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how to calculate hamming distance between vectors in matrix

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jim 2014-10-19

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评论： Deeksha Varshney 2017-11-2

采纳的回答： the cyclist

在 MATLAB Online 中打开

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)

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

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the cyclist 2014-10-19

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编辑：the cyclist 2014-10-19

在 MATLAB Online 中打开

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.

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

the cyclist 2014-10-19

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编辑：the cyclist 2014-10-19

在 MATLAB Online 中打开

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 2014-10-19

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编辑：jim 2014-10-19

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 2014-10-19

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编辑：the cyclist 2014-10-19

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 2014-10-19

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

the cyclist 2014-10-20

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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 2014-10-20

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编辑：the cyclist 2014-10-20

在 MATLAB Online 中打开

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 2014-10-20

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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 2014-10-20

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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 2014-10-21

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在 MATLAB Online 中打开

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