数字图像处理--空间变换
上次讲了数字图像处理的一题,今天再贴一题
Geometric transform (test image: fig3.tif)
Develope geometric transform program that will rotate, translate, and scale an imageby specified amounts, using the nearest neighbor and bilinear interpolationmethods, respectively.
背景
在对图像进行空间变换的过程中,典型的情况是在对图像进行放大,旋转处理的时候,图像会出现失真的现象。这是由于在变换之后的图像中,存在着一些变换之前的图像中没有的像素位置。处理这一问题的方法被称为图像灰度级插值。常用的插值方式有三种:最近邻域插值、双线性插值、双三次插值。理论上来讲,最近邻域插值的效果最差,双三次插值的效果最好,双线性插值的效果介于两者之间。不过对于要求不是非常严格的图像插值而言,使用双线性插值通常就足够了。
最近领域算法Matlab代码
sourcePic=imread('fig3.tif'); %以下为了彩色图像 %[m,n,o]=size(sourcePic); %grayPic=rgb2gray(sourcePic); grayPic=sourcePic; [m,n]=size(grayPic); %比例系数为0.2-5.0 K = str2double(inputdlg('请输入比例系数(0.2 - 5.0)', '输入比例系数', 1, {'0.5'})); %验证范围 if (K < 0.2) && (K > 5.0) errordlg('比例系数不在0.2 - 5.0范围内', '错误'); error('请输入比例系数(0.2 - 5.0)'); end figure; imshow(grayPic); width = K * m; height = K * n; resultPic = uint8(zeros(width,height)); widthScale = m/width; heightScale = n/height; for x = 5:width - 5 for y = 5:height - 5 xx = x * widthScale; yy = y * heightScale; if (xx/double(uint16(xx)) == 1.0) && (yy/double(uint16(yy)) == 1.0) % if xx and yy is integer,then J(x,y) <- I(x,y) resultPic(x,y) = grayPic(int16(xx),int16(yy)); else % xx or yy is not integer a = double(round(xx)); % (a,b) is the base-dot b = double(round(yy)); resultPic(x,y) = grayPic(a,b); % calculate J(x,y) end end end figure; rotate(resultPic,-20); imshow(resultPic);
双线性插值Matlab算法
sourcePic=imread('fig3.tif'); %以下为了彩色图像 %[m,n,o]=size(sourcePic); %grayPic=rgb2gray(sourcePic); grayPic=sourcePic; [m,n]=size(grayPic); %比例系数为0.2-5.0 K = str2double(inputdlg('请输入比例系数(0.2 - 5.0)', '输入比例系数', 1, {'0.5'})); %验证范围 if (K < 0.2) or (K > 5.0) errordlg('比例系数不在0.2 - 5.0范围内', '错误'); error('请输入比例系数(0.2 - 5.0)'); end figure; imshow(grayPic); %输出图片长宽 width = K * m; height = K * n; resultPic = uint8(zeros(width,height)); widthScale = n/width; heightScale = m/height; for x = 5:width-5 for y = 5:height-5 xx = x * widthScale; yy = y * heightScale; if (xx/double(uint16(xx)) == 1.0) && (yy/double(uint16(yy)) == 1.0) % if xx and yy is integer,then J(x,y) <- I(x,y) resultPic(x,y) = grayPic(int16(xx),int16(yy)); else % xx or yy is not integer a = double(uint16(xx)); % (a,b) is the base-dot b = double(uint16(yy)); x11 = double(grayPic(a,b)); % x11 <- I(a,b) x12 = double(grayPic(a,b+1)); % x12 <- I(a,b+1) x21 = double(grayPic(a+1,b)); % x21 <- I(a+1,b) x22 = double(grayPic(a+1,b+1)); % x22 <- I(a+1,b+1) resultPic(x,y) = uint8( (b+1-yy) * ((xx-a)*x21 + (a+1-xx)*x11) + (yy-b) * ((xx-a)*x22 +(a+1-xx) * x12) ); end end end figure; resultPic = imrotate(resultPic,-20); imshow(resultPic);
效果如下
最近领域算法放大2倍并顺时针旋转20度
![](http://hi.csdn.net/attachment/201107/11/0_1310390414TSBe.gif)
双线性插值算法放大2倍并顺时针旋转20度
体会
该实验表明双线性插值得到的图像效果是比较好的。能够避免采用最近领域插值方式时可能存在的图像模糊、块状失真等问题。但双线性插值也存在问题,在放大倍数比较高的时候,图像失真将会比较严重,此时应该考虑使用更高阶的插值算法。