From: David Jarvis <thangalin_at_gmail.com>

Date: Fri, 18 Jun 2010 16:54:48 -0700

R-help_at_r-project.org mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Fri 18 Jun 2010 - 23:57:22 GMT

Date: Fri, 18 Jun 2010 16:54:48 -0700

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

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