[R] converting result of substitute to 'ordidnary' expression

[Charles C. Berry]

On Fri, 25 Jun 2010, Vadim Ogranovich wrote:

> Dear R users,
>
>
> As substitute() help page points out:
>     Substituting and quoting often causes confusion when the argument
>     is 'expression(...)'. The result is a call to the 'expression'
>     constructor function and needs to be evaluated with 'eval' to give
>     the actual expression object.
>
> And indeed I am confused. Consider:
>
>> dat <- data.frame(x=1:10, y=1:10)
>
>> subsetexp <- substitute(a<x, list(a=5))
>
> ## this doesn't work
>> subset(dat, subsetexp)
> Error in subset.data.frame(dat, subsetexp) :
>  'subset' must evaluate to logical
>
> ## this does work (thanks to the help page), but one needs to remember to call eval
>> subset(dat, eval(subsetexp))
>
>
> Is there a way to create subsetexp that needs no eval inside the call to subset()?

I do not think so. See

 	page(subset.data.frame,'print')

Then think about this:

> eval(substitute(subsetexp))
5 < x
> eval(substitute(subsetexp),list(x=2))
5 < x
> eval(substitute(eval(subsetexp)),list(x=2))
[1] FALSE
>

The added layer of substitution is making things a bit tricky.

One alternative is to build up your own call like this:

> sss <- expression(subset(dat,sbst))
> sss[[1]][[3]] <- subsetexp
> sss
expression(subset(dat, 5 < x))
> eval(sss)
     x  y
6   6  6
7   7  7
8   8  8
9   9  9
10 10 10
>

HTH,

Chuck


>
> I experimented with the following, but it didn't work:
>> subsetexp <- eval(substitute(a<x, list(a=5)))
> Error in eval(expr, envir, enclos) : object 'x' not found
>
> Thank you very much for your help,
> Vadim
>
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Charles C. Berry                            (858) 534-2098
                                             Dept of Family/Preventive Medicine
E mailto:cberry at tajo.ucsd.edu	            UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901