[R] can't find 'predict.loess' in a loop

[Prof Brian Ripley]

On 22/12/2011 08:55, Uwe Ligges wrote:
> See methods(predict) and find that methods are hidden in the Namespaces
> and are ought to be accessed by the generic. If you really want to
> access it explicitly (which i not intended by design), see help(":::")
>
> Your code is not reproducible, hence we cannot help in detail.

Correct, but we can guess.  Look at

 >> + Lag.lo<-loess(RBj~RELBj,span=sp, family=("symmetric"), degree=2)
 >> + Lag.pr<-predict(Lag.lo, data.frame(x=seq (-80,80,1)))

Shame about the unreadability (broken space bar?), but your model is

RBj ~ RELBj

and your newdata contains 'x' not 'RELBj'.  Since you didn't use a data 
frame for the fit, it finds the globally visible RELBj ....

Nothing to do with loess, predict.loess or loops ....

> Uwe Ligges
>
>
>
>
>
> On 22.12.2011 09:39, Stuart Leask wrote:
>>
>> Season's Greetings.
>>
>> I am using 'predict' in a loop, but it is behaving unexpectedly, so I
>> am more explicit and use 'predict.loess', and R says it can't find it.
>> Even if I explicitly load the stats library, which is already in the
>> search path.
>>
>> Any ideas?
>> This is odd, because if I use RB& LB from a different dataset, it
>> works fine...
>>
>>
>> Stuart
>>
>> *********************************************************************
>> *************************
>> Original problem:
>>
>>> j<-0 # start an iteration counter
>>> while(j<30) # (max 30 iterations)
>> +
>> + {
>> +
>> + lenRB<-length(RB) # number of rows in original RB
>> + lenLagRB<-(j+1):lenRB # row numbers for 'lagged' RB
>> + lenLagRB0<-1:(lenRB-j) # row numbers for 'unlagged' RB
>> + RBj<-RB[lenLagRB] # create a 'lagged' RB of this length
>> + LBj<-LB[lenLagRB] # create a 'lagged' LB of this length
>> + RB0<-RB[lenLagRB0] # create an 'unlagged', shortened RB of this length
>> + LB0<-LB[lenLagRB0] # create an 'unlagged' LB of this length
>> +
>> + cor.j<-cor(RBj,LB0) # check RB lagged& LB have low correlation
>> + if(abs(cor.j)<0.01) # if so...
>> +
>> + {
>> +
>> + RELBj<-100*(RBj-LB0)/(RBj+LB0) # ...generate index for lagged pair
>> + RELB0<-100*(RB0-LB0)/(RB0+LB0) # generate index for unlagged pair
>> +
>> + Lag.lo<-loess(RBj~RELBj,span=sp, family=("symmetric"), degree=2) #
>> close approx of lowess!
>> + Real.lo<-loess(RB0~RELB0,span=sp, family=("symmetric"), degree=2)
>> + Lag.pr<-predict(Lag.lo, data.frame(x=seq (-80,80,1))) # predict for
>> uniform index scale
>> + Real.pr<-predict(Real.lo,data.frame(x=seq (-80,80,1)))
>> +
>> + Diff.pr<-Real.pr-Lag.pr # difference plot
>> + lines (Diff.pr~seq(-80,80,1),pch=1,cex=0.25,col=1)
>> +
>> + }
>> +
>> + j<-j+1 # increment counter and repeat
>> + }
>> Error in model.frame.default(formula = Diff.pr ~ seq(-80, 80, 1),
>> na.action = NULL) :
>> variable lengths differ (found for 'seq(-80, 80, 1)')
>> In addition: Warning messages:
>> 1: 'newdata' had 161 rows but variable(s) found have 250 rows
>> 2: 'newdata' had 161 rows but variable(s) found have 250 rows
>>
>> ****************************
>> Make it more explicit I want to use predict.loess:
>>
>>> j<-0 # start an iteration counter
>>> while(j<30) # (max 30 iterations)
>> +
>> + {
>> +
>> + lenRB<-length(RB) # number of rows in original RB
>> + lenLagRB<-(j+1):lenRB # row numbers for 'lagged' RB
>> + lenLagRB0<-1:(lenRB-j) # row numbers for 'unlagged' RB
>> + RBj<-RB[lenLagRB] # create a 'lagged' RB of this length
>> + LBj<-LB[lenLagRB] # create a 'lagged' LB of this length
>> + RB0<-RB[lenLagRB0] # create an 'unlagged', shortened RB of this length
>> + LB0<-LB[lenLagRB0] # create an 'unlagged' LB of this length
>> +
>> + cor.j<-cor(RBj,LB0) # check RB lagged& LB have low correlation
>> + if(abs(cor.j)<0.01) # if so...
>> +
>> + {
>> +
>> + RELBj<-100*(RBj-LB0)/(RBj+LB0) # ...generate index for lagged pair
>> + RELB0<-100*(RB0-LB0)/(RB0+LB0) # generate index for unlagged pair
>> +
>> + Lag.lo<-loess(RBj~RELBj,span=sp, family=("symmetric"), degree=2) #
>> close approx of lowess!
>> + Real.lo<-loess(RB0~RELB0,span=sp, family=("symmetric"), degree=2)
>> + Lag.pr<-predict.loess(Lag.lo, data.frame(x=seq (-80,80,1))) #
>> predict for uniform index scale
>> + Real.pr<-predict.loess(Real.lo,data.frame(x=seq (-80,80,1)))
>> +
>> + Diff.pr<-Real.pr-Lag.pr # difference plot
>> + lines (Diff.pr~seq(-80,80,1),pch=1,cex=0.25,col=1)
>> +
>> + }
>> +
>> + j<-j+1 # increment counter and repeat
>> + }
>> Error: could not find function "predict.loess"
>>
>> ****************
>>
>>> search()
>> [1] ".GlobalEnv" "nshd"
>> [3] "package:foreign" "package:stats"
>> [5] "package:graphics" "package:grDevices"
>> [7] "package:utils" "package:datasets"
>> [9] "package:methods" "Autoloads"
>> [11] "package:base"
>>> ?predict.loess
>>
>> ... Confirms that predict.loess is in the path, and can be accessed.
>> Only not within the loop?
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-- 
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)
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