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

From: Charles C. Berry <>
Date: Fri, 25 Jun 2010 19:16:12 -0700

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, 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


Then think about this:

> eval(substitute(subsetexp))

5 < x
> eval(substitute(subsetexp),list(x=2))
5 < x
> eval(substitute(eval(subsetexp)),list(x=2))

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	            UC San Diego La Jolla, San Diego 92093-0901 mailing list PLEASE do read the posting guide and provide commented, minimal, self-contained, reproducible code. Received on Sat 26 Jun 2010 - 02:21:13 GMT

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