[R] Cross-validation in R
Prof Brian Ripley
ripley at stats.ox.ac.uk
Tue Jun 10 13:05:15 CEST 2008
1) cv.glm is not 'in R', it is part of contributed package 'boot'. Please
give credit where it is due.
2) There is nothing 'cross' about your 'home-made cross validation'.
cv.glm is support software for a book, so please consult it for the
definition used of cross-validation, or MASS (the book: see the posting
guide) or another reputable source.
3) If you want to know how a function works please consult a) its help
page and b) its code. Here a) answers at least your first question, and
your fundamental misunderstanding of 'cross-validation' answers the other
On Mon, 9 Jun 2008, Luis Orlindo Tedeschi wrote:
> Folks; I am having a problem with the cv.glm and would appreciate someone
> shedding some light here. It seems obvious but I cannot get it. I did read
> the manual, but I could not get more insight. This is a database containing
> 3363 records and I am trying a cross-validation to understand the process.
> When using the cv.glm, code below, I get mean of perr1 of 0.2336 and SD of
> 0.000139. When using a home-made cross validation, code below, I get mean of
> perr2 of 0.2338 and SD of 0.02184. The means are similar but SD are
You are comparing apples and oranges.
> Questions are:
> (1) how the $delta is computed in the cv.glm? In the home-made version, I
> simply use ((Yobs - Ypred)^2)/n. The equation might be correct because the
> mean is similar.
> (2) in the cv.glm, I have the impression the system is using glm0.dmi that
> was generated using all the data points whereas in my homemade version I
> only use the "test" database. I am confused if the cv.glm generates new glm
> models for each simulation of if it uses the one provided?
> (3) is the cv.glm sampling using replacement = TRUE or not?
> Thanks in advance.
> ***** cv.glm method
> # Simulation for 50 re-samplings...
> for (j in 1:50)
> cv.dmi<-cv.glm(data.dmi, glm0.dmi, K = 10)
> ***** homemade method
> # Brute-force cross-validation. This should be similar to the cv.glm
> perr2.vect <- vector()
> for(j in 1:50)
> select.dmi <- sample(1:nrow(data.dmi), 0.9*nrow(data.dmi))
> train.dmi <- data.dmi[select.dmi,] #Selecting 90% of the data for
> training purpose
> test.dmi <- data.dmi[-select.dmi,] #Selecting 10% (remaining) of the
> data for testing purpose
> glm1.dmi <- glm(DMI_kg~Sex+DOF+Avg_Nem+In_Wt, na.action=na.omit, data =
> #Create fitted values using test.dmi data
> dmi_pred <- predict.glm(glm1.dmi, test.dmi)
> # Get the prediction error = MSE
> perr2 <- t(dmi_obs - dmi_pred)%*%(dmi_obs - dmi_pred)/nrow(test.dmi)
> perr2.vect <- c(perr2.vect, perr2)
Brian D. Ripley, ripley at stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
More information about the R-help