Matlab图像旋转

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 29岁程序员，3月因学历无情被辞！ 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）]）; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 结果： 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 endendimgshow(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); % heightW=size(input_image,2); % widthth=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 endendimshow(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 这篇关于Matlab图像旋转的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！