[Rd] Capturing the expression representing the body of a function
Hi all, What's the preferred way of capturing the expression representing the contents of a function? * body(write.csv) gives me a braced expression * body(write.csv)[-1] gives me a mangled call * as.list(body(write.csv)[-1]) gives me a list of calls * as.expression(as.list(body(write.csv)[-1])) is what I want but seems like too much work Any suggestions? Thanks, Hadley -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University http://had.co.nz/ __ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
Re: [Rd] Capturing the expression representing the body of a function
On 02/05/2011 3:21 PM, Hadley Wickham wrote:
Hi all,
What's the preferred way of capturing the expression representing the
contents of a function?
* body(write.csv) gives me a braced expression
* body(write.csv)[-1] gives me a mangled call
* as.list(body(write.csv)[-1]) gives me a list of calls
* as.expression(as.list(body(write.csv)[-1])) is what I want but seems
like too much work
Any suggestions?
The body of a function isn't an expression, it's a language object. A
language object is represented internally as a pairlist, while an
expression is represented as a generic vector, i.e. the thing that
list() gives.
Your 1st try gives you the language object.
The other ones only work when the body consists of a call to `{`, as the
body of most complex functions does, but not for simple ones like
f - function(x) 2*x
So I would say your question should be: What's the best way to
construct an expression vector s.t. evaluating its elements in order is
like evaluating the body of a function?
And the answer to that is something like
body2expr - function(f) {
body - body(f)
if (typeof(body) == language identical(body[[1]], quote(`{`)))
as.expression(as.list(body[-1]))
else as.expression(body)
}
Duncan Murdoch
__
R-devel@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel
Re: [Rd] Capturing the expression representing the body of a function
On Mon, May 2, 2011 at 4:11 PM, Duncan Murdoch murdoch.dun...@gmail.com wrote:
On 02/05/2011 3:21 PM, Hadley Wickham wrote:
Hi all,
What's the preferred way of capturing the expression representing the
contents of a function?
* body(write.csv) gives me a braced expression
* body(write.csv)[-1] gives me a mangled call
* as.list(body(write.csv)[-1]) gives me a list of calls
* as.expression(as.list(body(write.csv)[-1])) is what I want but seems
like too much work
Any suggestions?
The body of a function isn't an expression, it's a language object. A
language object is represented internally as a pairlist, while an expression
is represented as a generic vector, i.e. the thing that list() gives.
Your 1st try gives you the language object.
The other ones only work when the body consists of a call to `{`, as the
body of most complex functions does, but not for simple ones like
f - function(x) 2*x
So I would say your question should be: What's the best way to construct
an expression vector s.t. evaluating its elements in order is like
evaluating the body of a function?
And the answer to that is something like
body2expr - function(f) {
body - body(f)
if (typeof(body) == language identical(body[[1]], quote(`{`)))
as.expression(as.list(body[-1]))
else as.expression(body)
}
Also try as.expression(as.list(f)[[3]])
e.g.
f - function(x, y) { x + y }
as.expression(as.list(f)[[3]])
expression({
x + y
})
g - function(x, y) x + y
as.expression(as.list(g)[[3]])
expression(x + y)
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
__
R-devel@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel
Re: [Rd] Capturing the expression representing the body of a function
The body of a function isn't an expression, it's a language object. A
language object is represented internally as a pairlist, while an expression
is represented as a generic vector, i.e. the thing that list() gives.
That doesn't agree with the documentation of is.language which implies
a language object is a ‘name’, a ‘call’, or an ‘expression’, and
doesn't mention pairlist anywhere.
Your 1st try gives you the language object.
It gives me a call, doesn't it? And the documentation for body
implies that this special type of call is called a bracketed
expression (despite it not being an expression as defined in the
documentation for expression)
(I don't mean to be nit-picky, I'm just trying to understand what's going on)
So I would say your question should be: What's the best way to construct
an expression vector s.t. evaluating its elements in order is like
evaluating the body of a function?
And the answer to that is something like
body2expr - function(f) {
body - body(f)
if (typeof(body) == language identical(body[[1]], quote(`{`)))
as.expression(as.list(body[-1]))
else as.expression(body)
}
Ok, thanks.
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/
__
R-devel@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel
