Re: [R] Help needed understanding eval,quote,expression
You are missing eval(parse(text=)). E.g.
x - list(y=list(y1=hello,y2=world),z=list(z1=foo,z2=bar))
(what do you mean by the $ at the start of these lines?)
eval(parse(text=x$y$y1))
[1] hello
However, bear in mind
fortune(parse)
If the answer is parse() you should usually rethink the question.
-- Thomas Lumley
R-help (February 2005)
In your indicated example you could probably use substitute() as
effectively.
On Wed, 28 Jun 2006, [EMAIL PROTECTED] wrote:
I am trying to build up a quoted or character expression representing a
component in a list in order to reference it indirectly.
For instance, I have a list that has data I want to pull, and another list
that has character vectors and/or lists of characters containing the names
of the components in the first list.
It seems that the way to do this is as evaluating expressions, but I seem
to be missing something. The concept should be similar to the snippet
below:
For instance:
$x = list(y=list(y1=hello,y2=world),z=list(z1=foo,z2=bar))
$y = quote(x$y$y1)
$eval(y)
[1] hello
but, I'm trying to accomplish this by building up y as a character and
then evaluating it, and having no success.
$y1=paste(x$y$,y1,sep=)
$y1
[1] x$y$y1
How can I evaluate y1 as I did with y previously? or can I?
Much Thanks !
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--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] Help needed understanding eval,quote,expression
Prof Brian Ripley wrote: You are missing eval(parse(text=)). E.g. x - list(y=list(y1=hello,y2=world),z=list(z1=foo,z2=bar)) (what do you mean by the $ at the start of these lines?) eval(parse(text=x$y$y1)) [1] hello However, bear in mind fortune(parse) If the answer is parse() you should usually rethink the question. -- Thomas Lumley R-help (February 2005) In your indicated example you could probably use substitute() as effectively. On Wed, 28 Jun 2006, [EMAIL PROTECTED] wrote: I am trying to build up a quoted or character expression representing a component in a list in order to reference it indirectly. For instance, I have a list that has data I want to pull, and another list that has character vectors and/or lists of characters containing the names of the components in the first list. It seems that the way to do this is as evaluating expressions, but I seem to be missing something. The concept should be similar to the snippet below: For instance: $x = list(y=list(y1=hello,y2=world),z=list(z1=foo,z2=bar)) $y = quote(x$y$y1) $eval(y) [1] hello but, I'm trying to accomplish this by building up y as a character and then evaluating it, and having no success. $y1=paste(x$y$,y1,sep=) $y1 [1] x$y$y1 How can I evaluate y1 as I did with y previously? or can I? Much Thanks ! if I understand you correctly you can achieve your goal much easier than with eval, parse, substitute and the like: x - list(y=list(y1=hello,y2=world),z=list(z1=foo,z2=bar)) s1 - 'y' s2 - 'y1' x[[s1]][[s2]] i.e. using `[[' instead of `$' for list component extraction allows to use characters for indexing (in other words: x$y == x[['y']]) joerg __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Help needed understanding eval,quote,expression
On Thu, 29 Jun 2006, Joerg van den Hoff wrote: Prof Brian Ripley wrote: You are missing eval(parse(text=)). E.g. x - list(y=list(y1=hello,y2=world),z=list(z1=foo,z2=bar)) (what do you mean by the $ at the start of these lines?) eval(parse(text=x$y$y1)) [1] hello However, bear in mind fortune(parse) If the answer is parse() you should usually rethink the question. -- Thomas Lumley R-help (February 2005) In your indicated example you could probably use substitute() as effectively. On Wed, 28 Jun 2006, [EMAIL PROTECTED] wrote: I am trying to build up a quoted or character expression representing a component in a list in order to reference it indirectly. For instance, I have a list that has data I want to pull, and another list that has character vectors and/or lists of characters containing the names of the components in the first list. It seems that the way to do this is as evaluating expressions, but I seem to be missing something. The concept should be similar to the snippet below: For instance: $x = list(y=list(y1=hello,y2=world),z=list(z1=foo,z2=bar)) $y = quote(x$y$y1) $eval(y) [1] hello but, I'm trying to accomplish this by building up y as a character and then evaluating it, and having no success. $y1=paste(x$y$,y1,sep=) $y1 [1] x$y$y1 How can I evaluate y1 as I did with y previously? or can I? Much Thanks ! if I understand you correctly you can achieve your goal much easier than with eval, parse, substitute and the like: x - list(y=list(y1=hello,y2=world),z=list(z1=foo,z2=bar)) s1 - 'y' s2 - 'y1' x[[s1]][[s2]] i.e. using `[[' instead of `$' for list component extraction allows to use characters for indexing (in other words: x$y == x[['y']]) But what he actually asked for was I am trying to build up a quoted or character expression representing a component in a list in order to reference it indirectly. You just typed in x[[s1]][[s2]], not 'built [it] up'. Suppose the specification had been r - x s - c(y, y1) and s was of variable length? Then you need to construct a call similar to x[[y]][[y1]] from r and s. [There was another reason for sticking with $ rather than using [[: the latter makes unnecessary copies in released versions of R.] -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Help needed understanding eval,quote,expression
Great help, thanks to both of you. I actually think I get it.
