使用RANSAC進行線條擬合

Question

我正在做一個圖像處理項目，基本上是使用圖像處理技術的Vectorise手繪圖像。 我在我的項目中使用RANSAC。我面臨的挑戰是算法不能按需要執行最佳擬合，但它使用任意兩個隨機點並繪製一條連線，如下圖所示。

RANSAC結果

在我的算法Vectorise手繪的圖像，我還做了灰度化，圖像閾值（圖像二值化）， 並使用形態學算子在骨骼化。

我正在爲我的項目使用MATLAB。

以下是迄今爲止我做過

% Line fitting using RANSAC 
[x, y] =size(skeleton_image); 
point =[]; 
count =1; 

% figure; imshow(~data); hold on 

    for n =1:x 
    for m =1:y 
     if skeleton_image(n,m)==1 
     point(count,1)=m; 
     point(count,2)=n; 
     count= count+1; 
     end 
    end 
    end 
    data = point'; 
number = size(data,2); % Total number of points 
X = 1:number; 
iter=100; num=2; thresh = 1000;count_inlines=103; best_count=0; best_line=[]; 

for i=1:iter 
% Randomly select 2 points 
    ind = randi(number,num); % randperm(number,num); 
    rnd_points= data(:,ind); 
% Fitting line 
    Gradient = (rnd_points(2,2)-rnd_points(2,1))/(rnd_points(1,2)-rnd_points(1,1)); 
    Constant = rnd_points(2,1)-Gradient*rnd_points(1,1); 
    Line = Gradient*X+Constant; [j,k]=size(Line); 
% How many pixels are in the line? 
    for i=1:number 

    Distance = sqrt((Line(:,i)-data(1,i)).^2)+(Line(:,i)-data(2,i)).^2); 

    if Distance<=thresh 
    inlines = data(:,i); 
    count_inlines=countinlines+1; 
     best_line=Line; 
end 

Answer 1

我覺得你的問題可能是在您計算的距離和/那就是目前1000它可以選擇所有的方式或閾值碼在任何情況下點，並選擇第一個或最後一個ransac線。 ％使用RANSAC的線路配件

%create skeleton_image objects 
skeleton_image = zeros(50,50); 

% draw a circle 
circle_center = [15,15]; 
radius = 6; 
for i=1:50 
    for j = 1:50 
    if abs(radius - sqrt((i-circle_center(1))^2 + (j-circle_center(2))^2)) <0.5 % < controls the thickness of the circle 
     skeleton_image(i,j) = 1; 

    endif 
    end 
end 

% draw a line 
grad=0.5; 
dy = 20; 
for i=10:50 
    skeleton_image(ceil(dy + grad*i),i)=1; 
    if (i < 50) 
    skeleton_image(ceil(dy + grad*i)+1,i)=1; 
    endif 
end 

% a handful of random points to make it more realistic 
skeleton_image(20,22)=1; 
skeleton_image(30,7)=1; 
skeleton_image(18,45)=1; 
skeleton_image(10,10)=1; 
skeleton_image(20,23)=1; 
skeleton_image(31,6)=1; 
skeleton_image(19,45)=1; 
skeleton_image(9,13)=1; 
skeleton_image(20,24)=1; 
skeleton_image(31,5)=1; 
skeleton_image(18,46)=1; 

% [x, y] =size(skeleton_image); 
x = 50; 
y = 50; 
points =[]; 
count =1; 

    for n =1:x 
    for m =1:y 
     if skeleton_image(n,m)==1 
     points(count,1)=m; 
     points(count,2)=n; 
     count= count+1; 
     end 
    end 
    end 



best_line = []; 
best_count = 0; 
line_point_list = []; 
% how close the pixel has to be to the line to be accepted 
threshold = 1; 

% how many samples are taken 
steps = 10; 

for i=1:steps 
    % pick two points 
    ind1 = randi(number,1); 
    ind2 = randi(number,1); 
    point1 = points(ind1,:); 
    point2 = points(ind2,:); 

    %auxiliaries 
    line = [point1;point2]; 
    lpl = []; %line_point_list 
    count_i = 0; 

    if point1 != point2 
    vector1 = point2-point1; 
    % unit vector 
    vector1_normalized = vector1 ./ norm(vector1); 
    % normal direction of the line 
    normal_of_vector1 = [vector1_normalized(2), -vector1_normalized(1)]; 

    % loop over points 
    for j = 1:size(points) 
     % calculate distance 
     normal_of_vector1; 
     vector2 = points(j,:) - point1; 
     distance = abs(dot(vector2, normal_of_vector1)); 
     if (distance < threshold) 
     count_i +=1; 
     lpl(count_i,:) = points(j,:); 
     endif 
    end 
    endif 
    if (count_i > best_count) 
    best_count = count_i; 
    best_line = line; 
    line_point_list = lpl; 
    endif 
end 
%best_c 
%best_l 
%line_point_list 


% draw found points 
for i=1:size(line_point_list) 
    skeleton_image(line_point_list(i,2),line_point_list(i,1)) = 0.25; 
end 

%visualize 
figure(1) 
imshow(skeleton_image)