how to calculate hamming distance between vectors in matrix

i try this code as:

a = [1 0 1 0 1;
     0 1 1 1 0;
     1 1 0 0 1];

   D = pdist(a,'minkowski',1)

the answer came as: 4 2 4 while it should be : 0 4 4 how to solve it? and to make run for bigger matrix as (50x30)

[Deeksha Varshney]

firstly change minkowski to hamming then may be you will get correct answer.

Answer 1

[the cyclist]

Well, this doesn't give your expected output, but

D = pdist(a,'hamming')

gives the Hamming distance between each pair of rows.

I'm not sure why you used the input argument "minkowski".

You can see details in the documentation.

[jim]

it didn't work it give : 0.8000 0.4000 0.8000 instead of 0 4 4

[the cyclist]

OK. A couple of comments.

First, I see that for Minkowski distance, the third input argument is used, so I deleted that part of my answer.

Second, note that according to the documentation, the output will be the average distance, not the total. So, multiplying by 5 (because there are 5 points in each vector), we get

5*D = [4 2 4]

According to the Wikipedia page on Hamming distance, this is exactly what I would expect. It is the number of positions at which the vectors differ.

Why do expect the answer to be [0 4 4]?

[jim]

because for the first vector is zero case its vector one, the second vector = v2-v1 and third vector = v3 - v2

http://classroom.synonym.com/calculate-hamming-distance-2656.html

[the cyclist]

I would say that you simply did not read the documentation carefully enough.

MATLAB is not reporting the following (which is what you expected):

• D(v1,v1)
• D(v1,v2)
• D(v2,v3)

It is reporting every pairwise distance (except for self-distance).

• D(v1,v2)
• D(v1,v3)
• D(v2,v3)

It is straightforward to pluck the distances you need from the output.

[jim]

so how to manipulate the code to make it run as D(v1,v1) D(v1,v2) D(v2,v3)

[the cyclist]

I don't understand the pattern in the distances you want, out of the possible choices, so I don't understand how to generalize it to the larger matrix.

Maybe you could try it yourself, and post your code here if you get stuck?

[jim]

a= [1 0 1 0 1;0 1 1 1 0;1 1 0 0 1];

%C = nchoosek(1:4,2);
 %  D = sum(a(C(:,2),:)~=a(C(:,1),:),2)'; 

D = pdist(a,'minkowski',1)

%D = dec2bin (pdist2(a,b,'hamming'))
  %D = pdist(a,'minkowski',1)
%D = pdist(a)
%h_d = sum(abs(a-a))

%D = dec2bin(pdist(a,'hamming'))
% hamming_dist = @(a,b)sum(a(:)~=b(:));
% hamming_dist(eye(3), zeros(3))

i try it but it didn't go..

[the cyclist]

I am just trying to understand which distance pairs you want, in general. For example, if a is 4 rows (4 vectors) instead of 3, which pairs do you want to compare?

pdist will give you

• D(v1,v2)
• D(v1,v3)
• D(v1,v4)
• D(v2,v3)
• D(v2,v4)
• D(v3,v4)

What do you actually want as the output from a 4-row matrix?

[jim]

D(V1-V1) D(V2-V1) D(V3-V2) D(V4-V3) and so on for the rest of the matrix rows. any row with its previous row, and row one should give zero.

[the cyclist]

I think the following code does what you want:

a = [1 0 1 0 1;
     0 1 1 1 0;
     1 1 0 0 1];

D = pdist(a,'minkowski',1);

D_matrix = squareform(D);

consecutive_rows_D = [0; diag(D_matrix,1)];

Note that I use the squareform function (as mentioned in the documentation for pdist) to create a matrix form of the distances, and then the diag function to pull the values of that matrix at positions

• (1,2)
• (2,3)
• (3,4)
• and so on if there were more rows

Finally, I tag on the initial zero by hand.

[jim]

it works right now... thanks alot

[the cyclist]

The best form of thanks to accept the answer, which indicates to others (who may have a similar problem) that this resolved the question you posed.