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Multi-Task - Ranking concensus #32
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Dear @ReubenDo ,
To avoid confusion, you refer to rankThenAggregate. Yes, you are right, in the case of rankThenAggregate both approaches are valid. The reason for the current implementation is mainly that it is more general and independently works no matter whether rank-then-aggregate or aggregate-then-rank has been used before and follows the theory of consensus rankings. With respect to your example, you might argue that the implementation looses precision in not using aggregated ranks (but essentially rounding them to integers). On the other hand, you might also argue each algorithm is best in one of the tasks, so each of them won in one task. Goal of a consensus ranking would then be to summarize this information (each algorithm best in one task) and you don’t want to over-interpret the small difference in mean ranks. I guess this is also a question whether you want to „force“ to have a winner no matter how small the difference is between the algorithms or whether you only want to distinguish algorithms if one is clearly better (this is a similar question in significance ranking where there will be often multiple winners if performance is close and number of cases small). So thank you for your comment and this clarification, you are absolutely right. I appreciate such discussions! Best wishes, |
Hello,
Thank you for your excellent work and nice implementation!
I am currently using the the rank-then-aggregate scheme. The output of
aggregateThenRank
is the mean ranking for each task for each team. As far as I understood, theconsensus
function is used to merge the mean rankings, by 1/ computing the ranking of the mean ranking score associated to each task; 2/ averaging these new rankings. However, you could also directly average the mean ranking scores.For example, let consider two teams A and B and two tasks T1 and T2. Let assume that the mean ranking for A and B are respectively 1.2 and 1.8 on task T1 and 1.6 and 1.4 on task T2. In the current implementation, it seems that the consensus function uses the ranking of the mean ranking, i.e. 1 and 2 on T1 + 2 and 1 on T2, leading to final ranking scores of 1.5 for both teams. However, previous challenges (e.g., BraTS) seem to directly average the mean rankings.
This approach would give a different ranking: A and B would respectively have a ranking score of 1.4 and 1.6 and thus A will be the winner.
I think that both approaches are valid but I was wondering if there was a specific reason that explained why you chose to average the rankings of the mean rankings instead of averaging the mean rankings.
Cheers,
Reuben
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