ZiNgA BuRgA Wrote:Linear or not, it's directionally the same. 

Linear is just scaling by a set factor, old*factor=new, so always the same direction.  Standardizing can changes directions depending on certain factors (although in our case, probably also the same direction).  If you mean it won't change the resulting correlation coefficient... errr, actually you're right. Scaling (with yourself) won't change anything. .

ZiNgA BuRgA Wrote:But what you're proposing could have issues such as overstating directional inconsistency.
To give a basic example, say A rates "good show" 8 and "bad show" 4, B rates "good show" 9 and "bad show" 3.  If wee do a simple scale to the mean, and say the mean of "good show" is 8.5 and "bad show" is 3.5, A would have scaled scores of -0.5 and +0.5 and B would have +0.5 and -0.5 which would be negatively correlated - probably not what you want.

Well, that simply means their opinion is conflicting with respect to the mean.  A for whether you want that or not... depends on your point of view.  But yeah, I do agree with you it's probably not the best measure for compatibility.  Scrap that then.

Now that I think of it again, their system isn't too bad.  The major fault is that they decided to scale (-1,1) to (0,1), and labelled everything based on if it were (0,1) natively.  I mean they label 50% (actually 0 correlation) as "medium compatibility", and my 62% with Sensei is (0.24 correlation) as "medium to very compatible"... That's all fucked up, 0.24 is actually very weak correlation.  So yeah, either fix the scaling or fix the labeling.

ZiNgA BuRgA Wrote:EDIT: just tried, yes, possible to keep a running SD - you need to store the sum of differences squared - this can be updated with a few (constant time) calculations when a new vote arrives.

Spoiler for explanation:

You remember:
- total vote count (n)
- sum of all votes (t)
- sum of squared differences with mean (d)

When a new vote (v) arrives, calculate new total and mean (easy to do).  Now let m_o be the old mean and m_n be new mean.
Let i = m_n - m_o
d_n = d_o + n_o(2im_o + i²) - 2it_o + (v - m_n)²

Now that you have d_n, calculating the stdev is trivial

Seems do-able then.

(This post was last modified: 02/11/2009 07:11 PM by Assassinator.)