如何在matlab中以任意角度旋轉邊界框？

Question

我感到困惑，要沿任何角度來裁剪圖像，即pi/4，我該如何製作它？

Answer 1

以下示例演示瞭如何裁剪旋轉的邊界框。

該示例不顯示如何選擇邊界框和角度（排除用戶界面）。

爲了保持示例簡單，邊界框參數爲：中心，大小和角度（而不是4個角）。
旋轉時，中心保持原位。
大小是目標裁剪區域的寬度和高度（旋轉後的大小）。

如果你需要計算彎道改造，就可以使用旋轉矩陣：https://en.wikipedia.org/wiki/Rotation_matrix

代碼示例：

%Bounding box parameters: center, width, height and angle. 
%Center is selected, because center is kept the same when rotating. 
center_x = 256; %Center column index (applied center is 256.5) 
center_y = 192; %Center row index (applied center is 192.5) 
width = 300; %Number of columns of destination (cropped) area (even integer) 
height = 200; %Number of rows of destination (cropped) area (even integer) 
phi = 120; %Rotation angle in degrees 

%Read sample image. 
I = imread('peppers.png'); 

%Add center cross, for verifying center is kept. 
I(center_y-1:center_y, center_x-30:center_x+30, :) = 255; 
I(center_y-30:center_y+30, center_x-1:center_x, :) = 255; 

%Rotate the entire input image, dimensions of J will be the same as I. 
J = imrotate(I, phi, 'bicubic', 'crop'); 

%Crop rectangular are with size width by height around center_x, center_y from J. 
C = J(center_y-height/2+1:center_y+height/2, center_x-width/2+1:center_x+width/2, :); 

%Display result: 
figure;imshow(C); 

結果：