Matlab的霍夫曼編碼在矩陣

Question

我試圖編碼矩陣，我有與霍夫曼碼（計算幀差之後），但我有完成它

我希望與霍夫曼編碼矩陣的困難被稱爲「安培「

的東西，我發現是這樣的：

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 values. 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, uint8, uint16, uint32, int8, int16, or double, 
% with integer values, it can be input directly to MAT2HUFF. 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 0 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. 
% 
% See also HUFF2MAT. 

% Copyright 2002-2004 R. C. Gonzalez, R. E. Woods, & S. L. Eddins 
% Digital Image Processing Using MATLAB, Prentice-Hall, 2004 
% $Revision: 1.5 $ $Date: 2003/11/21 15:21:12 $ 

if ndims(x) ~= 2 | ~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 and store its minimum value biased 
% 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 matrix and store the result. 
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))'; 

... 

但在上面的代碼我如何申報使用我的矩陣稱爲‘放’？

在此先感謝

Answer 1

只是調用函數huf_amp = mat2huf(amp); huf_amp是您的代碼示例中詳述的結構。