R help - May 2010 - wildcard operator

Hi Ruser

As so usual I'm trying to replicate some SAS code. I wold like to know  
if there is a wildcard operators, as  " : " in SAS, in R?

When running:

lm(y ~ x1 + x2 + x3 + x4 + x5 + x6 .... x9860, data=mydata)

I would like to be able to get around it by just writing something  
like this:

lm(y ~ x1:x9860, data=mydata)

Anyone?

Sorry for no including a working example, but I figured that it wasn't  
necessary.

Thanks

Eric

Hi Eric,

Try

lm(y ~ . , data = mydata)

HTH,
Jorge


On Thu, May 27, 2010 at 12:58 AM, Eric Fail <> wrote:
> Hi Ruser
>
> As so usual I'm trying to replicate some SAS code. I wold like to know
if
> there is a wildcard operators, as  " : " in SAS, in R?
>
> When running:
>
> lm(y ~ x1 + x2 + x3 + x4 + x5 + x6 .... x9860, data=mydata)
>
> I would like to be able to get around it by just writing something like
> this:
>
> lm(y ~ x1:x9860, data=mydata)
>
> Anyone?
>
> Sorry for no including a working example, but I figured that it wasn't
> necessary.
>
> Thanks
>
> Eric
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
	[[alternative HTML version deleted]]

Eric Fail wrote:
> Hi Ruser
> 
> As so usual I'm trying to replicate some SAS code. I wold like to know 
> if there is a wildcard operators, as  " : " in SAS, in R?
> 
> When running:
> 
> lm(y ~ x1 + x2 + x3 + x4 + x5 + x6 .... x9860, data=mydata)
> 
> I would like to be able to get around it by just writing something like 
> this:
> 
> lm(y ~ x1:x9860, data=mydata)
>
See ?formula, specifically:

      There are two special interpretations of ?.? in a formula.  The
      usual one is in the context of a ?data? argument of model fitting
      functions and means ?all columns not otherwise in the formula?:
      see ?terms.formula?.  In the context of ?update.formula?, *only*,
      it means ?what was previously in this part of the formula?.


So assuming all those RHS variables + y make up your data.frame

lm(y ~ ., data = mydata) would be fine.  You can easily create such a data.frame
if you don't already have it using regexs.

--Erik

Hi:

On the off chance that your 9860 explanatory variables do not completely
take up the 'non-y'
part of your data frame, you can always write a formula string to pass to
the model
function. For example,
form <- paste('y ~ ', paste('x', 1:10, sep = '',
collapse = ' + '))
> form
[1] "y ~  x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10"
m <- lm(form, data = df)

HTH,
Dennis


On Wed, May 26, 2010 at 9:58 PM, Eric Fail <e@it.dk> wrote:
> Hi Ruser
>
> As so usual I'm trying to replicate some SAS code. I wold like to know
if
> there is a wildcard operators, as  " : " in SAS, in R?
>
> When running:
>
> lm(y ~ x1 + x2 + x3 + x4 + x5 + x6 .... x9860, data=mydata)
>
> I would like to be able to get around it by just writing something like
> this:
>
> lm(y ~ x1:x9860, data=mydata)
>
> Anyone?
>
> Sorry for no including a working example, but I figured that it wasn't
> necessary.
>
> Thanks
>
> Eric
>
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
	[[alternative HTML version deleted]]