Don't know about the correlations (never used them in a gam context
actually...), but you can "bin" the mean by :
> x <- 1:100
> tapply(x,cut(x,5),mean)
(0.901,20.7] (20.7,40.6] (40.6,60.4] (60.4,80.3] (80.3,100]
10.5 30.5 50.5 70.5 90.5
Cheers
Joris
On Sat, Jun 19, 2010 at 1:54 AM, David Jarvis <thanga...@gmail.com> wrote:
> Hi,
>
> Standard correlations (Pearson's, Spearman's, Kendall's Tau) do not
> accurately reflect how closely the model (GAM) fits the data. I was told
> that the accuracy of the correlation can be improved using a root mean
> square deviation (RMSD) calculation on binned data.
>
> For example, let 'o' be the real, observed data and 'm' be the model data. I
> believe I can calculate the root mean squared deviation as:
>
> sqrt( mean( o - m ) ^ 2 )
>
> However, this does not bin the data into mean sets. What I would like to do
> is:
>
> oangry <- c( mean(o[1:5]), mean(o[6:10]), ... )
> mangry <- c( mean(m[1:5]), mean(m[6:10]), ... )
>
> Then:
>
> sqrt( mean( oangry - mangry ) ^ 2 )
>
> That calculation I would like to simplify into (or similar to):
>
> sqrt( mean( bin( o, 5 ) - bin( m, 5 ) ) ^ 2 )
>
> I have read the help for ?cut, ?table, ?hist, and ?split, but am stumped for
> which one to use in this case--if any.
>
> How do you calculate c( mean(o[1:5]), mean(o[6:10]), ... ) for an arbitrary
> length vector using an appropriate number of bins (fixed at 5, or perhaps
> calculated using Sturges' formula)?
>
> I have also posted a more detailed version of this question on
> StackOverflow:
>
> http://stackoverflow.com/questions/3073365/root-mean-square-deviation-on-binned-gam-results-using-r
>
> Many thanks.
>
> Dave
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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>
--
Joris Meys
Statistical consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
tel : +32 9 264 59 87
joris.m...@ugent.be
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