On Jun 18, 2010, at 10:38 PM, David Jarvis wrote:
Hi, David.
accurately reflect how closely the model (GAM) fits the data. I was
told
This was my presumption; I could be mistaken.
that the accuracy of the correlation can be improved using a root mean
square deviation (RMSD) calculation on binned data.
By whom? ... and with what theoretical basis?
I talked with Christian Schunn. He mentioned that using RMSD would
produce a better result for goodness-of-fit (if that term is not
synonymous with correlation, I apologise -- I'm still rather new to
this level of statistics):
http://www.lrdc.pitt.edu/schunn/gof/index.html
It was regarding a chart similar to:
http://i.imgur.com/X0gxV.png
In the chart, the calculation for Pearson's, Spearman's, and
Kendall's Tau provide, in my opinion, an incorrect indicator as to
the strength of GAM's fit to the data. I could be wrong here, too.
His suggestion was to use bin the means (in groups of 5 or so) to
reduce the noise.
I doubt that your strategy offers any statistical advantage, but if
you want to play around with it then consider:
binned.x <- round( (x + 2.5)/5)
> d <-
c
(1,3,5,4,3,6,3,1,5,7,8,9,4,3,2,7,3,6,8,9,5,3,1,4,5,8,9,3,3,2,5,7,8,8,5,4,3,2,6,4,3,1,4,5,6,8,9,0,7,7,5,4,3,3,2,1,3,4,5,6,7,9,0,2,4,3,3
)
> binned.d <- round( (d + 2.5)/5)
> print(binned.d)
[1] 1 1 2 1 1 2 1 1 2 2 2 2 1 1 1 2 1 2 2 2 2 1 1 1 2 2 2 1 1 1 2 2
2 2 2 1 1 1
[39] 2 1 1 1 1 2 2 2 2 0 2 2 2 1 1 1 1 1 1 1 2 2 2 2 0 1 1 1 1
That doesn't make sense to me.
Then I blame your powers of exposition. Without some sort of explicit
example the parsing of English is very prone to error. If you want to
pick out elements of x in some pre-specified order in groups of five
then consider:
> x <- 1:100
>
> rep(1:20, each=5)
[1] 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 5
5 5
[24] 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8 8 9 9 9 9
9 10
[47] 10 10 10 10 11 11 11 11 11 12 12 12 12 12 13 13 13 13 13 14 14
14 14
[70] 14 15 15 15 15 15 16 16 16 16 16 17 17 17 17 17 18 18 18 18 18
19 19
[93] 19 19 19 20 20 20 20 20
> tapply(x, rep(1:20, each=5), mean)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 # this
row is just indices
3 8 13 18 23 28 33 38 43 48 53 58 63 68 73 78 83 88 93 98 # this
row is the means
If you wanted them in random groups of roughly 5, then you could use
sample(x, prob=rep(5/n, n/5))
My impression was that I should try to put every 5 values in a bin,
average that bin, then calculate the RMSD between the observed
values and the values from GAM. In other words (o is observed and m
is model):
Do you intend that m[n] would be the predicted value from a model? How
are you forming the groups of 5? Are they ordered? If so ordered by
observed of by predicted? (In R a "model" is a complex list structure,
but may in some cases have a simple predicted value for each case.
Again a specific example might work wonders.
--
David.
bins <- 5
while( length(o) %% bins != 0 ) {
o <- o[-length(o)]
}
omean <- apply( matrix(o, bins), 2, mean )
while( length(m) %% bins!= 0 ) {
m <- m[-length(m)]
}
mmean <- apply( matrix(m, bins), 2, mean )
sqrt( mean( omean - mmean ) ^ 2 )
But that feels sloppy, error prone, and fragile.
Joris mentioned that I could try using tapply with
cut(d,round(length(d)/5)). I couldn't figure out how to get the
means back from the factors.
Dave
David Winsemius, MD
West Hartford, CT
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