drbh
commented
4 years ago @makoConstruct
Thanks for checking out the project. However, when I run the example above I get back the expected median and mean scores.
Code
use ranking::{calculate_ranking, Metric};
fn main() {
let ranking_a = vec!["a", "b"];
let ranking_b = vec!["a", "b"];
let ranking_c = vec!["a", "b"];
let ranking_d = vec!["b", "c"];
let ranking_e = vec!["c", "d"];
// names are names of oceans
let everyones_rankings = vec![
("Artic", ranking_a),
("Southern", ranking_b),
("Pacific", ranking_c),
("Atlantic", ranking_d),
("Indian", ranking_e),
];
// here we use the median to calculate the final ranking
let m_metric = Metric::Median;
let rankings_by_median = calculate_ranking(everyones_rankings.clone(), m_metric);
println!("{:#?}", rankings_by_median);
// here we use the average to calculate the final ranking
let a_metric = Metric::Average;
let rankings_by_average = calculate_ranking(everyones_rankings.clone(), a_metric);
println!("{:#?}", rankings_by_average);
}Results
I get two sets of results from this script.
Ranking Metric: Median
[
( 0.0, "a"),
( 1.0, "b"),
( 1.0, "c"),
( 1.0, "d"),
]
Ranking Metric: Average
[
( 0.0, "a"),
( 0.5, "c"),
( 0.75, "b"),
( 1.0, "d"),
]Breakdown
The response above contain a sorted list from the top-ranked to the lowest-ranked items. The score (which is the numerical value in the tuple) is calculated by the metric passed to the ranker. (Median, Mean)
Median
In the first case, the score is the Median value, which if you do by hand you should get the same results.
a was in 0th position 3 times and the Median of [0,0,0] is 0.
Where c was in 0th position and 1st position and the Median of [0,1] is calculated as 1.
The median case may return semi unexpected results with such small inputs, it's unintuitive that c, b and d all have a median of 1, but a result of using such short ranking lists.
Mean
Above you mention that c does not rank higher than b but it does in this case.
The mean position for c is in 0.5th place. c was in 0th position and 1st position, the average of [0,1] is 0.5
The next value b has an average of 0.75th place. b was ranked in 0th position once and 1st position 3 times, the average of [0,1,1,1] is 0.75
The lower the score the "higher" the ranking. The same way 1st place is better than 8th place in a race.
Also note that the score is not a rank placement, and two values can have the same score in some cases which is essentially a tie
I hope this snippet and description was helpful and helps answer your question of what is this useful for.
Users 🤷♀️
It's useful for any case where you have many rankings from many individuals and you'd like to consolidate them to into a composite ranking.
makoConstruct
I'm very interested in ranking aggregation. I am a bit confused by this one. I tried it on the ranking set [a b] [a b] [a b] [a b] [b c] [c d] and neither Mean or Median gives c a higher score than b, even though the only judgement in the set about b and c is saying that c is better than b. Does it not do very well with partial rankings? I'm curious as to what this would be used for.