Fix operator precedence bug in lambda_ij calculation#60
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Nepomuk5665 wants to merge 1 commit intoNetflix:masterfrom
Open
Fix operator precedence bug in lambda_ij calculation#60Nepomuk5665 wants to merge 1 commit intoNetflix:masterfrom
Nepomuk5665 wants to merge 1 commit intoNetflix:masterfrom
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In resolve_model(), line 263 used 'n / pp*2' which evaluates as '(n/pp)*2' due to Python's left-to-right evaluation for operators with same precedence. The formula comment on line 256 clearly states: lambda_ij = n_ij / (p_i + p_j)^2, i != j And line 262 correctly uses 'pp**2' for the same formula in lbda_ii. This change fixes lbda_ij to use 'pp**2' (exponentiation) instead of 'pp*2' (multiplication), matching the mathematical formula specification.
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Summary
BradleyTerryMlePairedCompSubjectiveModel.resolve_model()wheren / pp*2was evaluated as(n/pp)*2instead of the intendedn / pp**2Problem
In
sureal/pc_subjective_model.pyline 263, the code reads:Due to Python's operator precedence (division and multiplication have equal precedence and are evaluated left-to-right), this computes
(n / pp) * 2, which is incorrect.The mathematical formula specified in the comment on line 256 states:
Line 262 correctly implements this for
lbda_ii:Fix
Changed line 263 from:
to:
Impact
This affects the covariance matrix calculation (
cova_p,cova_v) and the standard deviation (stdv_p,stdv_v) outputs of the Bradley-Terry MLE model. The quality scores themselves are not affected, only their uncertainty estimates.Note: The existing tests will need their expected values updated since they were calibrated against the buggy implementation.