Re: [Rd] as.name and namespaces
Hi Pat,
You could use substitute(),
> mycall <- quote(list(lm(Y ~ x1), lm(Y ~ x2)))
> do.call("substitute", list(mycall, list(lm=quote(stats::lm
list(stats::lm(Y ~ x1), stats::lm(Y ~ x2))
The do.call is necessary because substitute() does not evaluate
its first argument and we want 'mycall' evaluated to become
the call to list(...) before substitute works on it.
substitute replaces all instances of a name in an expression.
bquote lets you be more selective (only names in .() get replaced):
> mycall2 <- quote(list(lm(Y ~ x1), .(lm)(Y ~ x2)))
> do.call("bquote", list(mycall2, list(lm=quote(stats::lm),
list=quote(base::list
list(lm(Y ~ x1), stats::lm(Y ~ x2))
S+'s substitute() has an evaluate=FALSE/TRUE argument to control
whether its first argument is evaluated thus letting you avoid do.call():
S+> mycall <- quote(list(lm(Y ~ x1), lm(Y ~ x2)))
S+> substitute(mycall, list(lm=quote(stats::lm)), evaluate=TRUE)
list(stats::lm(Y ~ x1), stats::lm(Y ~ x2))
It is much harder if you want to find and replace expressions more general
than name, e.g., changing "stats::lm" to "lm" or "log(x+1)" to "logp1(x)" .
package::codetools and package::compiler might help. I have used rapply()
in S+ for this sort of thing but R's rapply() does not work on functions:
S+> f <- function(x, y) exp(log(x+1) + log(abs(y)+1) + log(z))
S+> fNew <- rapply(f, how="replace", classes="call",
function(e) {
if (identical(e[[1]], quote(log)) && is.call(logArg <- e[[2]])) {
if (identical(logArg[[1]], quote(`+`)) && identical(logArg[[3]],
1)) {
e <- call("logp1", logArg[[2]])
}
}
e
})
S+> fNew
function(x, y)
exp(logp1(x) + logp1(abs(y)) + log(z))
The is pretty ugly code but it did help us quickly install optimizations in
a large mass of code.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-devel-boun...@r-project.org [mailto:r-devel-boun...@r-project.org] On
> Behalf
> Of Patrick Burns
> Sent: Wednesday, April 24, 2013 2:30 AM
> To: Duncan Murdoch
> Cc: r-devel
> Subject: Re: [Rd] as.name and namespaces
>
> Here is an example problem:
>
> > mycall <- expression(lm(Y ~ x))[[1]]
> > mycall
> lm(Y ~ x)
> > newname <- "stats::lm"
>
> > desiredResult
> stats::lm(Y ~ x)
>
> I've solved the problem in the kludgy way of
> deparsing, fixing the string and then parsing.
>
> I like Duncan's third method, but it seems like
> it assumes the solution. Moving functions around
> is unappetizing for my use -- this is for testing
> and keeping things as faithful to real use is a
> good thing.
>
> Pat
>
>
> On 23/04/2013 21:18, Duncan Murdoch wrote:
> > On 13-04-23 3:51 PM, Patrick Burns wrote:
> >> Okay, that's a good reason why it shouldn't.
> >>
> >> Why it should is that I want to substitute
> >> the first element of a call to be a function
> >> including the namespace.
> >
> > Three ways:
> >
> > 1. Assign the function from the namespace locally, then call the local
> > one.
> > 2. Import the function in your NAMESPACE (if you know the name in
> > advance).
> > 3. Construct an expression involving ::, and substitute that in.
> >
> > For example:
> >
> > substitute(foo(x), list(foo=quote(baz::bar)))
> >
> > Duncan Murdoch
> >
> >>
> >> Pat
> >>
> >>
> >> On 23/04/2013 18:32, peter dalgaard wrote:
> >>>
> >>> On Apr 23, 2013, at 19:23 , Patrick Burns wrote:
> >>>
> >>>> 'as.name' doesn't recognize a name with
> >>>> its namespace extension as a name:
> >>>>
> >>>>> as.name("lm")
> >>>> lm
> >>>>> as.name("stats::lm")
> >>>> `stats::lm`
> >>>>> as.name("stats:::lm")
> >>>> `stats:::lm`
> >>>>
> >>>>
> >>>> Is there a reason why it shouldn't?
