R help - Jun 2010 - converting result of substitute to 'ordidnary' expression

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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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
>
> Note: This email is for the confidential use of the named addressee(s) only
and may contain proprietary, confidential or privileged information. If you are
not the intended recipient, you are hereby notified that any review,
dissemination or copying of this email is strictly prohibited, and to please
notify the sender immediately and destroy this email and any attachments.  Email
transmission cannot be guaranteed to be secure or error-free.  Jump Trading,
therefore, does not make any guarantees as to the completeness or accuracy of
this email or any attachments.  This email is for informational purposes only
and does not constitute a recommendation, offer, request or solicitation of any
kind to buy, sell, subscribe, redeem or perform any type of transaction of a
financial product.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
Charles C. Berry                            (858) 534-2098
                                             Dept of Family/Preventive Medicine
E mailto:cberry at tajo.ucsd.edu	            UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901

On Fri, Jun 25, 2010 at 8:59 PM, Vadim Ogranovich
<vogranovich at jumptrading.com> 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))
>
>
If its ok to call it this way then it works without the eval:

do.call(subset, list(dat, subsetexp))

I switched the thread to r-devel because here I am proposing a patch for
subset.data.frame().


Thank you Chuck, it was inspiring. It turns out that a simple modification to
subset.data.frame makes my example work:

subset.data.frame <-
  function (x, subset, select, drop = FALSE, ...)
{
  if (missing(subset))
    r <- TRUE
  else {
    r <- eval(substitute(subset), x, parent.frame())

    if (!is.logical(r)) {
      ## try w/o substitute
      r <- eval(subset, x, parent.frame())
    }

    if (!is.logical(r))
      stop("'subset' must evaluate to logical")

    r <- r & !is.na(r)
  }
  if (missing(select))
    vars <- TRUE
  else {
    nl <- as.list(1L:ncol(x))
    names(nl) <- names(x)
    vars <- eval(substitute(select), nl, parent.frame())
  }
  x[r, vars, drop = drop]
}


And now:
> dat <- data.frame(x=1:10, y=1:10)
> subset(dat, 5<x)
    x  y
6   6  6
7   7  7
8   8  8
9   9  9
10 10 10
> subsetexp <- expression(5<x)
> subset(dat, subsetexp)
    x  y
6   6  6
7   7  7
8   8  8
9   9  9
10 10 10

> do.call(subset, list(dat, subsetexp))
    x  y
6   6  6
7   7  7
8   8  8
9   9  9
10 10 10
> version
               _
platform       i386-pc-mingw32
arch           i386
os             mingw32
system         i386, mingw32
status
major          2
minor          9.1
year           2009
month          06
day            26
svn rev        48839
language       R
version.string R version 2.9.1 (2009-06-26)


Thank you,
Vadim



-----Original Message-----
From: Charles C. Berry [mailto:cberry at tajo.ucsd.edu]
Sent: Friday, June 25, 2010 9:16 PM
To: Vadim Ogranovich
Cc: 'r-help at r-project.org'
Subject: Re: [R] converting result of substitute to 'ordidnary'
expression

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
>
> Note: This email is for the confidential use of the named addressee(s) only
and may contain proprietary, confidential or privileged information. If you are
not the intended recipient, you are hereby notified that any review,
dissemination or copying of this email is strictly prohibited, and to please
notify the sender immediately and destroy this email and any attachments.  Email
transmission cannot be guaranteed to be secure or error-free.  Jump Trading,
therefore, does not make any guarantees as to the completeness or accuracy of
this email or any attachments.  This email is for informational purposes only
and does not constitute a recommendation, offer, request or solicitation of any
kind to buy, sell, subscribe, redeem or perform any type of transaction of a
financial product.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
Charles C. Berry                            (858) 534-2098
                                             Dept of Family/Preventive Medicine
E mailto:cberry at tajo.ucsd.edu               UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901



Note: This email is for the confidential use of the named addressee(s) only and
may contain proprietary, confidential or privileged information. If you are not
the intended recipient, you are hereby notified that any review, dissemination
or copying of this email is strictly prohibited, and to please notify the sender
immediately and destroy this email and any attachments.  Email transmission
cannot be guaranteed to be secure or error-free.  Jump Trading, therefore, does
not make any guarantees as to the completeness or accuracy of this email or any
attachments.  This email is for informational purposes only and does not
constitute a recommendation, offer, request or solicitation of any kind to buy,
sell, subscribe, redeem or perform any type of transaction of a financial
product.