Weight a unique function by its best match, not its last (#157) - #166
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danielplohmann merged 2 commits intoSep 8, 2026
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getUniqueFamilyMatchInfoForSample assigned the score-weighted bytes of a function on every match, so the value that survived was the match iterated last rather than the best one, while the families and samples beside it were accumulated. The "unique" percentage on the 1-vs-N result page was therefore order-dependent (danielplohmann#157). The assignment is a max() now, and the test checks the score against a fixture whose matches are spread over different scores, in both iteration orders.
A function matched in several samples of its one family credited every sample with its best score overall; the maximum is now per (function, sample), which is what each sample's bytes are made of. Test covers two differently scored samples of one family.
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Closes #157.
getUniqueFamilyMatchInfoForSampleassigned the score-weighted bytes of a function on every match, so the value that survived was whichever match was iterated last, while the family and sample sets beside it were accumulated. The "unique" percentage on the 1-vs-N page was order-dependent. The assignment ismax()now.The test spreads the fixture's matches over different scores (the captured report scores every match of a function alike, which is why this went unnoticed), computes the expected best-per-function bytes for functions unique to one family, and checks the result in both iteration orders; it fails on
main.ruff,tyand the full suite pass.