I think I was locked into eval(expression ... ) as the solution. I did
search the archives for this question, but it must not have clicked with
me.The Thomas Lumley R-help (February 2005) was on the money. I was
missing the power flexibility of the [[ operation. It is certainly more
direct.
I do believe this pattern will satisfy my original problem.
x = list(y=list(y1=hello,y2=world),z=list(z1=foo,z2=bar))
ids = list( zIds=c(z1,z2),yIds=c(y1,y2))
str=y1
x$y[[str]]
[1] hello
btw: I set my prompt to $, so the first post, the $ at the beginning of
the line was the prompt. apologies for the confusion.
cheers.
Prof Brian Ripley [EMAIL PROTECTED]
06/29/2006 04:06 AM
To
Joerg van den Hoff [EMAIL PROTECTED]
cc
[EMAIL PROTECTED], r-help@stat.math.ethz.ch
Subject
Re: [R] Help needed understanding eval,quote,expression
On Thu, 29 Jun 2006, Joerg van den Hoff wrote:
Prof Brian Ripley wrote:
You are missing eval(parse(text=)). E.g.
x - list(y=list(y1=hello,y2=world),z=list(z1=foo,z2=bar))
(what do you mean by the $ at the start of these lines?)
eval(parse(text=x$y$y1))
[1] hello
However, bear in mind
fortune(parse)
If the answer is parse() you should usually rethink the question.
-- Thomas Lumley
R-help (February 2005)
In your indicated example you could probably use substitute() as
effectively.
On Wed, 28 Jun 2006, [EMAIL PROTECTED] wrote:
I am trying to build up a quoted or character expression representing
a
component in a list in order to reference it indirectly.
For instance, I have a list that has data I want to pull, and another
list
that has character vectors and/or lists of characters containing the
names
of the components in the first list.
It seems that the way to do this is as evaluating expressions, but I
seem
to be missing something. The concept should be similar to the snippet
below:
For instance:
$x = list(y=list(y1=hello,y2=world),z=list(z1=foo,z2=bar))
$y = quote(x$y$y1)
$eval(y)
[1] hello
but, I'm trying to accomplish this by building up y as a character and
then evaluating it, and having no success.
$y1=paste(x$y$,y1,sep=)
$y1
[1] x$y$y1
How can I evaluate y1 as I did with y previously? or can I?
Much Thanks !
if I understand you correctly you can achieve your goal much easier than
with
eval, parse, substitute and the like:
x - list(y=list(y1=hello,y2=world),z=list(z1=foo,z2=bar))
s1 - 'y'
s2 - 'y1'
x[[s1]][[s2]]
i.e. using `[[' instead of `$' for list component extraction allows to
use
characters for indexing (in other words: x$y == x[['y']])
But what he actually asked for was
I am trying to build up a quoted or character expression representing
a
component in a list in order to reference it indirectly.
You just typed in x[[s1]][[s2]], not 'built [it] up'. Suppose the
specification had been
r - x
s - c(y, y1)
and s was of variable length? Then you need to construct a call similar
to x[[y]][[y1]] from r and s.
[There was another reason for sticking with $ rather than using [[: the
latter makes unnecessary copies in released versions of R.]
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
CONFIDENTIALITY NOTICE: This electronic mail transmission (i...{{dropped}}
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
