[R] using a condition given as string in subset function - how?

[William Dunlap]

> -----Original Message-----
> From: r-help-bounces at r-project.org 
> [mailto:r-help-bounces at r-project.org] On Behalf Of Albert-Jan Roskam
> Sent: Sunday, March 21, 2010 2:25 AM
> To: Mark Heckmann; jim holtman
> Cc: r-help at r-project.org
> Subject: Re: [R] using a condition given as string in subset 
> function - how?
> 
> Hi,
> 
> Is eval always used in conjunction with parse? Based on other 
> languages, I'd expect the expression already to work without 
> the use of parse(), but indeed it doesn't, or at least not as 
> intended. Just a newbie question..

When eval is given a parse tree (e.g., the output of
parse() or call()) it recursively ascends the tree,
effectively calling eval() on the subtrees.  E.g.,
parse(text='paste("No", 1)')[[1]] returns a "call" object
of length 3, the elements being the name object `paste`,
the character object "No", and the numeric object 1.
Evaluating a name object returns its value and evaluating
a character or numeric object returns the object (unchanged).
I.e., in this case
   eval(`paste`) -> function(..., sep=" ", collapse=NULL)
   eval("No") -> "No"
   eval(1) -> 1
The root of the parse tree is the call object and evaluating
that means applying the function given by the first argument
to the rest of the argumeents.

If eval("No") meant the same as eval(parse(text="No"))
then this scheme would break down, since you wouldn't be able
to distinguish between character objects and name objects,
so it would not be possible to say you wanted the string "No"
or the value of the object called "No".
 
You might say that eval("String") should do one thing when
called directly, whatever that means, and another when called
by a recursive call to eval(), but I think functions should
act the same no matter who calls them.

There are also times when you want the raw parse tree, as when
processing model formulae like log(response)~groupId/predictor
or trellis/lattice formulae like log(y)~predictor|groupId.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com  

> 
> Cheers!!
> 
> Albert-Jan
> 
> 
> 
> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
> 
> In the face of ambiguity, refuse the temptation to guess.
> 
> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
> 
> --- On Sun, 3/21/10, jim holtman <jholtman at gmail.com> wrote:
> 
> From: jim holtman <jholtman at gmail.com>
> Subject: Re: [R] using a condition given as string in subset 
> function - how?
> To: "Mark Heckmann" <mark.heckmann at gmx.de>
> Cc: r-help at r-project.org
> Date: Sunday, March 21, 2010, 2:33 AM
> 
> I know that if you have to resort to 'parse(text=...)', you 
> should look for
> another way (it is a 'fortune'), but it is getting late, and 
> at least it
> works:
> 
> > eval(parse(text="subset(df, A==1 & B==1)"))
> Â  A B
> 1 1 1
> 
> 
> On Sat, Mar 20, 2010 at 9:09 PM, Mark Heckmann 
> <mark.heckmann at gmx.de> wrote:
> 
> > df <- data.frame(A=c(1,2), B=c(1,1))
> >
> > I have a string containing a condition for a subset function, like:
> > conditionAsString <- paste(names(df), df[1,], sep="==", 
> collapse=" & ")
> >Â  > conditionAsString
> >Â  > "A==1 & B==1"
> >
> > Now I want to use this string in the subset call, like
> >
> > subset(df, conditionAsString)
> >
> > I do not exactly now how to combine substitute, expression, 
> parse and
> > so on to get what I want, which would be:
> >
> > subset(df, A==1 & B==1)
> >
> > but using the string conditionAsString.
> >
> > Thanks,
> > Mark
> > 
> ––––––––––––––––––––â€
“––––––––––––––––––
> > Mark Heckmann
> > Blog: www.markheckmann.de
> > R-Blog: http://ryouready.wordpress.com
> >
> >
> >
> >
> >
> >Â  Â  Â  Â  [[alternative HTML version deleted]]
> >
> >
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> > and provide commented, minimal, self-contained, reproducible code.
> >
> >
> 
> 
> -- 
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
> 
> What is the problem that you are trying to solve?
> 
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> 
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