> >>>
> >>> Any reason why it should? :: and ::: are operators. foo$bar is not
> >>> the same as `foo$bar` either.
> >>>
> >>
> >
> >
>
> --
> Patrick Burns
> pbu...@pburns.seanet.com
> twitter: @burnsstat @portfolioprobe
> http://www.portfolioprobe.com/blog
> http://www.burns-stat.com
> (home of:
> 'Impatient R'
> 'The R Inferno'
> 'Tao Te Programming')
>
> __
> R-devel@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel
__
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Re: [Rd] as.name and namespaces
On Wed, Apr 24, 2013 at 4:29 AM, Patrick Burns wrote: > Here is an example problem: > >> mycall <- expression(lm(Y ~ x))[[1]] >> mycall > lm(Y ~ x) >> newname <- "stats::lm" > >> desiredResult > stats::lm(Y ~ x) > > I've solved the problem in the kludgy way of > deparsing, fixing the string and then parsing. I'm working on a comprehensive write up of computing on the language at the moment. It's still a bit rough, but you might find https://github.com/hadley/devtools/wiki/Expressions#structure-of-expressions and https://github.com/hadley/devtools/wiki/Expressions#calls to be useful. Hadley -- Chief Scientist, RStudio http://had.co.nz/ __ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
Re: [Rd] as.name and namespaces
There's also the brute force option: > mycall <- expression(lm(Y ~ x))[[1]] > mycall[[1]] <- quote(stats::lm) > mycall stats::lm(Y ~ x) > eval(mycall, list(Y=rnorm(5),x=1:5)) Call: stats::lm(formula = Y ~ x) Coefficients: (Intercept)x 0.07430 0.02981 On Apr 24, 2013, at 11:29 , Patrick Burns wrote: > Here is an example problem: > > > mycall <- expression(lm(Y ~ x))[[1]] > > mycall > lm(Y ~ x) > > newname <- "stats::lm" > > > desiredResult > stats::lm(Y ~ x) > > I've solved the problem in the kludgy way of > deparsing, fixing the string and then parsing. > > I like Duncan's third method, but it seems like > it assumes the solution. Moving functions around > is unappetizing for my use -- this is for testing > and keeping things as faithful to real use is a > good thing. > > Pat -- Peter Dalgaard, Professor Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
Re: [Rd] as.name and namespaces
Here is an example problem:
> mycall <- expression(lm(Y ~ x))[[1]]
> mycall
lm(Y ~ x)
> newname <- "stats::lm"
> desiredResult
stats::lm(Y ~ x)
I've solved the problem in the kludgy way of
deparsing, fixing the string and then parsing.
I like Duncan's third method, but it seems like
it assumes the solution. Moving functions around
is unappetizing for my use -- this is for testing
and keeping things as faithful to real use is a
good thing.
Pat
On 23/04/2013 21:18, Duncan Murdoch wrote:
On 13-04-23 3:51 PM, Patrick Burns wrote:
Okay, that's a good reason why it shouldn't.
Why it should is that I want to substitute
the first element of a call to be a function
including the namespace.
Three ways:
1. Assign the function from the namespace locally, then call the local
one.
2. Import the function in your NAMESPACE (if you know the name in
advance).
3. Construct an expression involving ::, and substitute that in.
For example:
substitute(foo(x), list(foo=quote(baz::bar)))
Duncan Murdoch
Pat
On 23/04/2013 18:32, peter dalgaard wrote:
On Apr 23, 2013, at 19:23 , Patrick Burns wrote:
'as.name' doesn't recognize a name with
its namespace extension as a name:
as.name("lm")
lm
as.name("stats::lm")
`stats::lm`
as.name("stats:::lm")
`stats:::lm`
Is there a reason why it shouldn't?
Any reason why it should? :: and ::: are operators. foo$bar is not
the same as `foo$bar` either.
