thanks! using update( ) in a function?

On Mon, 21 Apr 2003, Roger Peng wrote:

 

I believe this is bug PR#1861.  Not sure what the workaround is.
   

No, the code as given doesn't get as far as that

 

On Mon, 21 Apr 2003, Mahmood ARAI wrote:

   

Hello,

using the function "update()" in the following function
does not work.

# a function for running a model:  x1 ~ x2
# by updating the original model: fm1 <- lm(y ~ x1 + x2,data=df)
# update(fm1, x1 ~ . -x1, data=df) works at the command line
# but not in the function below.


R>   foo <- function(fm,x,df)
+     {
+     update(fm, x ~ . -x,data=df)
+     }
R>
R>
R>  df1 <- data.frame(y= 1:10, x1=(1:10)^2, x2=sqrt(1:10)^3)
R>  fm1 <- lm(y ~ x1+x2, data=df1)
R>
R>  foo(fm1, x1, df1)
Error in eval(expr, envir, enclos) : Object "x" not found
     

The main problem here is that the argument `x' is only going to have the
*value* of x1, where you need the symbol x1.  On top of that,
update(fm, x ~ . -x,data=df)
uses the symbol `x' rather than its value, so even if its value were
correct it wouldn't work.

This works
 

foo<-function(model,term,data){
   

+ ff<-substitute(term~.-term)
+ update(model,ff,data=data)
+ }
 

data(trees)
a<-lm(Girth~Height+Volume,data=trees)
foo(a,Height,data=trees)
   

Call:
lm(formula = Height ~ Volume, data = data)

Coefficients:
(Intercept)       Volume
   69.0034       0.2319

but is not recommended style, as you are defeating the call-by-value
semantics of R.  A better version is
 

bar<-function(model,term,data){
   

+ ff<-substitute(x~.-x,list(x=as.name(term)))
+  update(model,ff,data=data)
+ }
 

bar(a,"Height",data=trees)
   

Call:
lm(formula = Height ~ Volume, data = data)

Coefficients:
(Intercept)       Volume
   69.0034       0.2319

and you could have versions that allowed `term' to be a quoted name or
expression rather than a string.

-thomas