ranking

I would like to rank or sort a collection of items (with size potentially greater than 100,000) where items in the collection have no intrinsic (comparable) value, instead all I have is the comparisons between any two items which have been provided by users in a subjective manner. Example: Consider a collection with elements [a, b, c, d] and comparisons by users b > a , a > d , d > c . The correct order of this collection would be [b, a, d, c] . This example is simple, however there could be more complicated cases: Since the comparisons are subjective, a user could also say that c > b . In

trec 2019 fair ranking track 最近实验室要求参加trec 2019新出的track：fair ranking track。这里整理一下该任务的思想和要求。这次track主要为学术论文数据的排序。 1 Protocol 会给定一个query集合Q，其中$q\in Q$。对于每个请求，会有一个query q和一个文档集合$D_q$。你需要做的就是根据q来重排序(rerank)$D_q$,重排序结果是$\pi$。最后把每一个请求都处理完返回$\pi$的集合的$\Pi$。过程如下： Algorithm 1 Evaluation protocol $\Pi$←{} for q,$D_q\in Q$ do $\pi$←SYSTEM(q,$D_q$) $\Pi$←$\Pi+[\pi]$ end for return $\Pi$ 2 Evaluation 衡量指标主要分为两部分，相关性(revelance)和公平性(fairness)。 所谓相关性就是document和query的相关性，公平性主要为Author Exposure即论文作者的曝光度。 先介绍如何衡量作者的曝光度： 2.1 Measuring Fairness 2.1.1 Measuring Author Exposure for a Single Ranking 先为单个请求的重排序结果$\pi

I need to create a view that automatically adds virtual row number in the result. the graph here is totally random all that I want to achieve is the last column to be created dynamically. > +--------+------------+-----+ > | id | variety | num | > +--------+------------+-----+ > | 234 | fuji | 1 | > | 4356 | gala | 2 | > | 343245 | limbertwig | 3 | > | 224 | bing | 4 | > | 4545 | chelan | 5 | > | 3455 | navel | 6 | > | 4534345| valencia | 7 | > | 3451 | bartlett | 8 | > | 3452 | bradford | 9 | > +--------+------------+-----+ Query: SELECT id, variety, SOMEFUNCTIONTHATWOULDGENERATETHIS() AS num

I'm looking for an efficient way to calculate the rank vector of a list in Python, similar to R's rank function. In a simple list with no ties between the elements, element i of the rank vector of a list l should be x if and only if l[i] is the x -th element in the sorted list. This is simple so far, the following code snippet does the trick: def rank_simple(vector): return sorted(range(len(vector)), key=vector.__getitem__) Things get complicated, however, if the original list has ties (i.e. multiple elements with the same value). In that case, all the elements having the same value should