非均匀插值

Question

我一直在阅读一个关于超分辨率图像重建的话题，该领域的目标是从多个移位（亚像素）低分辨率（LR）图像创建高分辨率（HR）图像。以下代码从一幅HR图像创建4个LR图像。然后使用非单向插值对高分辨率网格上的4个LR图像进行插值，以获得两侧大于LR的4个LR图像的HR图像。

的main.m

im=double(imread('lena.bmp')); 

figure,imshow(uint8(im)),title('original HR image'); 
shifts=[ 0,   0; 
     4.1, 2.68; 
     -3.7, 7.8; 
     -1.1, -6.5]; 

factor=4; 

im1=create_low(im,shifts(1,1),shifts(1,2),factor); 
im2=create_low(im,shifts(2,1),shifts(2,2),factor); 
im3=create_low(im,shifts(3,1),shifts(3,2),factor); 
im4=create_low(im,shifts(4,1),shifts(4,2),factor); 

LR_images={im1,im2,im3,im4}; 

estimated_image = interpolate(LR_images,shifts,factor); 
figure,imshow(uint8(estimated_image)),title('reconstructed image'); 

create_low.m这个函数创建4个LR图像。

function [ low ] = create_low(im,x_shift,y_shift,factor) 

low = shift(im,x_shift,y_shift); 

low=downsample(low,factor); 
low=low'; 
low = downsample(low,factor); 
low=low'; 

end 

shift.m该函数通过线性插值使子像素偏移。

interpolate.m将4个LR图像内插到HR网格上。

function rec = interpolate(s,shifts,factor)         

n=length(s); 
ss = size(s{1}); 
if (length(ss)==2) ss=[ss 1]; end 

% compute the coordinates of the pixels from the N images. 
for k=1:ss(3) % for each color channel 
    for i=1:n % for each image 
    s_c{i}=s{i}(:,:,k); 
    s_c{i} = s_c{i}(:);  
    r{i} = [1:factor:factor*ss(1)]'*ones(1,ss(2)); % create matrix with row indices 
    c{i} = ones(ss(1),1)*[1:factor:factor*ss(2)]; % create matrix with column indices 
    r{i} = r{i}+factor*shifts(i,2);  %% the problem is here. 
    c{i} = c{i}+factor*shifts(i,1);  %% the problem is here. 
    rn{i} = r{i}((r{i}>0)&(r{i}<=factor*ss(1))&(c{i}>0)&(c{i}<=factor*ss(2))); 
    cn{i} = c{i}((r{i}>0)&(r{i}<=factor*ss(1))&(c{i}>0)&(c{i}<=factor*ss(2))); 
    sn{i} = s_c{i}((r{i}>0)&(r{i}<=factor*ss(1))&(c{i}>0)&(c{i}<=factor*ss(2))); 
end 

s_ = []; r_ = []; c_ = []; sr_ = []; rr_ = []; cr_ = []; 
for i=1:n % for each image 
    s_ = [s_; sn{i}]; 
    r_ = [r_; rn{i}]; 
    c_ = [c_; cn{i}]; 
end 
clear s_c r c coord rn cn sn 

% interpolate the high resolution pixels using cubic interpolation 
rec_col = griddata(c_,r_,s_,[1:ss(2)*factor],[1:ss(1)*factor]','cubic'); 
rec(:,:,k) = reshape(rec_col,ss(1)*factor,ss(2)*factor); 
end 
rec(isnan(rec))=0; 

我用griddata函数插值（立方）和重建图像是太糟糕了，因为我认为，“griddata`的参数值是错误的。如何纠正它们？

注：当我这个代码

r{i} = r{i}+factor*shifts(i,2);  %% the problem is here. 
c{i} = c{i}+factor*shifts(i,1);  %% the problem is here. 

改变

r{i} = r{i}-shifts(i,2);  %% the problem is here. 
c{i} = c{i}-shifts(i,1);  %% the problem is here. 

我得到了良好的形象，但我不知道为什么！

编辑 lena.bmp

Answer 1

在create_low应用的变化在高分辨率坐标。所以在内插中，您也应该应用高分辨率坐标中的偏移。因此，您不应该将它们乘以因子，这是完全合理的，正如您已经发现的那样。