Hi Brima,
# Fit model model.lm <- lm(Sepal.Length ~ Petal.Length, data = iris[1:75,]) # Predict data for some new data pred.dat <- predict(model.lm, newdata = iris[76:150,]) # Calculate correlation between predicted values for new data # and actual values, then square cor(iris[1:75,"Sepal.Length"], pred.dat)^2 Cheers, Josh On Mon, Oct 4, 2010 at 7:06 PM, Brima <adamsteve2...@yahoo.com> wrote: > > Hi all, > > I have used a hold out sample to predict a model but now I want to compute > an R squared value for the prediction. Any help is appreciated. > > Best regards > -- > View this message in context: > http://r.789695.n4.nabble.com/R-squared-for-lm-prediction-tp2955328p2955328.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > R-help@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. > -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ ______________________________________________ R-help@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.
