[R] drop unused levels in sqldf

[Gabor Grothendieck]

Here are two approaches:

1. rename the column to circumvent the column classing heuristic.  In
the following
Species will be a 3-level factor and Species2 will be a 2-level factor

sqldf("select *, Species as Species2 from iris where Species <> 'setosa'")

2. use method = "raw".  This code below returns Species as a "character" vector
and then we make it a factor manually:

out <- sqldf("select * from iris where Species <> 'setosa'", method = "raw")
out$Species <- as.factor(out$Species)

There is more discussion in the Heuristic section of the home page:
http://sqldf.googlecode.com

On Thu, Aug 28, 2008 at 4:36 AM, glaporta <glaporta at freeweb.org> wrote:
>
> Hi,
> sqldf is a fantastic package, but when the SELECT procedure runs unused
> levels remain in the output. I tried with the drop function, but without
> success. Do you have any suggestions?
> Thanx, Gianandrea
>
> data(iris)
> require(sqldf)
> base<-sqldf("select * from iris where Species <> 'setosa'")
> str(base) # Species with 3 levels!
>
> --
> View this message in context: http://www.nabble.com/drop-unused-levels-in-sqldf-tp19196464p19196464.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at 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.
>