R help - Feb 2006 - "main" parameter in plot.default vs plot.formula

Folks:

R 2.2.0 on Windows.

I find the following somewhat puzzling:
> a<-1; x<-0:1; y<-x
## following works fine:
> plot(x,y ,main= bquote(n[1] == .(a) ))
## following produces an error:
> plot(y~x ,main= bquote(n[1] == .(a) ))
Error in paste(n[1] == 1, " and ", n[2] == 2) : object "n"
not found

*******************************

Note 1: I assume that this is due to the following documented behavior of
plot.formula():

"Both the terms in the formula and the ... arguments are evaluated in data
enclosed in parent.frame() if data is a list or a data frame."

Nevertheless, the behavior seems inconsistent to me. Am I missing something
(including the "I assume ..." comment)?

Note 2: If one uses substitute() instead, it works fine:

plot(y~x ,main= substitute(bquote(n[1] == a),list(a=a)))


Any illumination, public or private, would be appreciated.

Cheers,
Bert

 
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
 
"The business of the statistician is to catalyze the scientific learning
process."  - George E. P. Box

I cannot explain it but I must have come across it since
I noticed in various places in some of my code I have
used, in terms of your example, the following:

plot(y ~ x, main = as.expression(bquote(m[1] == .(a))))

plot(y ~ x, main = list(bquote(m[1] == .(a))))

both of which work as expected.

On 2/9/06, Berton Gunter <gunter.berton at gene.com>
wrote:
>
> Folks:
>
> R 2.2.0 on Windows.
>
> I find the following somewhat puzzling:
>
> > a<-1; x<-0:1; y<-x
>
> ## following works fine:
> > plot(x,y ,main= bquote(n[1] == .(a) ))
>
> ## following produces an error:
> > plot(y~x ,main= bquote(n[1] == .(a) ))
> Error in paste(n[1] == 1, " and ", n[2] == 2) : object
"n" not found
>
> *******************************
>
> Note 1: I assume that this is due to the following documented behavior of
> plot.formula():
>
> "Both the terms in the formula and the ... arguments are evaluated in
data
> enclosed in parent.frame() if data is a list or a data frame."
>
> Nevertheless, the behavior seems inconsistent to me. Am I missing something
> (including the "I assume ..." comment)?
>
> Note 2: If one uses substitute() instead, it works fine:
>
> plot(y~x ,main= substitute(bquote(n[1] == a),list(a=a)))
>
>
> Any illumination, public or private, would be appreciated.
>
> Cheers,
> Bert
>
>
> -- Bert Gunter
> Genentech Non-Clinical Statistics
> South San Francisco, CA
>
> "The business of the statistician is to catalyze the scientific
learning
> process."  - George E. P. Box
>
> ______________________________________________
> R-help at 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
>

On 2/9/06, Berton Gunter <gunter.berton at gene.com>
wrote:
>
> Folks:
>
> R 2.2.0 on Windows.
>
> I find the following somewhat puzzling:
>
> > a<-1; x<-0:1; y<-x
>
> ## following works fine:
> > plot(x,y ,main= bquote(n[1] == .(a) ))
>
> ## following produces an error:
> > plot(y~x ,main= bquote(n[1] == .(a) ))
> Error in paste(n[1] == 1, " and ", n[2] == 2) : object
"n" not found
>
> *******************************
>
> Note 1: I assume that this is due to the following documented behavior of
> plot.formula():
>
> "Both the terms in the formula and the ... arguments are evaluated in
data
> enclosed in parent.frame() if data is a list or a data frame."
>
> Nevertheless, the behavior seems inconsistent to me. Am I missing something
> (including the "I assume ..." comment)?
>
> Note 2: If one uses substitute() instead, it works fine:
>
> plot(y~x ,main= substitute(bquote(n[1] == a),list(a=a)))
>
Regarding your note 2,  the key thing that seems to
be necessary is not really substitute vs. bquote but
just doing it twice.  In your example
above you did not replace bquote with substitute but did a
substitute and a bquote so now its two levels deep.  But if we
just did two bquotes or two substitutes that would be ok too.
For example, we can just apply bquote twice and then it works:

plot(y~x ,main = bquote(bquote(n[1] == .(a))))

The various calls must be stripping off one layer so that
you have to protect it twice so that the underlying code
does not get evaluated.