Monday, 23 January 2017

Whiskas statistics and the pitfalls of mean rank

In the fingerprint similarity paper I published last year, I made the following observation: "However the use of the mean rank is itself problematic as the pairwise similarity of two methods can be altered (and even inverted) by adding additional methods to an evaluation." I thought about providing an example illustrating this, but I didn't want to digress too much. And besides, that's perfect blog fodder...

Let's say we want to compare N methods, and we have their scores for 10 diverse datasets. Because they are diverse, we think to ourselves that it wouldn't be an accurate evaluation if we used the scores directly, e.g. taking the mean scores or some such, because some of the datasets are easy (all of the scores are high but spread out) and some are hard (all of the scores are low, but close together). As I reference in the paper, Bob Sheridan has discussed this problem, and among other solutions suggested using the mean rank.

So here's the thing. Consider two methods, A and B, where A outperforms B on 9 out of the 10 datasets, i.e. A has rank 1 nine times and rank 2 once, while vice versa for B. So the mean rank for A is 1.1 while that for B is 1.9, implying that A is better than B. So far so good.

Now let's suppose I start tweaking the parameters of B to generate 9 additional methods (B1 to B9). Unfortunately, while they have similar performance to B, they are all slightly worse on each dataset. So the rank order for 9 of the 10 datasets in now A > B > B1...B9, and that for the 10th dataset is B > B1...B9 > A. So A has rank 1 nine times (as before) and rank 11 once (giving a mean rank of 2.0), while B has the same rank as before. So now B is better than A.

Wait a second. Neither A nor B has changed, so how can our evaluation of their relative performance have changed? And therein lies the problem.

So what's the solution? Even a cat knows the answer: as I stated at the start, on 9 out of 10 datasets A outperformed B. So A is better than B, or A>B. It's as simple as that. We don't need to calculate the mean anything. Our goal is not to find a summary value for the overall performance for a method, and then to compare those. It is to evaluate relative performance, and for diverse datasets each dataset has an equal vote towards that answer. So on dataset 1 (taking a different imaginary dataset), each method dukes it out giving A>B, A>C, A>D, B>C, B>D, C=D. Then round 2, on dataset 2. Toting up the values over all datasets will yield something like A>B 10 times, while B>A 5 times, and A=B twice, so A>B overall (and similar results for other pairwise comparisons).

The nice thing about all of these pairs is that you can then plot a Hasse diagram to summarise the info. There are some niggly details like handling incomparable values (e.g. A>B, B>C, C>A) but when it's all sorted out, the resulting diagram is quite information-rich.

I didn't mention statistical significance, but the assessment of whether you can say A>B overall requires a measurement of significance. In my case, I could generate multiple datasets from the same population, and so I used a a basic T-test over the final results from each of the datasets (and corrected for multiple testing).

I've gone on and gone on a bit about what's really quite a simple idea. In fact, it's so simple that I was sure it must be how people already carry out performance evaluations across diverse datasets. However, I couldn't find any evidence of this, nor could I find any existing description of this method. Any thoughts?

1 comment:

bernatv said...

Yeah, calculating the mean rank seems not a good idea because the rank is not a normally distributed random variable. Non-parametric test like e.g. Kruskal-Wallis, or maybe Bayesian inference with non-informative prior would be better fits for the task, I guess.
Or one could derive some kind of 'normalized' rank but it'd be a temporary fix.