如何合并有序preferences集合 [英] How to merge a collection of ordered preferences

问题描述

我有一个集合R评审谁是等级A组n个对象的。每个审阅独立地产生他或她选择排名对象的有序列表。的目标是产生一个列表，它是各种有序列表的排序规则。我们可以假设，认为每个评论者的观点是相同的权重。

I have a set of r reviewers who are rating a set of n objects. Each reviewer independently produces an ordered list of the objects he or she chooses to rank. The goal is to produce one list that is the collation of the various ordered lists. We can assume that each reviewer's point of view is equally weighted.

此不同于大多数合并并且在没有全局顺序进行排序名单的问题。一位评论家可以给A> B，而另一个可以给B> A。如前所述，每个对象不一定为来自每个审阅。

This differs from most merging and ordered list questions in that there is no global ordering. One reviewer can rate A > B while another can rate B > A. As mentioned, each object is not necessarily rated by each reviewer.

我目前的想法是分解每个审阅者的名单到每个M *（M-1）* 0.5个唯一对的列表，其中m为额定对象的条目数的一组有​​序元组。现在需要将所有审阅所有的记录。对于给定的组合（A，B）找到所有元组，并采取多数表决（那些投票）作为判定是否＆LT;湾

My current thought is to decompose each reviewer's list into a set of ordered tuples for each of the m * (m-1) * .5 unique pairs of entries in the list, where m is the number of objects rated. Now take all the tuples from all reviewers. For a given combination (a,b) find all such tuples and take the majority vote (of those voting) as the determinator of whether a < b.

现在我有一组有序元组重新presents所有的智慧。但我怎么把这些成一个有序的列表？我可以用随机选择的对对象开始，并责令其，然后再添一个在正确的顺序，但输出将取决于哪一个我选​​择开始。也有可能是环

Now I have a set of ordered tuples that represents the wisdom of all. But how do I turn these into one ordered list? I can start with a randomly chosen pair of objects, and order them, then add another one in the right order, but the output will depend on which one I choose to start with. Also there may be loops.

我想AP preciate任何想法。

I'd appreciate any ideas.

推荐答案

一个解决方案，看起来优雅，仍然没有什么需要做的是将每个顺序转换成分数从1到0，其中1是第一个（上图）在给定的审阅者的名单和0排名项目是他们最后的（底部项目），并在两者之间取得一个线性缩放分数的所有项目。因此，如果审稿1行列仅有3项，他们会得到的分数为1，0.5该名单，和0那你干脆把平均分每一个项目的生成整理列表。领带可以通过评审数项被打破（这样的项目一致标记为最好的3评审会出现较高的比一致被标记为最好的2审阅项目等方面的最终名单。）

A solution that seems elegant and still does what it needs to do would be to convert each ordering into a score from 1 to 0, where 1 is the first (top) ranked item in a given reviewer's list and 0 is their last (bottom item) and all items in between get a linearly scaled score. So if reviewer 1 ranks just 3 items, they would get scores for that list of 1, 0.5, and 0. Then you simply take the mean score of every item to generate a collated list. Ties could be broken by the number of "reviews" for an item (so that an item unanimously marked as best by 3 reviewers would appear higher in the final list than an item unanimously marked as best by 2 reviewers, etc.)

您的要求我们的目标是产生一个列表，它是各种有序列表的排序规则。我们可以假设，看每个审阅的观点是相同的权重。肯定是这个简单的算法实现，但往往这样的问题有比较复杂的要求，一旦你深入到他们。

Your requirement "The goal is to produce one list that is the collation of the various ordered lists. We can assume that each reviewer's point of view is equally weighted." is definitely met by this simple algorithm, but often problems like this have a bit more complex requirements once you dig into them.

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