JPEG compression algorithm implementation in MATLAB

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S.
S. 2014-8-23
回答： P.Anushri P.Anushri 2023-9-26
I'm working on an implementation of the JPEG compression algorithm in MATLAB. I've run into some issues when computing the discrete cosine transform(DCT) of the 8x8 image blocks(T = H * F * H_transposed, H is the matrix containing the DCT coefficients of an 8x8 matrix, generated with dctmtx(8) and F is an 8x8 image block). The code is bellow:
*jpegCompress.m*
function y = jpegCompress(x, quality)
% y = jpegCompress(x, quality) compresses an image X based on 8 x 8 DCT
% transforms, coefficient quantization and Huffman symbol coding. Input
% quality determines the amount of information that is lost and compression achieved. y is the encoding structure containing fields:
% y.size size of x
% y.numblocks number of 8 x 8 encoded blocks
% y.quality quality factor as percent
% y.huffman Huffman coding structure
narginchk(1, 2); % check number of input arguments
if ~ismatrix(x) || ~isreal(x) || ~ isnumeric(x) || ~ isa(x, 'uint8')
error('The input must be a uint8 image.');
end
if nargin < 2
quality = 1; % default value for quality
end
if quality <= 0
error('Input parameter QUALITY must be greater than zero.');
end
m = [16 11 10 16 24 40 51 61 % default JPEG normalizing array
12 12 14 19 26 58 60 55 % and zig-zag reordering pattern
14 13 16 24 40 57 69 56
14 17 22 29 51 87 80 62
18 22 37 56 68 109 103 77
24 35 55 64 81 104 113 92
49 64 78 87 103 121 120 101
72 92 95 98 112 100 103 99] * quality;
order = [1 9 2 3 10 17 25 18 11 4 5 12 19 26 33 ...
41 34 27 20 13 6 7 14 21 28 35 42 49 57 50 ...
43 36 29 22 15 8 16 23 30 37 44 51 58 59 52 ...
45 38 31 24 32 39 46 53 60 61 54 47 40 48 55 ...
62 63 56 64];
[xm, xn] = size(x); % retrieve size of input image
x = double(x) - 128; % level shift input
t = dctmtx(8); % compute 8 x 8 DCT matrix
% Compute DCTs pf 8 x 8 blocks and quantize coefficients
y = blkproc(x, [8 8], 'P1 * x * P2', t, t');
y = blkproc(y, [8 8], 'round(x ./ P1)', m); % <== nearly all elements from y are zero after this step
y = im2col(y, [8 8], 'distinct'); % break 8 x 8 blocks into columns
xb = size(y, 2); % get number of blocks
y = y(order, :); % reorder column elements
eob = max(x(:)) + 1; % create end-of-block symbol
r = zeros(numel(y) + size(y, 2), 1);
count = 0;
for j = 1:xb % process one block(one column) at a time
i = find(y(:, j), 1, 'last'); % find last non-zero element
if isempty(i) % check if there are no non-zero values
i = 0;
end
p = count + 1;
q = p + i;
r(p:q) = [y(1:i, j); eob]; % truncate trailing zeros, add eob
count = count + i + 1; % and add to output vector
end
r((count + 1):end) = []; % delete unused portion of r
y = struct;
y.size = uint16([xm xn]);
y.numblocks = uint16(xb);
y.quality = uint16(quality * 100);
y.huffman = mat2huff(r);
mat2huff is implemented as:
*mat2huff.m*
function y = mat2huff(x)
%MAT2HUFF Huffman encodes a matrix.
% Y = mat2huff(X) Huffman encodes matrix X using symbol
% probabilities in unit-width histogram bins between X's minimum
% and maximum value s. The encoded data is returned as a structure
% Y :
% Y.code the Huffman - encoded values of X, stored in
% a uint16 vector. The other fields of Y contain
% additional decoding information , including :
% Y.min the minimum value of X plus 32768
% Y.size the size of X
% Y.hist the histogram of X
%
% If X is logical, uintB, uint16 ,uint32 ,intB ,int16, or double,
% with integer values, it can be input directly to MAT2HUF F. The
% minimum value of X must be representable as an int16.
%
% If X is double with non - integer values --- for example, an image
% with values between O and 1 --- first scale X to an appropriate
% integer range before the call.For example, use Y
% MAT2HUFF (255 * X) for 256 gray level encoding.
%
% NOTE : The number of Huffman code words is round(max(X(:)))
% round (min(X(:)))+1. You may need to scale input X to generate
% codes of reasonable length. The maximum row or column dimension
% of X is 65535.
if ~ismatrix(x) || ~isreal(x) || (~isnumeric(x) && ~islogical(x))
error('X must be a 2-D real numeric or logical matrix.');
end
% Store the size of input x.
y.size = uint32(size(x));
% Find the range of x values
% by +32768 as a uint16.
x = round(double(x));
xmin = min(x(:));
xmax = max(x(:));
pmin = double(int16(xmin));
pmin = uint16(pmin+32768);
y.min = pmin;
% Compute the input histogram between xmin and xmax with unit
% width bins , scale to uint16 , and store.
x = x(:)';
h = histc(x, xmin:xmax);
if max(h) > 65535
h = 65535 * h / max(h);
end
h = uint16(h);
y.hist = h;
% Code the input mat rix and store t h e r e s u lt .
map = huffman(double(h)); % Make Huffman code map
hx = map(x(:) - xmin + 1); % Map image
hx = char(hx)'; % Convert to char array
hx = hx(:)';
hx(hx == ' ') = [ ]; % Remove blanks
ysize = ceil(length(hx) / 16); % Compute encoded size
hx16 = repmat('0', 1, ysize * 16); % Pre-allocate modulo-16 vector
hx16(1:length(hx)) = hx; % Make hx modulo-16 in length
hx16 = reshape(hx16, 16, ysize); % Reshape to 16-character words
hx16 = hx16' - '0'; % Convert binary string to decimal
twos = pow2(15 : - 1 : 0);
y.code = uint16(sum(hx16 .* twos(ones(ysize ,1), :), 2))';
Why is the block processing step generating mostly null values?
  3 个评论
显示 1更早的评论
Walter Roberson
Walter Roberson 2015-9-22
P1 is t and P2 is t' . This code uses blkproc which was replaced in 2009,years before the posting was written
Walter Roberson
Walter Roberson 2019-2-4
In newer releases, blkproc would have to be replaced
y = blkproc(x, [8 8], 'P1 * x * P2', t, t');
y = blkproc(y, [8 8], 'round(x ./ P1)', m); % <== nearly all elements from y are zero after this step
would become
y = blockproc(x, [8 8], @(block) t * block.data * t');
y = blockproc(y, [8 8], @(block) round(block.data ./ m));

