Re: similarity scores
- John Hollinger wrote:
> Using the (3FG * 3FG%)/MP is a way to rate players in thesePlease let me know what you discover. As I mention in the write-up on
> categories that I hadn't considered; I'll have to put these
> into my formulae and see how the output shapes up.
my site, I wanted a way to measure both three-point quantity and
quality. I thought this might work well.
Basketball Stats! http://www.basketball-reference.com
- Thank you. I am an idiot sometimes. It is a challenge sometimes to
write out correct and clear instructions. And to think this is what
I do for a living. :)
1. Compute the weighted mean.
2. Compute the squared deviations from the weighted mean.
3. Weight the squared deviations by minutes played.
4. Sum the weighted squared deviations.
5. Divide this sum by the sum of minutes played.
6. Take the square root of this weighted average of the squared
And I do not think there is any need to square the weights. I do
not believe this is what is done in most typical regression packages.
--- In APBR_analysis@yahoogroups.com, "Michael Tamada" <tamada@o...>
> Yup, although if it's a standard deviation that we'rehttp://www.itl.nist.gov/div898/software/dataplot/refman2/ch2/weightsd
> calculating, I think what you want in step 2 is to
> SQUARE the deviations.
> And in 5, although dividing by the sum of minutes played
> is good, arguably better might be to reduce that
> figure slightly to correct for degrees of freedom,
> by multiplying the minutes played by (N-1)/N.
> And given the squaring that I describe in step 2,
> then there of course needs to be a step 6: take
> the square root, after you finish step 5.
> The procedure that I describe is, e.g., the one
> described in the National Institute of Standards
> and Technology's nifty statistics website:
> Come to think of it, in a regression framework,
> wouldn't we square the weights too, in Step 2?
> Ah, I'll worry about that later. DanR's procedure
> is the one I'd follow, but with the amendments listed
> -----Original Message-----
> From: dan_t_rosenbaum [mailto:rosenbaum@u...]
> Sent: Saturday, December 18, 2004 10:04 AM
> I just compute standard deviations weighted by minutes played. I
> could not find where Excel does this, but what you could do is the
> 1. Compute the weighted mean.
> 2. Compute the deviations from the weighted mean.
> 3. Weight the deviations by minutes played.
> 4. Sum the weighted deviations.
> 5. Divide this sum by the sum of minutes played.
> --- In APBR_analysis@yahoogroups.com, "thedawgsareout"
> <kpelton08@h...> wrote:
> > > The notion of the average representing the range of
> > > players that you'd actually see play is an interesting
> > > one.
> > >
> > > I think what it comes down to is this: do we want an
> > > average of what happens during NBA games, or an
> > > average of what NBA players do? You're advocating the
> > > former, and I guess I am asking about the latter.
> > >
> > > Either way is fine, I guess it comes down to
> > > semantics.
> > Maybe someone's mentioned this and I've missed it, but what do
> > guys plan to do about standard deviation if you use some sort of
> > weighted system?
> > I would argue that in this case, standard deviation is far more
> > important than average. You're not going to change average very
> > depending on what population you use, but standard deviation
> > quite significantly. The reason you don't use low-minutes guys
> > because they're not NBA players; it's because their stats are
> > obviously not significant.
> > Let's use rebounds per 48 minutes last year as an example.
> > If you take the pure average of everyone in the league, you get
> > 8.38. If you weight by minutes, you get 8.38. If you cut off at
> > minutes and take the pure average (which is what I do), you get
> > There's a difference there, but not an enormous one.
> > If you take the standard deviation of guys with 250 minutes or
> > it's 3.52. The standard deviation of everyone is 3.76. That's a
> > bigger difference (though you could argue that because changing
> > average takes guys from above average to below it, it's more
> > significant).
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