Why image doesnt show correctly while reading a binary file?

im trying to take an image saved in a binary formate and change it to a 2D matrix and print it as its orginal image but what ig get is a gray image with black strips , and i check that 2D array is holding atctual pixel numbers of the image.

here is the code for reading binary file:

function [M, M1]=read_raw(filename1, filename2)
%Storing size of image 
% if filename1(1)=='c'
%     row =352; col=288;
%     row1 = 352; col1=288;
% else
%     row = 720; col=480; 
%     row1 = 720; col1=480;
% end
row=160;
col=90;
%Reading the two files inputted in raw format in a 2D matrix
%For First File
if true  
X = fopen(filename1, 'r');
I = fread(X, row*col, 'uint8=>uint8');
Z = reshape(I, row, col);
Z = Z';
Z
figure(1);
subplot(211);
imshow(Z);
title(filename1);
M = Z;
end

%For Second File
if true
X1 = fopen(filename2, 'r');
I1 = fread(X1, row*col, 'uint8=>uint8');
Z1 = reshape(I1, row, col);
Z1 = Z1';
figure(1);
subplot(212);
imshow(Z1);
title(filename2);
M1 = Z1;
end
end

output:

[yanqi liu]

yes，sir，may be upload some file to debug

[Walter Roberson]

Especially if you can also show us an idea of what the intended image would look like.

[fady sameh]

It just a 160x90 image capture from laptop camera

[Walter Roberson]

Without the bin file and without the original image to guide us, there is not a lot we can do.

You should zip the .bin files and attach the .zip

[fady sameh]

I hope that helps

Answer 1

[DGM]

The files are just literal text files with a .bin extension. The easy way to deal with that is just rename the file. If you want to avoid renaming the files, check out Stephen's comment below.

row = 160;
col = 90;
filename1 = 'Rsaved1.txt';
filename2 = 'Rsaved2.txt';

I = readmatrix(filename1);
Z1 = reshape(uint8(I), row, col).';
I = readmatrix(filename2);
Z2 = reshape(uint8(I), row, col).';

imshow(Z1)

figure
imshow(Z2)

The clue was that the file contained way more bytes than would be necessary for a uint8 image of that geometry. A check with a hex editor helps a lot in figuring out what exactly a .bin file is.

The file represents the uint8 pixel values in decimal, but since there is no fractional part, the file is filled with a bunch of extraneous zeros.

EDIT

See comment below

row = 160;
col = 30;
filename1 = 'Rsaved1.txt';
filename2 = 'Rsaved2.txt';

I = readmatrix(filename1);
Z1 = permute(reshape(uint8(I), row, col, 3),[2 1 3]);
I = readmatrix(filename2);
Z2 = permute(reshape(uint8(I), row, col, 3),[2 1 3]);

figure
imshow(Z1)

figure
imshow(Z2)

[Stephen23]

Or perhaps tell READMATRIX that they are text files, without renaming:

I = readmatrix(.., 'FileType','text');

But replacing a misleading file extension is probably a good idea.

[DGM]

Oh, I didn't realize that was an option!

I would think that if there were any desire to keep the images in any form, it would be as a proper raster image file instead of an inflated text representation.

If the raw files are to be processed and discarded anyway, then your suggestion would probably simplify the task quite elegantly.

[fady sameh]

Why is the image in gray and not in its original colors?

[Image Analyst]

What are the colors? What is the format of the text file? Is it BGRBGRBGR etc. or something like that?

[fady sameh]

That how i turn the image into a text file

fileID=fopen(strcat("C:\\Users\\fadys\\Desktop\\Digital audio\\week9\\compressed\\Rsaved",int2str(index),".txt"),'w');
for r=1:size(Rframe,1)
    for c=1:size(Rframe,2)
        fprintf(fileID,"%f\n",Rframe(r,c));
    end
    fprintf(fileID,"\n");
end
fclose(fileID);

[Image Analyst]

@fady sameh, did you seem my answer below where I get a color image?

[DGM]

It's grayscale because that's what you indicated it would be with the geometry. The code you posted indicating how the file was generated dictates that the file has no color information. You only wrote the red channel to the file.

Rframe = ones(4,4,3).*permute([1 2 3],[1 3 2])

Rframe = 
Rframe(:,:,1) =

     1     1     1     1
     1     1     1     1
     1     1     1     1
     1     1     1     1


Rframe(:,:,2) =

     2     2     2     2
     2     2     2     2
     2     2     2     2
     2     2     2     2


Rframe(:,:,3) =

     3     3     3     3
     3     3     3     3
     3     3     3     3
     3     3     3     3

numel(Rframe)

ans = 48

fileID=fopen('test.txt','w');
for r=1:size(Rframe,1) % rows
    for c=1:size(Rframe,2) % columns
        fprintf(fileID,"%f\n",Rframe(r,c)); % only the first channel
    end
    fprintf(fileID,"\n");
end
fclose(fileID);

A = readmatrix('test.txt');
numel(A)

ans = 16

[min(A) max(A)]

ans = 1×2

1 1

Dim 3 of the array is never addressed. You'd probably have a third loop.

If it were pagewise RGB, then you could use the edit above. It's as simple as changing the specified geometry of the output array. The call to permute() is the same as elementwise transposition, but the .' operator only works on 2D arrays.

That said, I would question why any of this is necessary. Why convert an image to a text file and then back? Why write integers using %f instead of %d? Is this to feed to another program that takes formatted text images? I'd have to ask, because if that's so, then that would dictate how the color information needs to be arranged in the file. It may typically be pixelwise (RGBRGBRGB) or pagewise (RRRGGGBBB), though color order may vary.

Answer 2

[yanqi liu]

close all;
clear all;  clc;
a = load('Rsaved1.bin');
a = reshape(a, 160, 90);
b = load('Rsaved2.bin');
b = reshape(b, 160, 90);
figure; imshow(a', [])
figure; imshow(b', [])

Answer 3

[Image Analyst]

Try this:

col = 30;
fileName = 'Rsaved1.txt';
I = readmatrix(fileName);
% r = I(1:3:end);
% g = I(2:3:end);
% b = I(3:3:end);
% Assume all red first, then all the green, and finally all the blue.
r = I(1:4800);
g = I(4801:2*4800);
b = I(2*4800+1:end)
r = reshape(uint8(r), [], col).';
g = reshape(uint8(g), [], col).';
b = reshape(uint8(b), [], col).';
rgbImage = cat(3, r, g, b);
imshow(rgbImage, 'InitialMagnification', 300);
axis('on', 'image')