my question is this output is correct or not

4 次查看（过去 30 天）

Aarsha mv 2017-5-2

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/338316-my-question-is-this-output-is-correct-or-not

评论： Walter Roberson 2017-5-4

>>first image is laplaced image and second is that image is reconstruted with laplacian pyramid.this output is correct or not

6 个评论
显示 4更早的评论隐藏 4更早的评论

Aarsha mv 2017-5-4

编辑：Walter Roberson 2017-5-4

在 MATLAB Online 中打开

my code is given below.please say this code is correct or not .if any mistake are in there help to correct that code.

function [ img ] = pyr_Reconstruct( d )
%PYRRECONSTRUCT Uses a Laplacian pyramid to reconstruct a image
%   IMG = PYRRECONSTRUCT(PYR) PYR should be a 1*level cell array containing
%   the pyramid, SIZE(PYR{i}) = SIZE(PYR{i-1})*2-1
for p = length(d)-1:-1:1
  d(p) = d(p)+pyr_expand(d(p+1));
end
img = d(1);
end
function [ imgout ] = pyr_expand( img )
%PYR_EXPAND  Image pyramid expansion
%   B = PYR_EXPAND( A )  If A is M-by-N, then the size of B 
%  is (2*M-1)-by-(2*N-1). Support gray or rgb image.
%  B will be transformed to double class.
%  Results the same w/ MATLAB func impyramid.
% Yan Ke @ THUEE, xjed09@gmail.com
kw =5; % default kernel width
cw = .375; % kernel centre weight, same as MATLAB func impyramid. 0.6 in the Paper
ker1d = [.25-cw/2 .25 cw .25 .25-cw/2];
kernel = kron(ker1d,ker1d')*4;
% expand [a] to [A00 A01;A10 A11] with 4 kernels
ker00 = kernel(1:2:kw,1:2:kw); % 3*3
ker01 = kernel(1:2:kw,2:2:kw); % 3*2
ker10 = kernel(2:2:kw,1:2:kw); % 2*3
ker11 = kernel(2:2:kw,2:2:kw); % 2*2
img = im2double(img);
sz = size(img(:,:,1));
osz = sz*2-1;
imgout = zeros(osz(1),osz(2),size(img,3));
for p = 1:size(img,3)
  img1 = img(:,:,p);
  img1ph = padarray(img1,[0 1],'replicate','both'); % horizontally padded
  img1pv = padarray(img1,[1 0],'replicate','both'); % horizontally padded
  
  img00 = imfilter(img1,ker00,'replicate','same');
  img01 = conv2(img1pv,ker01,'valid'); % imfilter doesn't support 'valid'
  img10 = conv2(img1ph,ker10,'valid');
  img11 = conv2(img1,ker11,'valid');
  
  imgout(1:2:osz(1),1:2:osz(2),p) = img00;
  imgout(2:2:osz(1),1:2:osz(2),p) = img10;
  imgout(1:2:osz(1),2:2:osz(2),p) = img01;
  imgout(2:2:osz(1),2:2:osz(2),p) = img11;
end
end

Walter Roberson 2017-5-4

You are only passing in a scalar to the reconstruction, and only calculating a scalar as the result. The final output is going to be a single pixel, not even color.

Remember, when p is a scalar index, then d(p) is going to be referring to a scalar location.

请先登录，再进行评论。

请先登录，再回答此问题。