Re: [R] stumped by eval
Berwin A Turlach wrote:
G'day Peter,
On Wed, 13 Feb 2008 08:03:07 +0100
Peter Dalgaard [EMAIL PROTECTED] wrote:
Ross Boylan wrote:
In the following example, the inner evaluation pulls in the global
value of subset (a function) rather than the one I thought I was
passing in (a vector). Can anyone help me understand what's going
on, and what I need to do to fix the problem?
[...]
The point is that subset (and offset) arguments are subject to the
same evaluation rules as the terms inside the formula: First look in
data, then in the environment of the formula, which in this case is
the global environment.
Perhaps I have a [senior|blonde]-day today, but this does not seem to
be the full explanation about what is going on to me. According to this
explanation the following should not work:
lm(Reading~0+Spec+Reader, netto, subset=c(1) )
Call:
lm(formula = Reading ~ 0 + Spec + Reader, data = netto, subset = c(1))
Coefficients:
Spec Reader
1 NA
since the value passed to subset is not part of data and not in the
global environment. But, obviously, it works.
It is, however, an expression that can be evaluated in the global
environment, and that works.
OTOH, if we change f0 to
f0
function(formula, data, subset, na.action)
{
lm(formula, data, subset=subset, na.action=na.action)
}
then we get the same behaviour as with Ross's use of f1 inside of f0:
t3 - f0(Reading~0+Spec+Reader, netto, c(1) )
Error in xj[i] : invalid subscript type 'closure'
I told you it was elusive... The thing is that lm() is using nonstandard
evaluation, so it sees the _symbol_ subset --- er, wait a minute, too
many subsets, let us define it like this instead
f0 - function(formula, data, s, na.action)
{
lm(formula, data, subset=s, na.action=na.action)
}
Ok, lm() sees the _symbol_ s passed as subset and then looks for it in data
and then in environment(formula). It never gets the idea to look for s in the
evaluation frame of f0.
One workaround is this:
f0
function(formula, data, s, na.action)
{
eval(bquote(lm(formula, data, subset=.(s), na.action=na.action)))
}
Another, I think better, way is
f0
function(formula, data, s, na.action)
{
eval(bquote(lm(formula, data, subset=.(substitute(s)), na.action=na.action)))
}
The latter also allows
f0(Reading~0+Spec+Reader, netto, Spec0 )
Call:
lm(formula = formula, data = data, subset = Spec 0, na.action = na.action)
Coefficients:
Spec Reader
-1.2976 0.7934
More over, with the original definition of f0:
f0
function(formula, data, subset, na.action)
{
f1(formula, data, subset, na.action)
}
(f1(Reading~0+Spec+Reader, netto, subset= Spec==1 ))
Reading Spec Reader
1 11 1
f0(Reading~0+Spec+Reader, netto, subset= Spec==1 )
Error in xj[i] : invalid subscript type 'closure'
Given your explanation, I would have expected this to work.
I think the issue here is still that model.matrix ends up being called
with subset=subset rather than subset= Spec==1.
Reading up on `subset' in ?model.frame also does not seem to shed light
onto what is going on.
Remaining confused.
Cheers,
Berwin
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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Re: [R] stumped by eval
G'day Peter,
On Wed, 13 Feb 2008 08:03:07 +0100
Peter Dalgaard [EMAIL PROTECTED] wrote:
Ross Boylan wrote:
In the following example, the inner evaluation pulls in the global
value of subset (a function) rather than the one I thought I was
passing in (a vector). Can anyone help me understand what's going
on, and what I need to do to fix the problem?
[...]
The point is that subset (and offset) arguments are subject to the
same evaluation rules as the terms inside the formula: First look in
data, then in the environment of the formula, which in this case is
the global environment.
