上次讲了数字图像处理的一题，今天再贴一题html
Geometric transform (test image: fig3.tif)算法
Develope geometric transform program that will rotate, translate, and scale an imageby specified amounts, using the nearest neighbor and bilinear interpolationmethods, respectively.post
背景ui
在对图像进行空间变换的过程当中，典型的状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。这是因为在变换以后的图像中，存在着一些变换以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。经常使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。理论上来说，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二者之间。不过对于要求不是很是严格的图像插值而言，使用双线性插值一般就足够了。rest
最近领域算法Matlab代码code
sourcePic=imread('fig3.tif');
%如下为了彩色图像
%[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orm
sourcePic=imread('fig3.tif');
%如下为了彩色图像
%[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);
效果以下
最近领域算法放大2倍并顺时针旋转20度htm
双线性插值算法放大2倍并顺时针旋转20度blog
体会图片
该实验代表双线性插值获得的图像效果是比较好的。可以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。