Matlab图像旋转 [英] Matlab image rotation
本文介绍了Matlab图像旋转的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我是图像处理的新手，我已经实现了图像变形的代码，它完美无缺。我想通过使用线性插值来旋转图像而不使用内置函数（interp）来改进代码。这是我的代码：
全部关闭;
清除所有;
img ='woods.jpg';
input_image = double（imread（img））./ 255;
H = size（input_image，1）;
W = size（input_image，2）;
th = pi / 4;
s0 = 2;
s1 = 2;
x0 = -W / 2;
x1 = -H / 2;
T = [1 0 x0; ...
0 1 x1; ......
0 0 1];
RST = [（s0 * cos（th））（ - s1 * sin（th））（（s0 * x0 * cos（th）） - （s1 * x1 * sin（th）） ）; ...
（s0 * sin（th））（s1 * cos（th））（（s0 * x0 * sin（th））+（s1 * x1 * cos（th）））; ......
0 0 1];
M = inv（T）* R;
N = inv（M）;
output_image =零（H，W，3）; i = 1的
：j = 1的W
：H
x = [i; j; 1];
y = N * x;
a = y（1）/ y（3）;
b = y（2）/ y（3）;
a = round（a）;
b = round（b）;
if（a> 0& a< = W&& b> 0&& b< = H）
output_image（j，i，：）= input_image（b，A，:);
结束
结束
结束
imgshow（output_image）;
解决方案
检查以下解决方案：
我通过比较函数中的Matalb构建验证了实现 [imwarp] [1] 。
全部关闭;
清除所有;
img ='peppers.png';
input_image = double（imread（img））./ 255;
H = size（input_image，1）; ％height
W = size（input_image，2）; ％width
th = pi / 4;
s0 = 2;
s1 = 2;
x0 = -W / 2;
x1 = -H / 2;
T = [1 0 x0; ...
0 1 x1; ......
0 0 1];
RST = [（s0 * cos（th））（ - s1 * sin（th））（（s0 * x0 * cos（th）） - （s1 * x1 * sin（th）） ）; ...
（s0 * sin（th））（s1 * cos（th））（（s0 * x0 * sin（th））+（s1 * x1 * cos（th）））; ......
0 0 1];
M = inv（T）* RST;
N = inv（M）;
output_image =零（H，W，3）; i = 1的
：j = 1的W
：H
x = [i; j; 1];
y = N * x;
a = y（1）/ y（3）;
b = y（2）/ y（3）;
％最近邻居
％a = round（a）;
％b = round（b）;
x1 = floor（a）;
y1 = floor（b）;
x2 = x1 + 1;
y2 = y1 + 1;
％双线性插值ilsutration：
％图像坐标样式（水平索引优先）
％
％（x1，y1）| （x2，y1）
％| 1-dy
％1-dx | dx
％------（a，b）------------
％|
％|
％|
％| dy
％|
％|
％（x1，y2）| （x2，y2）
if（（x1> = 1）&&（y1> = 1）&&（x2< = W）&&（y2 < = H））
％加载2x2像素
i11 = input_image（y1，x1，:); ％左上角像素
i21 = input_image（y2，x1，:); ％左下角像素
i12 = input_image（y1，x2，:); ％右上角像素
i22 = input_image（y2，x2，:); ％右下角像素
％插值wieghts
dx = x2 - a;
dy = y2 - b;
％Bi-lienar interpolation
output_image（j，i，:) = i11 * dx * dy + i21 * dx *（1-dy）+ i12 *（1-dx）* dy + i22 *（1-dx）*（1-dy）;
结束
结束
结束
imshow（output_image）;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% b $ b tform = affine2d（M'）;
ref_image = imwarp（input_image，tform，'OutputView'，imref2d（size（input_image）），'Interp'，'linear'）;
figure; imshow（ref_image）
figure; imshow（output_image - ref_image）
max_diff = max（abs（output_image（:) - ref_image（:)））;
disp（['最大差异来自imwarp ='，num2str（max_diff）]）;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%<% b
结果：
< hr>
备注：
我错过了以下使用双线性插值调整图像大小的execelnt帖子（更好地解释了双线性插值）：
使用双线性插值调整图像大小而不会影响
I am new to image processing, I have implemented a code for image warping and it works perfectly. I would like to improve the code by using linear interpolation to rotate the image WITHOUT using the built-in function (interp). Here is my code:
close all;
clear all;
img = 'woods.jpg';
input_image =double(imread(img))./255;
H=size(input_image,1);
W=size(input_image,2);
th=pi/4;
s0 = 2;
s1 = 2;
x0 = -W/2;
x1 = -H/2;
T=[1 0 x0 ; ...
