From: Gabor Grothendieck <ggrothendieck_at_gmail.com>

Date: Fri, 18 May 2007 11:18:30 -0400

R-help_at_stat.math.ethz.ch 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. Received on Fri 18 May 2007 - 15:37:09 GMT

Date: Fri, 18 May 2007 11:18:30 -0400

Try this:

e <- quote(summary(y + z))

all.vars(e)

On 5/18/07, Vadim Ogranovich <vogranovich_at_jumptrading.com> wrote:

> Sorry, I didn't explain myself clear enough. I knew about the select arg in

*> subset(). My question was, given the expression expression(summary(x+y)),
**> how to extract all names that will be looked up during its evaluation.
**>
**> As to checking performance assumptions, you are right, in most cases the
**> overhead is negligible, but sometimes I work with really big data sets.
**>
**> Thanks a lot for your help,
**> Vadim
**>
**>
**> ----- Original Message -----
**> From: "Gabor Grothendieck" <ggrothendieck_at_gmail.com>
**> To: "Vadim Ogranovich" <vogranovich_at_jumptrading.com>
**> Cc: r-help_at_stat.math.ethz.ch
**> Sent: Friday, May 18, 2007 9:53:26 AM (GMT-0600) America/Chicago
**> Subject: Re: [R] subset arg in (modified) evalq
**>
**> I would check your performance assumption with an actual test before
**> concluding such but at any rate subset does have a select argument. See
**> ?subset
**>
**> On 5/18/07, Vadim Ogranovich <vogranovich_at_jumptrading.com> wrote:
**> > Thanks Gabor! This does exactly what I wanted.
**> >
**> > One follow-up question, how to extract the var names, in this case y, z,
**> > from the expression? The subset function creates a new object and this may
**> > be expensive when the data has a lot of irrelevant collumns. So I thougth
**> > that I could reduce this to the columns I actually need.
**> >
**> > Thanks,
**> > Vadim
**> >
**> >
**> >
**> > ----- Original Message -----
**> > From: "Gabor Grothendieck" <ggrothendieck_at_gmail.com>
**> > To: "Vadim Ogranovich" <vogranovich_at_jumptrading.com>
**> > Cc: r-help_at_stat.math.ethz.ch
**> > Sent: Friday, May 18, 2007 9:19:49 AM (GMT-0600) America/Chicago
**> > Subject: Re: [R] subset arg in (modified) evalq
**> >
**> > Try this:
**> >
**> > with(subset(data, x > 0), summary(y + z))
**> >
**> >
**> > On 5/18/07, Vadim Ogranovich <vogranovich_at_jumptrading.com> wrote:
**> > > Hi,
**> > >
**> > > When using evalq to evaluate expressions within a say data.frame context
**> I
**> > often wish there was a 'subset' argument, much like in lm() or any ather
**> > advanced regression model. I would be grateful for a tip how to do this.
**> > >
**> > > Here is an illustration of what I want:
**> > >
**> > > n <- 100
**> > > data <- data.frame(x=rnorm(n), y=rnorm(y), z=rnorm(z))
**> > >
**> > > # this works
**> > > evalq({ i <- 0<x; summary(y[i] + z[i]) }, data)
**> > >
**> > > # I want to do the above w/o explicit subscripting, e.g.
**> > > myevalq(summary(y + z), subset=0<x, data)
**> > >
**> > > Thanks,
**> > > Vadim
**> > >
**> > > [[alternative HTML version deleted]]
**> > >
**> > > ______________________________________________
**> > > R-help_at_stat.math.ethz.ch 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.
**> > >
**> >
**>
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