Perhaps I have a [senior|blonde]-day today, but this does not seem to
be the full explanation about what is going on to me. According to this
explanation the following should not work:
lm(Reading~0+Spec+Reader, netto, subset=c(1) )
Call:
lm(formula = Reading ~ 0 + Spec + Reader, data = netto, subset = c(1))
Coefficients:
Spec Reader
1 NA
since the value passed to subset is not part of data and not in the
global environment. But, obviously, it works. OTOH, if we change f0 to
f0
function(formula, data, subset, na.action)
{
lm(formula, data, subset=subset, na.action=na.action)
}
then we get the same behaviour as with Ross's use of f1 inside of f0:
t3 - f0(Reading~0+Spec+Reader, netto, c(1) )
Error in xj[i] : invalid subscript type 'closure'
More over, with the original definition of f0:
f0
function(formula, data, subset, na.action)
{
f1(formula, data, subset, na.action)
}
(f1(Reading~0+Spec+Reader, netto, subset= Spec==1 ))
Reading Spec Reader
1 11 1
f0(Reading~0+Spec+Reader, netto, subset= Spec==1 )
Error in xj[i] : invalid subscript type 'closure'
Given your explanation, I would have expected this to work.
Reading up on `subset' in ?model.frame also does not seem to shed light
onto what is going on.
Remaining confused.
Cheers,
Berwin
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Re: [R] stumped by eval
On Wed, 2008-02-13 at 11:18 +0100, Peter Dalgaard wrote:
Berwin A Turlach wrote:
G'day Peter,
On Wed, 13 Feb 2008 08:03:07 +0100
Peter Dalgaard [EMAIL PROTECTED] wrote:
Ross Boylan wrote:
In the following example, the inner evaluation pulls in the global
value of subset (a function) rather than the one I thought I was
passing in (a vector). Can anyone help me understand what's going
on, and what I need to do to fix the problem?
[...]
The point is that subset (and offset) arguments are subject to the
same evaluation rules as the terms inside the formula: First look in
data, then in the environment of the formula, which in this case is
the global environment.
Oh my! Thank you; in several hours of staring/debugging I didn't get
that, and I'm not sure how many more it would have taken. Is there any
way I could have discovered that through debugging? The fact that eval
was a primitive made it hard to see what it was doing, although I guess
the relevant behavior was in model.frame.
I thought at one point that it mattered that subset was not a named
argument of model.frame, although it is an argument for the default
model.frame method. Does that make any difference?
Also, when I return to my original problem in which the subset argument
(to f0) is missing, will I encounter more difficulties?
Maybe the safest thing to do would be to testing for missing at the top
level and eliminate the corresponding entries in the calll.
Finally, the sample code I gave was a distillation of some code that
started with the code of lm.
Ross
I told you it was elusive... The thing is that lm() is using nonstandard
evaluation, so it sees the _symbol_ subset --- er, wait a minute, too
many subsets, let us define it like this instead
f0 - function(formula, data, s, na.action)
{
lm(formula, data, subset=s, na.action=na.action)
}
Ok, lm() sees the _symbol_ s passed as subset and then looks for it in data
and then in environment(formula). It never gets the idea to look for s in
the evaluation frame of f0.
One workaround is this:
f0
function(formula, data, s, na.action)
{
eval(bquote(lm(formula, data, subset=.(s), na.action=na.action)))
}
Another, I think better, way is
f0
function(formula, data, s, na.action)
{
eval(bquote(lm(formula, data, subset=.(substitute(s)),
na.action=na.action)))
}
The latter also allows
f0(Reading~0+Spec+Reader, netto, Spec0 )
Call:
lm(formula = formula, data = data, subset = Spec 0, na.action = na.action)
Coefficients:
Spec Reader
-1.2976 0.7934
More over, with the original definition of f0:
f0
function(formula, data, subset, na.action)
{
f1(formula, data, subset, na.action)
}
(f1(Reading~0+Spec+Reader, netto, subset= Spec==1 ))
Reading Spec Reader
1 11 1
f0(Reading~0+Spec+Reader, netto, subset= Spec==1 )
Error in xj[i] : invalid subscript type 'closure'
Given your explanation, I would have expected this to work.
I think the issue here is still that model.matrix ends up being called
with subset=subset rather than subset= Spec==1.
Reading up on `subset' in ?model.frame also does not seem to shed light
onto what is going on.
Remaining confused.
