Read the the pixel value and rebuild an image
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Hello there! I want to read the pixel value of an image and rebuild it later, but the performance of the new image is much different from the original one, although the shape is similar.
Could you help me solve this issue? Thank you very much.
Attached is my code.
clear;
clc;
[File, Path]=uigetfile('*.tif','Please select your files:', ...
'MultiSelect', 'on');
fid = imread(File);
[m,n,k]=size(fid);
Num_scan=1;
for x=1:m
for y=1:n
RGB=impixel(fid,x,y);
Pix(Num_scan,1)=x;
Pix(Num_scan,2)=y;
A_Pix0(Num_scan,1)=RGB(1);
A_Pix0(Num_scan,2)=RGB(2);
A_Pix0(Num_scan,3)=RGB(3);
Num_scan=Num_scan+1;
end
end
A_Pix=reshape(A_Pix0,[n,m,3]);
image(A_Pix)
Below are the original image and the new image.
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Image Analyst
2018-7-14
编辑:Image Analyst
2018-7-14
I don't know why you're opening a tiff format file and trying to read it and rebuilt it a pixel at a time. Simply open it as a tif image and make a copy:
[baseFileName, folder] = uigetfile('*.tif','Please select your files:', ... 'MultiSelect', 'off');
fullFileName = fullfile(folder, baseFileName);
image1 = imread(fullFileName);
A_pix = image1;
Also, fid is commonly used as a FileID, or the file handle returned by fopen(). You should choose a name like rgbImage instead of fid, because it's confusing, at least it confused me until I finally figured out what you were doing.
By the way you're making the common beginner mistake of confusing rows and columns with x and y. Arrays are indexed (row, column), NOT (x, y).
This kind of mistake arises from another common beginner mistake of choosing non-descriptive single letter variable names like m,n,x,y,i,j, etc. Notice how m is really rows but you're having x (a horizontal designation) go from 1 to rows instead of 1 to columns? Fix is below:
[rows, columns, numberOfColorChannels] = size(rgbImage)
for col = 1 : columns
for row = 1 : rows
3 个评论
Imran Riaz
2022-8-3
@Hao Shi Can u share the complete code to rebuild new image from he pixel values of old image, I also want to do same thing. I have to discard the some portion which is informationless.
DGM
2022-8-3
编辑:DGM
2022-8-3
If you want to do the same thing as what the original post does, then you're in luck, because it basically does nothing.
% this is the original code.
% this code is so slow that it will take over 15 minutes
% to process a 384x512 image
fid = imread('peppers.png');
fid = fid(1:10,1:10,:); % no sense wasting all that time to accomplish nothing
[m,n,k]=size(fid);
Num_scan=1;
for x=1:m
for y=1:n
RGB=impixel(fid,x,y);
Pix(Num_scan,1)=x;
Pix(Num_scan,2)=y;
% reshape the image into a 3-column matrix
A_Pix0(Num_scan,1)=RGB(1);
A_Pix0(Num_scan,2)=RGB(2);
A_Pix0(Num_scan,3)=RGB(3);
Num_scan=Num_scan+1;
end
end
A_Pix=reshape(A_Pix0,[n,m,3]); % the image is just reshaped back again
% this is essentially the same thing
inpict = imread('peppers.png');
inpict = inpict(1:10,1:10,:); % use the same image region
[m,n,~] = size(inpict);
[xx yy] = meshgrid(1:n,1:m);
allpixindices = [xx(:) yy(:)]; % a pile of indices for some reason
outpict = inpict; % the code accomplishes no change to the image
isequal(Pix,allpixindices) % the index arrays are identical
isequal(double(A_Pix),outpict) % the images are identical
I'm assuming most people don't need help writing an image processing script that accompishes exactly zero alteration of image data, so you must not be after the same thing that the OP wanted. To create a new image from an old image is perhaps one of the most vague descriptions of image processing that could be conceived. Nobody can give you bespoke code to do a specific task if the task cannot be described.
If your goal is to merely reshape the image into a list of pixel tuples, then just use reshape() and permute() as necessary to get whatever orientation you want.
pixtable = reshape(inpict,[],size(inpict,3))
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