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采纳的回答

Matz Johansson Bergström
I think that you might as well replace
t = dctmtx(8);
y = blkproc(x, [8 8], 'P1 * x * P2', t, t');
with
y = blkproc(x, [8 8], 'dct2(x)')
and the quality is best for small values and very few coefficients are kept for larger values like 20. I hope this helps.
  3 个评论
显示 1更早的评论
AngryBird
AngryBird 2019-2-4
I am new to matlab, can anyone tell me how to run this code? I have png true color image called Lena.png.

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更多回答（4 个）

Arindom Mondal
Arindom Mondal 2018-6-1
编辑：Arindom Mondal 2018-6-1
I am new in Mat lab coding. I cant run this code. I dont understand how can declare x?
Output shows below: Error: File: image_compresssion1.m Line: 2 Column: 1 Function definitions are not permitted in this context.
Ashish Parmar
Ashish Parmar 2019-1-6
im new in matlab coding ....will anbody please provide me the excat working code for JPEG compression?
h hjjhk
h hjjhk 2021-6-15
what is huffmn function
map = huffman(double(h));
where huffman function code
  1 个评论
Walter Roberson
Walter Roberson 2021-6-15
The code is from the book "Digital Image Processing Using MATLAB" according to an earlier poster.

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P.Anushri P.Anushri

jpegCompress.m function y = jpegCompress(x, quality) % y = jpegCompress(x, quality) compresses an image X based on 8 x 8 DCT % transforms, coefficient quantization and Huffman symbol coding. Input % quality determines the amount of information that is lost and compression achieved. y is the encoding structure containing fields: % y.size size of x % y.numblocks number of 8 x 8 encoded blocks % y.quality quality factor as percent % y.huffman Huffman coding structure

narginchk(1, 2); % check number of input arguments if ~ismatrix(x) ~isreal(x) ~ isnumeric(x) ~ isa(x, 'uint8') error('The input must be a uint8 image.'); end if nargin < 2 quality = 1; % default value for quality end if quality <= 0 error('Input parameter QUALITY must be greater than zero.'); end

m = [16 11 10 16 24 40 51 61 % default JPEG normalizing array 12 12 14 19 26 58 60 55 % and zig-zag reordering pattern 14 13 16 24 40 57 69 56 14 17 22 29 51 87 80 62 18 22 37 56 68 109 103 77 24 35 55 64 81 104 113 92 49 64 78 87 103 121 120 101 72 92 95 98 112 100 103 99] * quality;

order = [1 9 2 3 10 17 25 18 11 4 5 12 19 26 33 ... 41 34 27 20 13 6 7 14 21 28 35 42 49 57 50 ... 43 36 29 22 15 8 16 23 30 37 44 51 58 59 52 ... 45 38 31 24 32 39 46 53 60 61 54 47 40 48 55 ... 62 63 56 64];

[xm, xn] = size(x); % retrieve size of input image x = double(x) - 128; % level shift input t = dctmtx(8); % compute 8 x 8 DCT matrix

% Compute DCTs pf 8 x 8 blocks and quantize coefficients y = blkproc(x, [8 8], 'P1 * x * P2', t, t'); y = blkproc(y, [8 8], 'round(x ./ P1)', m); % <== nearly all elements from y are zero after this step y = im2col(y, [8 8], 'distinct'); % break 8 x 8 blocks into columns xb = size(y, 2); % get number of blocks y = y(order, :); % reorder column elements

eob = max(x(:)) + 1; % create end-of-block symbol r = zeros(numel(y) + size(y, 2), 1); count = 0;

for j = 1:xb % process one block(one column) at a time i = find(y(:, j), 1, 'last'); % find last non-zero element if isempty(i) % check if there are no non-zero values i = 0; end p = count + 1; q = p + i; r(p:q) = [y(1:i, j); eob]; % truncate trailing zeros, add eob count = count + i + 1; % and add to output vector end

r((count + 1):end) = []; % delete unused portion of r

y = struct; y.size = uint16([xm xn]); y.numblocks = uint16(xb); y.quality = uint16(quality * 100); y.huffman = mat2huff(r);

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