# How to design a weighted formula with multiple inputs?

Apologies if this is the completely wrong verbiage for this sort of thing... kinda new to this.

We're designing a social website, where users on the site have several statistics associated with them. We want to assign each user a "score," which is a weighted interpretation of all their statistics.

We don't have a solid relationship between our stats yet, so I'm not looking for an exact formula, just general advice on how to build a formula like this (especially as I'm not terribly mathematically inclined). You know, something like User 1 who has 10,000 for Stat 2 might have a score of 34, and User 2 might who has 10 for Stat 1 might have a score of 16 (with 1 being the top-rated User), because of the weighted formula.

Our system has 3-4 user statistics we want to mix into this formula (basically, metrics involving the user's involvement on the site, like comments written and posts viewed, etc.)

Does that make sense? Please let me know if I need to clarify anything. Thanks!

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I'm confused -- user 1 has a value for stat 2, while user 2 has a value for stat 1? It it such that user $i$ has a value for stat $j$ whenever $i\ne j$ (that is, a number for each pair of users that you want to condense into a single number per user?) – Henning Makholm Apr 5 '12 at 19:22
Sorry, that was an unclear example... what I meant to show is that stat 1 would be weighted more heavily than stat 2... but that both users would have values for both stats (and 2 more). Does that clarify? – neezer Apr 5 '12 at 19:32
To add to Henning's answer. For each user, the $score(u)$ is a function of $stat_1(u), stat_2(u), \ldots, stat_n(u)$. The exact combination, and how you weight each $stat_i(u)$ will drastically change the scoring scheme. The linear weighted sum is a simple and popular scheme. But it really depends on the application in mind. You can also introduce non-linear scheme. For example: $score = age/weight$ gives high score to thin old people, and a low score for fat kids. I hope you got the idea. – user2468 Apr 5 '12 at 21:24

In your example the weight for statistic 2 might be around $0.003$ (very small, so it takes a lot of stat-2 points to earn a single increase in score), whereas the weight for statistic 1 should probably be around $1$.
Then compute the score for each user as $$\mathit{score}(u) = \mathit{stat}_1(u)\cdot w_1 + \mathit{stat}_2(u)\cdot w_2 + \cdots + \mathit{stat}_n(u)\cdot w_n$$ where there are $n$ separatet statistics, and $w_1, w_2, \ldots, w_n$ are the weights. You'll probably want to round the scores to integers for display.