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 @ gmail .com>
To: " Vadim Ogranovich " < vogranovich @ jumptrading .com>
Cc: r-help @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 @ 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 @ gmail .com>
> To: " Vadim Ogranovich " < vogranovich @ jumptrading .com>
> Cc: r-help @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 @ 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 @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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