>> clear; %此题是用双线性插值法实现图像缩放
I=imread('pepper.bmp'); %读入原图像
[rows,cols]=size(I);
K1 = str2double(inputdlg('请输入行缩放倍数', 'INPUT scale factor', 1, {'0.5'}));%行默认变为原来的0.5倍
K2 = str2double(inputdlg('请输入列缩放倍数', 'INPUT scale factor', 1, {'0.4'}));%列默认变为原来的0.4倍
width = K1 * rows;
height = K2 * cols;
Out = uint8(zeros(width,height)); %创建输出图像矩阵
widthScale = rows/width;
heightScale = cols/height;
for x = 6:width - 6 % 6是为了防止矩阵超出边界溢出
for y = 6:height - 6
oldX = x * widthScale; % oldX,oldY为原坐标,x,y为新坐标
oldY = y * heightScale;
if (oldX/double(uint16(oldX)) == 1.0) & (oldY/double(uint16(oldY)) == 1.0)
Out(x,y) = I(int16(oldX),int16(oldY));%若oldX,oldY为整数,直接赋值
else
a = double(uint16(oldX));
b = double(uint16(oldY));
x11 = double(I(a,b)); % x11 赋值为 I(a,b)
x12 = double(I(a,b+1)); % x12 赋值为 I(a,b+1)
x21 = double(I(a+1,b)); % x21 赋值为 I(a+1,b)
x22 = double(I(a+1,b+1)); % x22 赋值为 I(a+1,b+1)
Out(x,y) = uint8( (b+1-oldY) * ((oldX-a)*x21 + (a+1-oldX)*x11) + (oldY-b) * ((oldX-a)*x22 +(a+1-oldX) * x12) ); % 用双线性插值计算公式计算
end
end
end
figure(1);
imshow(I);
figure(2);
imshow(Out);
shuangxianxing.rar_matlab课程设计
版权申诉
183 浏览量
2022-09-14
19:54:48
上传
评论
收藏 859B RAR 举报
![avatar](https://profile-avatar.csdnimg.cn/9116002996824fde940a716bee54aca8_weixin_42663213.jpg!1)
钱亚锋
- 粉丝: 90
- 资源: 1万+