Difficulty in applying huffman encoding for image compression

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Nidhi Kumari
Nidhi Kumari on 9 Oct 2018
Commented: Walter Roberson on 14 Oct 2018
I got the given code to apply Huffman Encoding for image compression. But i am getting the error-
Error using huffmandict (line 71)
The symbol input must be a vector
Error in new (line 5)
[dict,avglen]=huffmandict(symbols,p)
The code is:
A=imread('xyz.jpg');
[symbols,p]=hist(A,double(unique(A)))
p=p/sum(p)
[dict,avglen]=huffmandict(symbols,p)
comp=huffmanenco(A,dict)
Help me as i am new to all these concepts.

Answers (2)

KALYAN ACHARJYA
KALYAN ACHARJYA on 10 Oct 2018
Edited: KALYAN ACHARJYA on 10 Oct 2018
When I have tested your code with the test image, both symbols and p have different sizes (see in the attached image) Read the documentation about huffmandict Clearly mentioned that-
The length of p must equal the length of symbols.
-------------------------------------------------------------------------------------------------------------------------------------------- Updated Answer
I have suspected another issue if you consider any 1 D array, the following code is working (Note that an image is 2D)
%A=double(imread('test.jpg'));
p=[.5 .125 .125 .125 .0625 .0625];
A=randsrc(100,1,[1:6; p]);
%[symbols,p]=hist(A,double(unique(A)));
[p,symbols]=hist(A,double(unique(A)));
p=p/sum(p);
[dict,avglen]=huffmandict(symbols,p);
comp=huffmanenco(A,dict);
  3 Comments
Nidhi Kumari
Nidhi Kumari on 11 Oct 2018
I am using R2014a ,so histcounts() is not present. Can you suggest any other alternative?

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Walter Roberson
Walter Roberson on 10 Oct 2018
[symbols,p] = hist(A(:), double(unique(A)));
  6 Comments
Walter Roberson
Walter Roberson on 14 Oct 2018
The output is not an image: it is a double vector containing the values 0 and 1.

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