MATLAB中的Houghlines [英] Houghlines in MATLAB
问题描述
使用 ，它会自动按逆时针顺序排列（或者是相反的方向！）。上面显示的数字表示顺序。
最后，我们使用 cp2tform 以获取转换矩阵，我们用来对齐图像和提取翻译，旋转和缩放。
以下是完整的代码：
%%＃第1步：读取和准备图像
％＃（因为你只提供一个，我通过旋转第一个创建了另一个）。
I1 = imread（'http://i.stack.imgur.com/Se6zX.jpg'）;
I1 = rgb2gray（imcrop（I1，[85 35 445 345]））; ％＃去掉白色边框
I2 =反转（I1，-10，'bilinear'，'crop'）; ％＃通过旋转10度创建第二个
%%＃步骤2：检测交叉点端点（按相同顺序排序）
p1 = getCross（I1）;
p2 = getCross（I2）;
%%＃步骤3：执行图像注册
％＃找到使用每个
t = cp2tform（p2，p1，'affine）的4个控制点将I2映射到I1的转换'）;
％＃将I2与I1对齐
II2 = imtransform（I2，t，'XData'，[1 size（I1,2）]，'YData'，[1 size （I1,1）]）;
％＃Plot
figure（'menu'，'none'）
subplot（131），imshow（I1），title（'I1'）
subplot （132），imshow（I2），title（'I2'）
subplot（133），imshow（II2），title（'I2（aligned）'）
％变换参数（平移，旋转，缩放）
ss = t.tdata.Tinv（2,1）;
sc = t.tdata.Tinv（1,1）;
tx = t.tdata.Tinv（3,1）;
ty = t.tdata.Tinv（3,2）;
scale = sqrt（ss * ss + sc * sc）
rotation = atan2（ss，sc）* 180 / pi
translation = [tx ty]
这里是提取线端点的函数：
code> function points = getCross（I）
％＃获取边缘（简单地通过阈值）
I = imfilter（I，fspecial（'gaussian'，[7 7]，1）对称'）;
BW = imclearborder（〜im2bw（I，0.5））;
％＃Hough transform
[H，T，R] = hough（BW）;
％＃检测峰值
P = houghpeaks（H，2）;
％＃检测行
lines = houghlines（BW，T，R，P）;
％＃按逆时针顺序排序2D点
points = [vertcat（lines.point1）; vertcat（lines.point2）];
idx = convhull（points（：，1），points（：，2））;
points = points（idx（1：end-1），:);
end
结果：
scale =
1.0025
rotation =
-9.7041
translation =
32.5270 -38.5021
旋转恢复为几乎10度（有一些不可避免的错误），缩放实际上是1（意味着没有缩放）。注意，在上面的例子中有一个翻译组件，因为旋转不是围绕十字标志的中心执行的）。
After detecting the lines in an image using Hough lines, how can I use it to calculate the change in angle (rotation) of the lines of a reference image?
Note to readers: This is a follow-up question, refer to these for background:
- How to select maximum intensity in Hough transform in MATLAB?
- Calculating displacement moved in MATLAB
The process is similar to what I showed before. Below I am using the images from your previous question (since you provided only one, I created the other by rotating the first by 10 degrees).
We start by detecting the lines for the two images. We do this with the help of the Hough transform functions. This what it looks like applied to both images:
Next, we want to perform image registration using the line endpoints as control-points. First, we make sure the points correspond to each other in the two images. I do this by computing the convex hull using convhull which automatically sorts them in counterclockwise-order (or is it in the opposite direction!). The numbers shown above indicate the order.
Finally, we use the function cp2tform to get the transformation matrix, which we use to align the images and extract the translation, rotation, and scaling.
The following is the complete code:
%% # Step 1: read and prepare images
%# (since you provided only one, I created the other by rotating the first).
I1 = imread('http://i.stack.imgur.com/Se6zX.jpg');
I1 = rgb2gray( imcrop(I1, [85 35 445 345]) ); %# Get rid of white border
I2 = imrotate(I1, -10, 'bilinear', 'crop'); %# Create 2nd by rotating 10 degrees
%% # Step 2: detect the cross sign endpoints (sorted in same order)
p1 = getCross(I1);
p2 = getCross(I2);
%% # Step 3: perform Image Registration
%# Find transformation that maps I2 to I1 using the 4 control points for each
t = cp2tform(p2,p1,'affine');
%# Transform I2 to be aligned with I1
II2 = imtransform(I2, t, 'XData',[1 size(I1,2)], 'YData',[1 size(I1,1)]);
%# Plot
figure('menu','none')
subplot(131), imshow(I1), title('I1')
subplot(132), imshow(I2), title('I2')
subplot(133), imshow(II2), title('I2 (aligned)')
%# Recover affine transformation params (translation, rotation, scale)
ss = t.tdata.Tinv(2,1);
sc = t.tdata.Tinv(1,1);
tx = t.tdata.Tinv(3,1);
ty = t.tdata.Tinv(3,2);
scale = sqrt(ss*ss + sc*sc)
rotation = atan2(ss,sc)*180/pi
translation = [tx ty]
And here's the function that extract the lines endpoints:
function points = getCross(I)
%# Get edges (simply by thresholding)
I = imfilter(I, fspecial('gaussian', [7 7], 1), 'symmetric');
BW = imclearborder(~im2bw(I, 0.5));
%# Hough transform
[H,T,R] = hough(BW);
%# Detect peaks
P = houghpeaks(H, 2);
%# Detect lines
lines = houghlines(BW, T, R, P);
%# Sort 2D points in counterclockwise order
points = [vertcat(lines.point1); vertcat(lines.point2)];
idx = convhull(points(:,1), points(:,2));
points = points(idx(1:end-1),:);
end
with the result:
scale =
1.0025
rotation =
-9.7041
translation =
32.5270 -38.5021
The rotation is recovered as almost 10 degrees (with some inevitable error), and scaling is effectively 1 (meaning there was no zooming). Note that there was a translation component in the above example, because rotation was not performed around the center of the cross sign).
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