1 
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“第一个索引超出数组”。我的代码有什么问题吗？

4

2 回答 2

1

这是一个有许多变化的工作版本。主要区别在于它在将坐标添加到旋转图像之前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 中的图像将 y 作为第一个维度，将 x 作为第二个维度，因此可以作为I(y, x)or访问I(row, column)。

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));
于 2020-06-07T10:26:50.943 回答
0

我不知道您是否一定要拥有自己的实现。但如果没有，你总是可以使用imrotate：

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

90=> 旋转度数

'bilinear'=> 双线性插值（替代方案：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。

于 2020-06-07T19:42:19.187 回答