How to find Hamming Distance ?
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I have a set of different codewords , how I separate those code words having the same hamming distance? Also D=pdist(A,'hamming') does not work in my case , If A is 1 1 1 2; 1 2 1 2 I want to calculate no. of position where they differ.
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Walter Roberson
2018년 1월 25일
Does it happen to be the case that all of your values are either 1 or 2 ? (Or, more generally, that you have exactly two different values and the two values are numerically 1 apart ?)
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the cyclist
2018년 1월 25일
편집: the cyclist
2018년 1월 25일
The Hamming distance is the fraction of positions that differ. If you want the number of positions that differ, you can simply multiply by the number of pairs you have:
numberPositionsDifferent = size(A,2)*pdist(A,'hamming');
If that's not what you meant, you might want to give more information (including the answer to Walter's questions in his comment.)
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