How to find Hamming Distance ?

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.

[Walter Roberson]

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 ?)

Answer 1

[the cyclist]

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.)

[Ammy]

Thank you very much.