非均勻插值

Question

我一直在閱讀一個關於超分辨率圖像重建的話題，該領域的目標是從多個移位（亞像素）低分辨率（LR）圖像創建高分辨率（HR）圖像。以下代碼從一幅HR圖像創建4個LR圖像。然後使用非單向插值對高分辨率網格上的4個LR圖像進行插值，以獲得兩側大於LR的4個LR圖像的HR圖像。

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這個函數創建4個LR圖像。

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該函數通過線性插值使子像素偏移。

interpolate.m將4個LR圖像內插到HR網格上。

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函數插值（立方）和重建圖像是太糟糕了，因爲我認爲，「griddata`的參數值是錯誤的。如何糾正它們？

注：當我這個代碼

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

改變

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，但我不知道爲什麼！

編輯 lena.bmp

Answer 1

在create_low應用的變化在高分辨率座標。所以在內插中，您也應該應用高分辨率座標中的偏移。因此，您不應該將它們乘以因子，這是完全合理的，正如您已經發現的那樣。