Re: [R] Predict with loess

[Prof Brian Ripley]

On Tue, 13 Jun 2006, Jouanin Celine wrote:

   I want to do a  nonparametric regression. I’m using the function loess.
     The variable are  the year from 1968 to 1977 and the dependant variable is 
 a proportion P. The dependant variable have  missing value (NA).
     The script is :
     year <- 1969:2002
     length(year)
     [1] 34

     P <- c(NA,0.1,0.56,NA,NA,0.5,0.4,0.75,0.9,
     0.98,0.2,0.56,0.7,0.89,0.3,0.1,0.45,0.46,0.49,0.78,
     0.25,0.79,0.23,0.26,0.46,0.12,0.56,0.8,0.55,0.41,
     0.36,0.9,0.22,0.1)
     length(P)
     [1] 34

     lo1 <- loess(P~year,span=0.3,degree=1)
     summary(lo1)
     yearCo <- 1969:2002
     year_lo <- data.frame(year = yearCo )
     length(year_lo)
     [1] 34

I get 1 here, and so should you.

     mlo <- predict(loess(P~year,span=0.3,degree=1),new.data=year_lo,se=T)

It should be newdata, not new.data

     mlo$fit
     mlo$se.fit

Notice that these are of length 31, not 34

You are trying to predict at the values used for fitting (possibly not 
what you intended), so you don't actually need this.  Try

lo1 <- loess(P~year,span=0.3,degree=1, na.action=na.exclude)
fitted(lo1)
plot(year,P,type='o')
lines(year, fitted(lo1))

Or if you want to try interpolation

lines(year, predict(lo1, newdata=year_lo))

This will not extrapolate to 1969, and as far as I recall the version of 
loess in R does not allow extrapolation.

     plot(year,P,type='o')
     lines(year,predict(loess(P~year,span=0.15,degree=1),new.data=year_lo,
     se=T,na.action=na.omit)$fit,col='blue',type='l')

     The message error  indicates that x and y don’t have the same length.

     In fact in m$fit and m$se.fit there are 3 values who don’t have a 
fitted value.

Correct, and that's because you used na.action=na.omit and did not specify 
newdata.

[...]

--
Brian D. Ripley,                  [EMAIL PROTECTED]
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
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