Using pdist with a custom correlation function the right way to avoid kappa score discrepancies?

Question

I read this but this is not solving my problem: I have this initial_comparison_frame

id  GO1 GO10    GO11    GO12    GO2 GO3 GO4 GO5 GO6 GO7 GO8 GO9
GO1 1   0   0   0   0   0   1   1   1   1   1   1
GO2 0   0   1   0   1   0   1   1   1   1   1   1
GO3 1   0   0   1   0   1   0   1   1   1   1   1
GO4 1   0   0   0   0   0   1   0   1   1   0   0
GO5 0   0   0   0   0   0   0   1   0   0   0   0
GO6 1   0   0   0   0   1   0   1   1   1   0   1
GO7 0   1   1   1   1   0   0   1   1   1   1   0
GO8 0   0   0   0   1   1   1   1   1   1   1   1
GO9 0   0   0   0   0   0   0   0   0   0   0   1

that you can generate this way:

initial_comparison_frame = pd.DataFrame([[1,0,0,0,0,0,1,1,1,1,1,1],[0,0,1,0,1,0,1,1,1,1,1,1],[1,0,0,1,0,1,0,1,1,1,1,1],[1,0,0,0,0,0,1,0,1,1,0,0],[0,0,0,0,0,0,0,1,0,0,0,0],[1,0,0,0,0,1,0,1,1,1,0,1],[0,1,1,1,1,0,0,1,1,1,1,0],[0,0,0,0,1,1,1,1,1,1,1,1],[0,0,0,0,0,0,0,0,0,0,0,1]], index =["GO1","GO2","GO3","GO4","GO5","GO6","GO7","GO8","GO9"],columns = ["GO1","GO10","GO11","GO12","GO2","GO3","GO4","GO5","GO6","GO7","GO8","GO9"])   

(I do another way generating from get_dummies but that doesn't change anything)

and I want to build the pairwise distance matrix of these vectors, using cohen_kappa_score

from sklearn.metrics import cohen_kappa_score
dist_matrix = pdist(initial_comparison_frame, cohen_kappa_score)
square_dist_matrix = squareform(dist_matrix) 
square_dist_frame = pd.DataFrame (square_dist_matrix, index = count_frame['id'], columns = count_frame['id'])

like this:

    id  GO1 GO2 GO3 GO4 GO5 GO6 GO7 GO8 GO9
GO1 0.0 0.47058823529411764 0.47058823529411764 0.5263157894736843  0.12195121951219523 0.5 -0.23529411764705865    0.47058823529411764 0.12195121951219523
GO2 0.47058823529411764 0.0 -0.125  0.09999999999999987 0.08695652173913038 0.0 0.25    0.625   0.08695652173913038
GO3 0.47058823529411764 -0.125  0.0 0.09999999999999987 0.08695652173913038 0.6666666666666667  -0.125  0.25    0.08695652173913038
GO4 0.5263157894736843  0.09999999999999987 0.09999999999999987 0.0 -0.15384615384615397    0.33333333333333337 -0.20000000000000018    0.09999999999999987 -0.15384615384615397
GO5 0.12195121951219523 0.08695652173913038 0.08695652173913038 -0.15384615384615397    0.0 0.16666666666666663 0.08695652173913038 0.08695652173913038 -0.09090909090909105
GO6 0.5 0.0 0.6666666666666667  0.33333333333333337 0.16666666666666663 0.0 -0.33333333333333326    0.33333333333333337 0.16666666666666663
GO7 -0.23529411764705865    0.25    -0.125  -0.20000000000000018    0.08695652173913038 -0.33333333333333326    0.0 -0.125  -0.17391304347826098
GO8 0.47058823529411764 0.625   0.25    0.09999999999999987 0.08695652173913038 0.33333333333333337 -0.125  0.0 0.08695652173913038
GO9 0.12195121951219523 0.08695652173913038 0.08695652173913038 -0.15384615384615397    -0.09090909090909105    0.16666666666666663 -0.17391304347826098    0.08695652173913038 0.0

that you can generate like this for convenience:

 square_dist_frame = pd.DataFrame([[0.0 0.47058823529411764 0.47058823529411764 0.5263157894736843  0.12195121951219523 0.5 -0.23529411764705865    0.47058823529411764 0.12195121951219523],[0.47058823529411764   0.0 -0.125  0.09999999999999987 0.08695652173913038 0.0 0.25    0.625   0.08695652173913038],[0.47058823529411764   -0.125  0.0 0.09999999999999987 0.08695652173913038 0.6666666666666667  -0.125  0.25    0.08695652173913038],[0.5263157894736843    0.09999999999999987 0.09999999999999987 0.0 -0.15384615384615397    0.33333333333333337 -0.20000000000000018    0.09999999999999987 -0.15384615384615397],[0.12195121951219523  0.08695652173913038 0.08695652173913038 -0.15384615384615397    0.0 0.16666666666666663 0.08695652173913038 0.08695652173913038 -0.09090909090909105],[0.5  0.0 0.6666666666666667  0.33333333333333337 0.16666666666666663 0.0 -0.33333333333333326    0.33333333333333337 0.16666666666666663],[-0.23529411764705865  0.25    -0.125  -0.20000000000000018    0.08695652173913038 -0.33333333333333326    0.0 -0.125  -0.17391304347826098],[0.47058823529411764  0.625   0.25    0.09999999999999987 0.08695652173913038 0.33333333333333337 -0.125  0.0 0.08695652173913038],[0.12195121951219523   0.08695652173913038 0.08695652173913038 -0.15384615384615397    -0.09090909090909105    0.16666666666666663 -0.17391304347826098    0.08695652173913038 0.0]], index =["GO1","GO2","GO3","GO4","GO5","GO6","GO7","GO8","GO9"],columns = ["GO1","GO2","GO3","GO4","GO5","GO6","GO7","GO8","GO9"])

My actual problem is that the values match actual kappa_score for individual values, except for diagonal ones. Since it is a distance matrix, the distance of one individual to itself should be 0 indeed, but then, why isn't the score in the matrix equal to 1-the kappa score?
Else, two logics are applied to the same matrix and I don't understand that. So I believe I made a mistake.

Thanks for enlighting me.

Answer 1

Actually I became aware that the problem lies in scipy squareform function; https://docs.scipy.org/doc/scipy-0.14.0/reference/generated/scipy.spatial.distance.squareform.html which states "Given a d*d(-1)/2 sized v for some integer d>=2 encoding distances as described, X=squareform(v) returns a d by d distance matrix X. The X[i, j] and X[j, i] values are set to v[{n choose 2}-{n-i choose 2} + (j-u-1)] and all diagonal elements are zero."

while kappa score of 0 means perfect decorrelation and 1 perfect correlation (-1 reverse correlation)

so I guess to merely modify the frame created by squareform by the right operations (i will edit later once i figured these the right way for negative values)