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:cbe...@tajo.ucsd.edu UC San Diego
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