[R] 'substitute' question
# I use this code to label a graph with the R2: # graph x - rnorm(100) y - x + rnorm(100) lm1 - lm(y~x) plot(x,y) # label R2text - substitute(paste(R^2, = ,r2),list(r2=r2)) text(1,-3,R2text, col=red) # i have modified this a bit, so that i have a vector with other labels, each of which # will be labelled on the graph. Example: texts - c(And the R2 is, R2text) x - c(-2,-2) y - c(2,1) for(i in 1:length(texts))text(x[i],y[i],texts[i],pos=4, col=blue) # As you can see the label R2 = 48.7 remains at paste(R^2, = , 48.7) # What to do? Thanks for your help, Remko ..-~-.-~-.-~-.-~-.-~-.-~-.-~-.-~-.-~- Remko Duursma Post-doctoral researcher Dept. Forest Ecology University of Helsinki, Finland _ With tax season right around the corner, make sure to follow these few simple tips. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] 'substitute' question
hi, your first argument to substitute should be an expression, not a character string (which the output of paste() will give you) so... # first assign to variable r2 (think you forgot to do that) # in your example r2 - summary(lm1)$r.squared # then to your label R2text - substitute(R^2==r2,list(r2=round(r2,2))) # then you can annotate with text() as you did alternatively, you can use R2text - bquote(R^2==.(round(r2,2))) --- remko duursma [EMAIL PROTECTED] wrote: # I use this code to label a graph with the R2: # graph x - rnorm(100) y - x + rnorm(100) lm1 - lm(y~x) plot(x,y) # label R2text - substitute(paste(R^2, = ,r2),list(r2=r2)) text(1,-3,R2text, col=red) # i have modified this a bit, so that i have a vector with other labels, each of which # will be labelled on the graph. Example: texts - c(And the R2 is, R2text) x - c(-2,-2) y - c(2,1) for(i in 1:length(texts))text(x[i],y[i],texts[i],pos=4, col=blue) # As you can see the label R2 = 48.7 remains at paste(R^2, = , 48.7) # What to do? Thanks for your help, Remko ..-~-.-~-.-~-.-~-.-~-.-~-.-~-.-~-.-~- Remko Duursma Post-doctoral researcher Dept. Forest Ecology University of Helsinki, Finland _ With tax season right around the corner, make sure to follow these few simple tips. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Finding fabulous fares is fun. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 'substitute' question
Dear Renko To modify the script plot(x,y) r2-summary(lm1)$r.squared*100 # label R2text - substitute(paste(R^2, = ,r2),list(r2=r2)) text(-1,1,R2text, col=red) # To see of the coordinates of the graph. Grettings Felipe -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of remko duursma Sent: Tuesday, March 13, 2007 8:02 AM To: r-help@stat.math.ethz.ch Subject: [R] 'substitute' question # I use this code to label a graph with the R2: # graph x - rnorm(100) y - x + rnorm(100) lm1 - lm(y~x) plot(x,y) # label R2text - substitute(paste(R^2, = ,r2),list(r2=r2)) text(1,-3,R2text, col=red) # i have modified this a bit, so that i have a vector with other labels, each of which # will be labelled on the graph. Example: texts - c(And the R2 is, R2text) x - c(-2,-2) y - c(2,1) for(i in 1:length(texts))text(x[i],y[i],texts[i],pos=4, col=blue) # As you can see the label R2 = 48.7 remains at paste(R^2, = , 48.7) # What to do? Thanks for your help, Remko ..-~-.-~-.-~-.-~-.-~-.-~-.-~-.-~-.-~- Remko Duursma Post-doctoral researcher Dept. Forest Ecology University of Helsinki, Finland _ With tax season right around the corner, make sure to follow these few simple tips. __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] 'substitute' question
I do not understand why you play with 'substitute' instead of something
like this:
# CAUTION: this will work only for bivariate case
# plotmaking function
plotModel - function(m)
{
x - m$model$x
y - m$model$y
r2 - summary(m)$r.squared
plot(x,y)
abline(m)
text( min(x), max(y), paste(And the R^2 is, round(r2,3)),
pos=4)
invisible(NULL)
}
# your data
x - rnorm(100)
y - x + rnorm(100)
lm1 - lm(y~x)
# make the plot
plotModel(lm1)
