[Rd] Constructing logical expressions dynamically
murdoch at stats.uwo.ca
Thu Jul 3 02:38:00 CEST 2008
On 02/07/2008 7:27 PM, Tom Murray wrote:
> I am trying to construct a logical expression dynamically, for use in the
> subset() function. I am puzzled by problems with the code that follows
> Probably there's an easier way to select rows of a data frame according to
> some dynamic criteria--and I'd love to hear about it--but I'd also like to
> understand what I'm doing incorrectly in constructing the expression in my
> Thanks in advance,
> # These two expressions print the same, at least.
> e1 = expression(name == 'a')
e1 is an object of class "expression":
It prints looking like a function call because R doesn't know any other
way to print an expression.
> e2 = substitute(expression(cname == 'a'), list(cname=as.symbol('name')))
e2 is not, it's an unevaluated function call:
It really is a function call, as class() tells us.
If you evaluate it, you get something just like e1:
> print (e1)
> print (e2)
> # But they do not evaluate the same!
> # This works.
> data = data.frame(name=c('a', 'b', 'c', 'a'), age=c(1,2,3,4))
> d1 = subset(data, eval(e1))
> print (d1)
> # This gives the error:
> # Error in subset.data.frame(data, eval(e2)) :
> # 'subset' must evaluate to logical
> d2 = subset(data, eval(e2))
> print (d2)
The problem here is that eval(e2) is an expression; you need to evaluate
it to get the logical vector, i.e.
d2 <- subset(data, eval(eval(e2)))
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