使用双线性插值在 matlab 中的图像旋转

Question

clear
I = imread('256.jpg');
%imshow(I);
center = 128;
[x, y] = size(I); % declare image size array
Original = [x, y];
Rotated_I = zeros(x,y); %declare size of array to store pixel
theta = 90;
for row = 1:y
for column = 1:x
x_original = (column - 128) * cos(theta) - (row - 128)*sin(theta);
y_original = (column - 128) * sin(theta) + (row - 128)*cos(theta); % reverse rotate
p = floor(x_original);
q = floor(y_original);
a = y_original - p;
b = x_original - q; %
Rotated_I(column, row) = (1-a)*((1-b)*Original(p,q)+b*Original(p,q+1))+a*((q-b)*Original(p+1,q)+b*Original(p+1,q+1)); % Find pixel using bilinear interpolation
end
end
imshow(Rotated_I);  

我尝试使用反向旋转和双线性插值来旋转图像，但只有我看到的是错误消息。它说"第一个索引超过数组"。我的代码有什么问题吗？

Answer 1

这是一个包含许多更改的工作版本。主要区别在于，在将坐标添加到旋转图像之前，它会检查原始图像中是否存在坐标。这允许任意旋转，例如 45 度。此外，MATLAB 中的图像以 y 作为第一维，x 作为第二维，因此可以作为I(y, x)或I(row, column)访问。

clear
I = imread('256.jpg');
% imshow(I);
center = 128;
[y, x] = size(I); % in MATLAB, images are y-by-x in size (ie. y is dimension 1)
Original = I; % Original needs to be the image I
Rotated_I = zeros(y, x);
theta = 90;
for row = 1:y
for column = 1:x
x_original = (column - center) * cosd(theta) - (row - center)*sind(theta) + center; % theta is in degrees so use cosd and sind
y_original = (column - center) * sind(theta) + (row - center)*cosd(theta) + center; % also add center back on
p = floor(y_original); % x_original and y_original were swapped here
q = floor(x_original); % x_original and y_original were swapped here
a = y_original - p; 
b = x_original - q;
% check if the coordinate is in the original image to prevent errors
if p > 0 && p <= y && q > 0 && q <= x
Rotated_I(row, column) = Rotated_I(row, column) + (1-a)*(1-b)*Original(p,q);
end
if p > 0 && p <= y && q+1 > 0 && q+1 <= x
Rotated_I(row, column) = Rotated_I(row, column) + (1-a)*b*Original(p,q+1);
end
if p+1 > 0 && p+1 <= y && q > 0 && q <= x
Rotated_I(row, column) = Rotated_I(row, column) + a*(1-b)*Original(p+1,q);
end
if p+1 > 0 && p+1 <= y && q+1 > 0 && q+1 <= x
Rotated_I(row, column) = Rotated_I(row, column) + a*b*Original(p+1,q+1);
end
end
end
% convert to uint image so it displays properly (double expects values from 0 to 1)
imshow(uint8(Rotated_I));  

Answer 2

我不知道你是否一定要有自己的实现。但如果没有，您可以随时使用imrotate：

Rotated_I = imrotate(I, 90, 'bilinear', 'crop');

90=> 旋转角度

'bilinear'=> 双线性插值(替代项：nearest、bicubic(

'crop'=> 保持旋转图像的像素大小与输入图像相同

imrotate是图像处理工具箱的一部分。