i want to know how to find the size of original image & compressed image..

Question

azizullah khan 2013-12-7

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回答： Daniel Glez 2016-11-17

I want to find the size of compressed and original images size after applying FFT or DCT or PCA.Kindly if anyone have idea about it please share it with me.Thanks in advance.

p=imread('picture.jpg');
a=imresize(p,[100,100]);
%DCT
Z(:,:,1)=dct2(a(:,:,1));
Z(:,:,2)=dct2(a(:,:,2));
Z(:,:,3)=dct2(a(:,:,3));
for i=1:100
      for j=1:100
          if((i+j)>60)
               Z(i,j,1)=0;
               Z(i,j,2)=0;
               Z(i,j,3)=0;
           end
      end
  end
  K(:,:,1)=idct2(Z(:,:,1));
K(:,:,2)=idct2(Z(:,:,2));
K(:,:,3)=idct2(Z(:,:,3));
subplot(2,4,1);
%imshow(Z);
imshow(uint8(K));
title('70% compression');

How can we find the size of the '70%' compressed image i.e K ?

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Walter Roberson 2013-12-7

duplicates http://www.mathworks.co.uk/matlabcentral/answers/105073-i-want-to-know-how-to-find-the-size-of-original-image-compressed-image

azizullah khan 2013-12-7

在 MATLAB Online 中打开

p=imread('picture.jpg');

a=imresize(p,[100,100]);

%DCT

Z(:,:,1)=dct2(a(:,:,1));

Z(:,:,2)=dct2(a(:,:,2));

Z(:,:,3)=dct2(a(:,:,3));

for i=1:100
    for j=1:100
        if((i+j)>60)
             Z(i,j,1)=0;
             Z(i,j,2)=0;
             Z(i,j,3)=0;
         end
    end
end
K(:,:,1)=idct2(Z(:,:,1));
K(:,:,2)=idct2(Z(:,:,2));
K(:,:,3)=idct2(Z(:,:,3));
subplot(2,4,1);
%imshow(Z);
imshow(uint8(K));
title('70% compression');
How can we find the size of the '70%' compressed image i.e K ?

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Answer 1

Image Analyst 2013-12-7

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Look at the file size in your operating system. For example, use Windows Explorer. Or if you want to do it all in MATLAB, use imfinfo(), if you have the Image Processing Toolbox.

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Image Analyst 2013-12-10

在 MATLAB Online 中打开

Try this:

info = imfinfo('peppers.png')
bytes = info.Width * info.Height * info.BitDepth
compressionRatio = bytes / info.FileSize
message = sprintf('The size is %d rows by %d columns by %d color channels = %d bytes.\n',...
  info.Width, info.Height, info.BitDepth/3, bytes);
message = sprintf('%sThe size on disk is %d bytes.\n', message, info.FileSize);
message = sprintf('%sSo the file is %.3f%%as big as it could be\n(compressed by a factor of %.3f)\n', ...
  message, 100/compressionRatio, compressionRatio)
uiwait(helpdlg(message));

In command window:

info = 
                  Filename: 'C:\Program Files\MATLAB\R2013a\toolbox\images\imdemos\peppers.png'
               FileModDate: '16-Dec-2002 06:10:58'
                  FileSize: 287677
                    Format: 'png'
             FormatVersion: []
                     Width: 512
                    Height: 384
                  BitDepth: 24
                 ColorType: 'truecolor'
           FormatSignature: [137 80 78 71 13 10 26 10]
                  Colormap: []
                 Histogram: []
             InterlaceType: 'none'
              Transparency: 'none'
    SimpleTransparencyData: []
           BackgroundColor: []
           RenderingIntent: []
            Chromaticities: []
                     Gamma: []
               XResolution: []
               YResolution: []
            ResolutionUnit: []
                   XOffset: []
                   YOffset: []
                OffsetUnit: []
           SignificantBits: []
              ImageModTime: '16 Jul 2002 16:46:41 +0000'
                     Title: []
                    Author: []
               Description: 'Zesty peppers'
                 Copyright: 'Copyright The MathWorks, Inc.'
              CreationTime: []
                  Software: []
                Disclaimer: []
                   Warning: []
                    Source: []
                   Comment: []
                 OtherText: []
bytes =
     4718592
compressionRatio =
          16.4023957424473
message =
The size is 512 rows by 384 columns by 8 color channels = 4718592 bytes.
The size on disk is 287677 bytes.
So the file is 6.097% as big as it could be
(compressed by a factor of 16.402)

azizullah khan 2014-2-1

Dear sir,in the above code i have compressed the image by taking the higher component to zero.and then take the inverse of that compressed image...then how you can say that images is not compressed.... please if you have code on compressing image using dct/fft/dft please share it with me....please...... simply i want compressed image using fft/dct/dft.I need code.... please help me...please....................

Image Analyst 2014-2-1

When you compress it, it's compressed. If you inverse/reverse/restore/undo the compression, you recover your original image and it won't be compressed anymore. Don't you agree? I don't have any compression demos using fft or dct available to give you.

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Answer 2

chitresh 2013-12-7

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original file size = (width* height*bit depth)/8;

and you can get compressed file size from imfinfo('imagefile'); and rest parameters to like width , height and bitdepth

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chitresh 2013-12-15

dear if your problem is solved then please accept the answer

azizullah khan 2014-1-29

编辑：azizullah khan 2014-1-29

dear bro... its not clear..kindly find the size of compressed images of the above.... please take a example of image and solve my problem i shall be very thankful to you.........best luck

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Answer 3

Daniel Glez 2016-11-17

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/108999-i-want-to-know-how-to-find-the-size-of-original-image-compressed-image#answer_243672

在 MATLAB Online 中打开

Hello! I would like to add a note here:

Be careful and check that you haven't registered into your workspace a variable named "sum". In that case this error also appears. It happened to me while computing MSE and PSNR in a matrix substraction like follows:

mse = sum(sum((I-I64).^2))/(m*n)
PSNR = 10*log(max(I(:))/sqrt(mse))

OUT:

Subscript indices must either be real positive
integers or logicals.
Error in mse_psnr (line 11)
mse = sum(sum((I-I64).^2))/(m*n)

Thanks to all.

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