RankAggregation.jl

Rank aggregation in Julia
Popularity
3 Stars
Updated Last
2 Years Ago
Started In
October 2019


Given a set of objects (e.g. rows of a table) with scores given by different scoring methods (e.g. columns), how to rank the objects? This problem is known in the literature as the rank aggregation problem.

The problem is trivial when there is only one score for each object (one column), but ranking objects on the basis of multiple (conflicting) scores is challenging. This package provides algorithms to aggregate multiple scores stored in a tabular format (see Tables.jl) into a final rank vector.

Installation

Get the latest stable release with Julia's package manager:

] add RankAggregation

Usage

Given a table with scores score1 and score2 for objects a, b, and c:

julia> using DataFrames
julia> using RankAggregation

julia> objects = DataFrame(object=[:a,:b,:c], score1=[0.9, 0.7, 0.5], score2=[0.8, 0.9, 0.4])
3×3 DataFrame
│ Row │ object │ score1  │ score2  │
│     │ Symbol │ Float64 │ Float64 │
├─────┼────────┼─────────┼─────────┤
│ 1   │ a      │ 0.90.8     │
│ 2   │ b      │ 0.70.9     │
│ 3   │ c      │ 0.50.4

rank the objects using:

julia> rank(objects, :score1)
3-element Array{Int64,1}:
 1
 2
 3

 julia> rank(objects, :score2)
3-element Array{Int64,1}:
 2
 1
 3

 julia> rank(objects, (:score1,:score2))
3-element Array{Int64,1}:
 1
 2
 3

Optionally, specify the aggregation method:

julia> rank(objects, (:score1,:score2), TauModel())
3-element Array{Int64,1}:
 1
 2
 3

and the reverse option:

julia> rank(objects, (:score1,:score2), rev=true)
3-element Array{Int64,1}:
 3
 2
 1

Aggregation Methods

Method References
TauModel Journel 2002. Combining Knowledge From Diverse Sources: An Alternative to Traditional Data Independence Hypotheses.

Contributing

Contributions are very welcome, as are feature requests and suggestions.

Please open an issue if you encounter any problems.

Used By Packages

No packages found.