[R] update.lm question

[Duncan Murdoch]

On 15/11/2009 11:28 AM, Duncan Murdoch wrote:
> On 15/11/2009 9:23 AM, Karsten Weinert wrote:
>> Hello,
>> at the Rgui command line I can easily remove a term from a fitted lm
>> object, like
>>
>> fit <- lm(y~x1+x2+x3, data=myData)
>> update(fit, .~.-x1)
>>
>> However, I would like to do this in a function with term given as string, like
>>
>> removeTerm <- function(linModel, termName) { ??? }
>> removeTerm(fit, "x1")
>>
>> but I can not fill the ???. I already tried
>>
>> removeTerm <- function(linModel, termName) { update(linModel, .~. - termName },
>> removeTerm <- function(linModel, termName) { update(linModel, .~. -
>> as.name(termName) },
>> removeTerm <- function(linModel, termName) { update(linModel, .~. -
>> eval(termName) },
>> removeTerm <- function(linModel, termName) { update(linModel, .~. -
>> eval.parent(termName) },
>> removeTerm <- function(linModel, termName) { update(linModel, .~. -
>> get(termName) },
>>
>> but these attempts produce error messages.
>>
>> Can you advise me here?
> 
> There are two problems:
> 
> 1. ".~." is different from ". ~ .".

Oops, wrong.  Those are the same.  Sorry...

Duncan Murdoch

> 
> 2.  You need to construct the formula ". ~ . - x1", and none of your 
> expressions do that.  You need to use substitute() or bquote() to edit a 
> formula. For example, I think both of these should work:
> 
> removeTerm <- function(linModel, termName)
>     update(linModel, bquote(. ~ . - .(as.name(termName))))
> 
> 
> removeTerm <- function(linModel, termName)
>     update(linModel, substitute(. ~ . - x, list(x=as.name(termName))))
> 
> Duncan Murdoch
> 
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