不均匀插值

Question

我一直在阅读有关超分辨率图像重建的主题，这个领域的目标是从多个移位（亚像素）低分辨率（LR）图像创建一个高分辨率（HR）图像。以下代码从一个HR图像创建4个LR图像。然后使用非均匀插值将4个LR图像插值到高分辨率网格上，以获取比LR图像两侧大4倍的HR图像。

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 这个函数用来创建4张低分辨率图像。

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这个函数通过线性插值来进行亚像素位移。

interpolate.m将4个低分辨率图像插值到高分辨率网格上。

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函数进行插值(三次样条)，但是重建的图像很差，我认为是因为'griddata`参数的值错误。如何更正？

注意：当我更改这段代码时

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

to

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

我得到了一张好的图片，但不知道为什么！

编辑lena.bmp

- HforHisham

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你能发布一下你正在使用的 lena.bmp 吗？ - chappjc

Answer 1

在 create_low 中，您应用高分辨率坐标中的移位。因此，在 interpolate 中，您也应该应用高分辨率坐标中的移位。因此，不将它们乘以因子完全是有道理的，正如您已经发现的那样。

- Klamer Schutte