非均匀插值

Question

我一直在阅读一个回合超分辨率图像重建主题，该领域的目的是从多个移位（子像素）低分辨率（LR）图像创建高分辨率（HR）图像。以下代码从一个HR图像创建4个LR图像。然后使用非unifrom插值在高分辨率网格上插入4个LR图像，以获得两侧大于LR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 此功能可创建4张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此函数通过线性插值使子像素移位。

interpolate.m 将4个LR图像插值到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;

我使用griddata函数进行插值（立方体），重建的图像太糟糕了，因为我认为'griddata`参数的值是错误的。如何纠正它们？

注意：当我更改此代码时

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

到

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

我得到了一个好的形象，但我不知道为什么！

修改 lena.bmp

Answer 1

在create_low中，您应用高分辨率坐标的移位。所以在插值中你也应该应用高分辨率坐标的移位。因此，完全有道理的是，你不应该将它们乘以因子 - 正如你已经发现的那样。