[R] Coercing a list of variables in a function call
Dimitris Rizopoulos
d.rizopoulos at erasmusmc.nl
Mon Dec 1 13:21:49 CET 2008
try the following:
model.it <- function (form, data, factor.id) {
if (!missing(factor.id) && all(factor.id %in% names(data)))
data[factor.id] <- lapply(data[factor.id], factor)
glm(form, data = data)
}
y <- c(1,2.1,3.3,4.3,5,6.5)
x1 <- c(1,1,1,2,2,2)
x2 <- c(1,2,3,1,2,3)
x3 <- c(1,3,2,4,2,4)
d <- data.frame(y, x1, x2, x3)
model.it(y ~ x1 + x2 + x3, d, c("x1", "x2"))
model.it(y ~ x1 + x2 + x3, d)
If you only plan to fit linear regression models, then it is advisable
that you lm() instead of glm(). I hope it helps.
Best,
Dimitris
Philip Whittall wrote:
> This is hopefully a trivial problem for list subscribers, but I am very
> new to writing R functions.
> I wish to call an R function written by myself from another program to
> fit a model. I need
> to tell it which of the independent variables are factors. I need to do
> this in a generic way,
> so that when the list is passed, R will work through the variables in
> the data frame and coerce them into being factors.
>
> This is what I tried ...
>
> # Fictitious data for illustration
> y<-c(1,2.1,3.3,4.3,5,6.5)
> x1<-c(1,1,1,2,2,2)
> x2<-c(1,2,3,1,2,3)
> x3<-c(1,3,2,4,2,4)
> d<-as.data.frame(cbind(y,x1,x2,x3))
> #Function definition
> model_it=function(data=" ", model=" ", factors=list()){
> attach(d) #Attach the data
> formula<-as.formula(model) #Set up the model
> #Identify the factors
> if (length(factors)>0) for(i in 1:length(factors)){
> factors[i]<-as.factor(factors[i])
> }
> #Fit the model
> glm(formula, data=data)
> }
> #Function call attempted
> model_it(data=d, model="y~x1+x2+x3", factors=list(x1,x2))
>
> And this is the error message I received ...
>
> Error in sort.list(unique.default(x), na.last = TRUE) :
> 'x' must be atomic for 'sort.list'
> Have you called 'sort' on a list?
>
> The line factors[i]<-as.factor(factors[i]) is the one responsible for
> the error.
>>From the word 'atomic', my guess is that I have asked it to set each
> value as a factor rather than the variable, but I don't know
> what syntax to use to get it to set the variable as a factor. Any
> suggestions gratefully appreciated.
>
> For the record, the above was run on windows XP using R2.8.0
>
>
>
>
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--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
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