)~=0))
                   %first_point
                   %biaoji(:,:,bianjie_length)
                   first_point=fun_find(bw1,biaoji,first_point(1,1),first_point(1,2),bianjie_length,0,0);
                   
                   if(first_point(1,1))~=0&(first_point(1,2)~=0)
                   tempPos=biaoji(:,:,bianjie_length);
                   temp1_im(tempPos(1),tempPos(2))=255;
                   bianjie_length=bianjie_length+1;
                   biaoji(:,:,bianjie_length)=first_point(:,:);
                   pause(0.01);
                   imshow(temp1_im);
                   disp(first_point);
                   end
                   
            end
            disp(bianjie_length)
            bianjie_length=0;
        end
    end
end
 
 end_im=temp1_im;
 figure,imshow(end_im);
 title('检测结果');

代码:

function re_point=fun_find(bw,bji,a,b,len,c,d)
%//////////////////////////////////////
%  bw为图象数据
%  m,n为图象数据的维数
%  bji为已搜索点记录
%  a,b为搜索的起始点坐标
%  len为已扫描的点数
%  EDIT BY　PSL　2008　08 20
%//////////////////////////////////////
    re_point=[c,d];
    [m,n]=size(bw);
    bool=0;
    if(a==m)
        a1=m;
    else
        a1=a+1;
    end
    if(a==0)
        a2=0;
    else
        a2=a-1;
    end
    if(b==n)
        b1=n;
    else
        b1=b+1;
    end
    if(b==0)
        b2=0;
    else
        b2=b-1;
    end
    for k=a1:-1:a2
        for l=b1:-1:b2
            if (bw(k,l)==255&(k~=a|l~=b))
               for bj=1:len
                   if (bji(:,:,bj)==[k,l])
                       bool=1;
                   end
               end
               if (bool==0);
                       re_point(:,:)=[k,l];
                       break;
               end
            end
        end  
    end
end