Cheers,
Berwin
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[R] stumped by eval
In the following example, the inner evaluation pulls in the global value
of subset (a function) rather than the one I thought I was passing in (a
vector). Can anyone help me understand what's going on, and what I need
to do to fix the problem?
f0 - function(formula, data,
subset, na.action
)
{
f1(formula, data,
subset, na.action
)
}
f1 - function (formula, data,
subset, na.action
)
{
mf - match.call()
mf[[1]] - as.name(model.frame)
mf - eval(mf, parent.frame())
}
netto - data.frame(Reading=seq(3), Spec=seq(3), Reader=seq(3))
t3 - f0(Reading~0+Spec+Reader, netto, c(1))
Error in xj[i] : invalid subscript type 'closure'
traceback()
7: `[.data.frame`(list(Reading = 1:3, Spec = 1:3, Reader = 1:3),
function (x, ...)
UseMethod(subset), , FALSE)
6: model.frame.default(formula = formula, data = data, subset = subset,
na.action = na.action)
5: model.frame(formula = formula, data = data, subset = subset,
na.action = na.action)
4: eval(expr, envir, enclos)
3: eval(mf, parent.frame())
2: f1(formula, data, subset, na.action)
1: f0(Reading ~ 0 + Spec + Reader, netto, c(1))
I started with a case in which f0 was called with only 2 arguments
(i.e., subset was missing), and that is the case I'm ultimately
interested in. However, even the situation above is failing.
According to resume, the class of subset is numeric in frames 2 and 4,
but a function in frames 3 and 5. That may be correct for frame 3,
since it is the evaluation function without a named argument subset (and
thus it picks up the global value). But it's wrong for 5 (and higher).
eval itself is primitive; I don't understand exactly what is going on
with the 2 entries for it (frames 3 + 4).
R package 2.6.1-1 on Debian GNU/Linux.
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Re: [R] stumped by eval
Ross Boylan wrote:
In the following example, the inner evaluation pulls in the global value
of subset (a function) rather than the one I thought I was passing in (a
vector). Can anyone help me understand what's going on, and what I need
to do to fix the problem?
f0 - function(formula, data,
subset, na.action
)
{
f1(formula, data,
subset, na.action
)
}
f1 - function (formula, data,
subset, na.action
)
{
mf - match.call()
mf[[1]] - as.name(model.frame)
mf - eval(mf, parent.frame())
}
netto - data.frame(Reading=seq(3), Spec=seq(3), Reader=seq(3))
t3 - f0(Reading~0+Spec+Reader, netto, c(1))
Error in xj[i] : invalid subscript type 'closure'
traceback()
7: `[.data.frame`(list(Reading = 1:3, Spec = 1:3, Reader = 1:3),
function (x, ...)
UseMethod(subset), , FALSE)
6: model.frame.default(formula = formula, data = data, subset = subset,
na.action = na.action)
5: model.frame(formula = formula, data = data, subset = subset,
na.action = na.action)
4: eval(expr, envir, enclos)
3: eval(mf, parent.frame())
2: f1(formula, data, subset, na.action)
1: f0(Reading ~ 0 + Spec + Reader, netto, c(1))
I started with a case in which f0 was called with only 2 arguments
(i.e., subset was missing), and that is the case I'm ultimately
interested in. However, even the situation above is failing.
According to resume, the class of subset is numeric in frames 2 and 4,
but a function in frames 3 and 5. That may be correct for frame 3,
since it is the evaluation function without a named argument subset (and
thus it picks up the global value). But it's wrong for 5 (and higher).
eval itself is primitive; I don't understand exactly what is going on
with the 2 entries for it (frames 3 + 4).
Yes, this is elusive, but it is not actually eval() that is doing you
in. It is the notion of a model environment, see the last bit of ?formula.
The point is that subset (and offset) arguments are subject to the same
evaluation rules as the terms inside the formula: First look in data,
then in the environment of the formula, which in this case is the global
environment.
This behaviour is generally a good thing because it prevents you from
accidentally picking up internal variables of f1, but working around it
can be a little painful. As far as I recall, you can use an explicit
substitute of the subset argument.
R package 2.6.1-1 on Debian GNU/Linux.
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R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
__
R-help@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.
