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 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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