0 1 x1 ; ...
0 0 1];
RST = [ (s0*cos(th)) (-s1*sin(th)) ((s0*x0*cos(th))-(s1*x1*sin(th))); ...
(s0*sin(th)) (s1*cos(th)) ((s0*x0*sin(th))+(s1*x1*cos(th))); ...
0 0 1];
M=inv(T)*R;
N = inv(M);
output_image=zeros(H,W,3);
for i=1:W
for j=1:H
x = [i ; j ; 1];
y = N * x;
a = y(1)/y(3);
b = y(2)/y(3);
a = round(a);
b = round(b);
if (a>0 && a<=W && b>0 && b<=H)
output_image(j,i,:)=input_image(b,a,:);
end
end
end
imgshow(output_image);
解决方案 Check the following solution:
I verified implementation by comparing to Matalb build in function [imwarp][1].
close all;
clear all;
img = 'peppers.png';
input_image =double(imread(img))./255;
H=size(input_image,1); % height
W=size(input_image,2); % width
th=pi/4;
s0 = 2;
s1 = 2;
x0 = -W/2;
x1 = -H/2;
T=[1 0 x0 ; ...
0 1 x1 ; ...
0 0 1];
RST = [ (s0*cos(th)) (-s1*sin(th)) ((s0*x0*cos(th))-(s1*x1*sin(th))); ...
(s0*sin(th)) (s1*cos(th)) ((s0*x0*sin(th))+(s1*x1*cos(th))); ...
0 0 1];
M=inv(T)*RST;
N = inv(M);
output_image=zeros(H,W,3);
for i=1:W
for j=1:H
x = [i ; j ; 1];
y = N * x;
a = y(1)/y(3);
b = y(2)/y(3);
%Nearest neighbor
%a = round(a);
%b = round(b);
x1 = floor(a);
y1 = floor(b);
x2 = x1 + 1;
y2 = y1 + 1;
%Bi-linear interpolation ilsutration:
%Image coordinates style (horizontal index first)
%
%(x1,y1) | (x2,y1)
% | 1-dy
% 1-dx | dx
% ------(a,b)------------
% |
% |
% |
% | dy
% |
% |
%(x1,y2) | (x2,y2)
if ((x1 >= 1) && (y1 >= 1) && (x2 <= W) && (y2 <= H))
%Load 2x2 pixels
i11 = input_image(y1, x1, :); %Top left pixel
i21 = input_image(y2, x1, :); %Bottom left pixel
i12 = input_image(y1, x2, :); %Top right pixel
i22 = input_image(y2, x2, :); %Bottom right pixel
%Interpolation wieghts
dx = x2 - a;
dy = y2 - b;
%Bi-lienar interpolation
output_image(j, i, :) = i11*dx*dy + i21*dx*(1-dy) + i12*(1-dx)*dy + i22*(1-dx)*(1-dy);
end
end
end
imshow(output_image);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Verify implementation by comparing with Matalb build in function imwarp:
tform = affine2d(M');
ref_image = imwarp(input_image, tform, 'OutputView', imref2d(size(input_image)), 'Interp', 'linear');
figure;imshow(ref_image)
figure;imshow(output_image - ref_image)
max_diff = max(abs(output_image(:) - ref_image(:)));
disp(['Maximum difference from imwarp = ', num2str(max_diff)]);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Result:
Remark:
I missed the following execelnt post that resizes an image using bilinear interpolation (bilinear interpolation is explained better):
Resize an image with bilinear interpolation without imresize
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