--
Patrick Burns
pbu...@pburns.seanet.com
twitter: @burnsstat @portfolioprobe
http://www.portfolioprobe.com/blog
http://www.burns-stat.com
(home of:
'Impatient R'
'The R Inferno'
'Tao Te Programming')
__
R-devel@r-project.org mailing list
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Re: [Rd] as.name and namespaces
On 13-04-23 3:51 PM, Patrick Burns wrote:
Okay, that's a good reason why it shouldn't.
Why it should is that I want to substitute
the first element of a call to be a function
including the namespace.
Three ways:
1. Assign the function from the namespace locally, then call the local one.
2. Import the function in your NAMESPACE (if you know the name in advance).
3. Construct an expression involving ::, and substitute that in.
For example:
substitute(foo(x), list(foo=quote(baz::bar)))
Duncan Murdoch
Pat
On 23/04/2013 18:32, peter dalgaard wrote:
On Apr 23, 2013, at 19:23 , Patrick Burns wrote:
'as.name' doesn't recognize a name with
its namespace extension as a name:
as.name("lm")
lm
as.name("stats::lm")
`stats::lm`
as.name("stats:::lm")
`stats:::lm`
Is there a reason why it shouldn't?
Any reason why it should? :: and ::: are operators. foo$bar is not the same as
`foo$bar` either.
__
R-devel@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel
Re: [Rd] as.name and namespaces
On Apr 23, 2013, at 21:51 , Patrick Burns wrote:
> Okay, that's a good reason why it shouldn't.
>
> Why it should is that I want to substitute
> the first element of a call to be a function
> including the namespace.
Offhand, I'd say that it shouldn't be a problem, but do you have a more
concrete example?
-pd
>
> Pat
>
>
> On 23/04/2013 18:32, peter dalgaard wrote:
>>
>> On Apr 23, 2013, at 19:23 , Patrick Burns wrote:
>>
>>> 'as.name' doesn't recognize a name with
>>> its namespace extension as a name:
>>>
as.name("lm")
>>> lm
as.name("stats::lm")
>>> `stats::lm`
as.name("stats:::lm")
>>> `stats:::lm`
>>>
>>>
>>> Is there a reason why it shouldn't?
>>
>> Any reason why it should? :: and ::: are operators. foo$bar is not the same
>> as `foo$bar` either.
>>
>
> --
> Patrick Burns
> pbu...@pburns.seanet.com
> twitter: @burnsstat @portfolioprobe
> http://www.portfolioprobe.com/blog
> http://www.burns-stat.com
> (home of:
> 'Impatient R'
> 'The R Inferno'
> 'Tao Te Programming')
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk Priv: pda...@gmail.com
__
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Re: [Rd] as.name and namespaces
Okay, that's a good reason why it shouldn't.
Why it should is that I want to substitute
the first element of a call to be a function
including the namespace.
Pat
On 23/04/2013 18:32, peter dalgaard wrote:
On Apr 23, 2013, at 19:23 , Patrick Burns wrote:
'as.name' doesn't recognize a name with
its namespace extension as a name:
as.name("lm")
lm
as.name("stats::lm")
`stats::lm`
as.name("stats:::lm")
`stats:::lm`
Is there a reason why it shouldn't?
Any reason why it should? :: and ::: are operators. foo$bar is not the same as
`foo$bar` either.
--
Patrick Burns
pbu...@pburns.seanet.com
twitter: @burnsstat @portfolioprobe
http://www.portfolioprobe.com/blog
http://www.burns-stat.com
(home of:
'Impatient R'
'The R Inferno'
'Tao Te Programming')
__
R-devel@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel
Re: [Rd] as.name and namespaces
On Apr 23, 2013, at 19:23 , Patrick Burns wrote:
> 'as.name' doesn't recognize a name with
> its namespace extension as a name:
>
> > as.name("lm")
> lm
> > as.name("stats::lm")
> `stats::lm`
> > as.name("stats:::lm")
> `stats:::lm`
>
>
> Is there a reason why it shouldn't?
Any reason why it should? :: and ::: are operators. foo$bar is not the same as
`foo$bar` either.
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk Priv: pda...@gmail.com
__
R-devel@r-project.org mailing list
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