[R] subset arg in (modified) evalq

[Gabor Grothendieck]

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]]
> >
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