4

I'm very new with MatLab, I have Run Length Encoding code but it seems to not work, can you help me?

I have this input :

ChainCode  = 11012321170701000700000700766666666666665555555544443344444333221322222322 

and I want make it into RLE output :

(1,2), (0,1), (1,1), (2,1), (3,1), (2,1), (1,2), (7,1), (0,1), (7,1), (0,1), 
(1,1), (0,3), (7,1), (0,5), (7,1), (0,2), (7,1), (6,13), (5,8), (4,4), (3,2), 
(4,5), (3,3), (2,2), (1,1), (3,1), (2,5), (3,1), (2,2) 

This is my code :

lengthcode = 1;
N = 1;

for i = 2:length(ChainCode)

    if x(i)==x(i-1)
        N = N + 1; 
        valuecode(N)  = x(i);
        lengthcode(N) = lengthcode(N) + 1;
    else 
        N = 1;
        lengthcode = 1;
    end

    i = i + 1;

end

But this is not working, and I am still confused about how can I print the output like that.

I hope you can help me. Thank you.

Rody Oldenhuis
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user1146895
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  • if anyone can help how to calculate the frequency of occurrence of the existing output. eg: (1.2), (0.1), (1.1), (2.1), (3.1), (2.1), (1.2), (7.1), (0 , 1) (1,2) = 2 times (0,1) = 2 times (1,1) = 1 times and so on. Thank you – user1146895 Aug 25 '12 at 12:12
  • You should modify your original post/question (or post as new question) in case you need something different than what posted. See my updated answer below for finding occurrence frequencies. – gevang Aug 26 '12 at 03:00

4 Answers4

11

Here is a compact solution without loop, cellfun or arrayfun:

chainCode = '11012321170701000700000700766666666666665555555544443344444333221322222322';
numCode = chainCode - '0'; % turn to numerical array

J=find(diff([numCode(1)-1, numCode]));
relMat=[numCode(J); diff([J, numel(numCode)+1])];
Mohsen Nosratinia
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1

By sticking to your original implementation, the following simple changes should work.

chainCode = '11012321170701000700000700766666666666665555555544443344444333221322222322';
numCode = chainCode - '0'; % turn to numerical array
relMat = [];
numCode = [numCode nan]; % dummy ending

N = 1;
for i = 1:length(numCode)-1   
    if numCode(i)==numCode(i+1)
        N = N + 1;
    else
        valuecode = numCode(i);
        lengthcode =  N;
        relMat = [relMat; valuecode lengthcode];
        N = 1;
    end
end

You can format the output however you like. For example as a sequence:

relMatT = relMat';
relSeq = relMatT(:)';

or format a string to the suggested output:

relString = [];
for i = 1:length(relMat)
    relString = [relString, sprintf('(%d, %d), ', relMat(i,1), relMat(i,2))];
end

As an extension, if you have alphanumerics in your source sequence you should modify the above in order to compare strings instead of numbers.

UPDATE: To count the occurrences of unique code pairs in the original relMat try finding the pairs and count zero-diffs row-wise. For example:

relMatUnique = unique(relMat, 'rows'); % find unique pairs 
nPairs = length(relMatUnique);
nOccur = zeros(nPairs, 1);
for i = 1:nPairs
    pairInMat = bsxfun(@minus, relMat, relMatUnique(i,:)); % find pair in relMat
    nOccur(i) = sum(~sum(pairInMat, 2));
end
relMatOccur = [relMatUnique nOccur]; % unique pairs and number of occurrences 
gevang
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1

You can avoid the for loop:

chainCode = '11012321170701000700000700766666666666665555555544443344444333221322222322';
numCode = chainCode - '0'; % turn to numerical array

% detect edges (changes)
edges = arrayfun( @(x,y) x ~= y,    ...
                  numCode(1:end-1), ...
                  numCode(2:end));
% get indexes
idx = find(edges);

% create tuples
relMat = cell2mat(arrayfun(         ...
  @(b,e) [ numCode(b) ; e-b+1 ],    ...
  [ 1 (idx + 1) ],                  ...
  [ idx length(numCode) ],          ...
  'UniformOutput', false));
gzm0
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0
% Convert string to numeric array
ChainCodeString  = '11012321170701000700000700766666666666665555555544443344444333221322222322';
ChainCodeArray = ChainCodeString - '0';

% Initialize
CurrentRleValue = ChainCodeArray(1);
CurrentRleCount = 1;
RleCodeIndex = 1;

for i = 2 : length(ChainCodeArray)
    if ChainCodeArray(i)==ChainCodeArray(i-1)
        % Increment current run-length count
        CurrentRleCount = CurrentRleCount + 1;
    else
        % Store current run-length
        valuecode(RleCodeIndex) = CurrentRleValue;
        lengthcode(RleCodeIndex) = CurrentRleCount;
        RleCodeIndex = RleCodeIndex + 1;

        % Initialize next run-length
        CurrentRleValue = ChainCodeArray(i);
        CurrentRleCount = 1;
    end;
end;
mola
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