*** Note that my e-mail address has changed to [EMAIL PROTECTED]
*** Please update your address books accordingly. Thank you!
_
Michal Bojanowski
ICS / Sociology
Utrecht University
Heidelberglaan 2; 3584 CS Utrecht
Room 1428
[EMAIL PROTECTED]
http://www.fss.uu.nl/soc/bojanowski
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of remko duursma
Sent: Tuesday, March 13, 2007 2:02 PM
To: r-help@stat.math.ethz.ch
Subject: [R] 'substitute' question
# I use this code to label a graph with the R2:
# graph
x - rnorm(100)
y - x + rnorm(100)
lm1 - lm(y~x)
plot(x,y)
# label
R2text - substitute(paste(R^2, = ,r2),list(r2=r2)) text(1,-3,R2text,
col=red)
# i have modified this a bit, so that i have a vector with other labels,
each of which # will be labelled on the graph. Example:
texts - c(And the R2 is, R2text)
x - c(-2,-2)
y - c(2,1)
for(i in 1:length(texts))text(x[i],y[i],texts[i],pos=4, col=blue)
# As you can see the label R2 = 48.7 remains at paste(R^2, = ,
48.7)
# What to do?
Thanks for your help,
Remko
..-~-.-~-.-~-.-~-.-~-.-~-.-~-.-~-.-~-
Remko Duursma
Post-doctoral researcher
Dept. Forest Ecology
University of Helsinki, Finland
_
With tax season right around the corner, make sure to follow these few
simple tips.
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Re: [R] 'substitute' question
I'm sorry for my first post as I did not properly understood your problem. The reason why you were getting paste(R^2, = , 48.7) instead of the properly formatted R^2=48.7 is that in creating 'texts' in your code you were mixing character string with expressions resulting in 'texts' being a list not a vector. The only modification to make is to index texts with double square brackets as below # the figure x - rnorm(100) y - x + rnorm(100) lm1 - lm(y~x) plot(x,y) # label R2text - substitute(paste(R^2, = ,r2),list(r2=r2)) text(1,-3,R2text, col=red) # i expanded your list of labels to four elements... texts - c(And the R2 is, R2text, third label, fourth label) # texts is a list not an atomic vector class(texts) str(texts) # ... and added more coordinates accordingly x - c(-2,-2, -1, -1) y - c(2,1, 2, 1) # here 'texts[[i]]' instead of 'texts[i]' for(i in 1:length(texts))text(x[i],y[i], texts[[i]],pos=4, col=blue) hth, Michal *** Note that my e-mail address has changed to [EMAIL PROTECTED] *** Please update your address books accordingly. Thank you! _ Michal Bojanowski ICS / Sociology Utrecht University Heidelberglaan 2; 3584 CS Utrecht Room 1428 [EMAIL PROTECTED] http://www.fss.uu.nl/soc/bojanowski -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of remko duursma Sent: Tuesday, March 13, 2007 2:02 PM To: r-help@stat.math.ethz.ch Subject: [R] 'substitute' question # I use this code to label a graph with the R2: # graph x - rnorm(100) y - x + rnorm(100) lm1 - lm(y~x) plot(x,y) # label R2text - substitute(paste(R^2, = ,r2),list(r2=r2)) text(1,-3,R2text, col=red) # i have modified this a bit, so that i have a vector with other labels, each of which # will be labelled on the graph. Example: texts - c(And the R2 is, R2text) x - c(-2,-2) y - c(2,1) for(i in 1:length(texts))text(x[i],y[i],texts[i],pos=4, col=blue) # As you can see the label R2 = 48.7 remains at paste(R^2, = , 48.7) # What to do? Thanks for your help, Remko ..-~-.-~-.-~-.-~-.-~-.-~-.-~-.-~-.-~- Remko Duursma Post-doctoral researcher Dept. Forest Ecology University of Helsinki, Finland _ With tax season right around the corner, make sure to follow these few simple tips. __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] substitute question
Tony Plate [EMAIL PROTECTED] writes: Earlier in this discussion, Peter Dalgard stated you can only do substitutions on language objects and then used the function is.language() in an example, which I took at that time to imply that substitute() would go inside objects for which is.language() returned true. However, from experimenting, it seems that is.call() rather than is.language() is the appropriate test. Right. That surprised me too, although perhaps it shouldn't have. Of course we might just have looked at the source, where we see that PROMSXP, SYMSXP, and LANGSXP are handled, and everything else, including EXPRSXPs are returned as is. I think this is deliberate and that expression objects are occasionally used as a quoting mechanism to prevent substitutions from taking place, but my recollection is a bit hazy on this point. So at the R level it is is.name() or is.call() or promises evaluating to either of the two. Notice, BTW, that expression() is weird in the same way as the function(...)... constructs that tripped Gabor: substitute(expression(a+b),list(b=2)) expression(a + 2) eval(substitute(substitute(e,list(b=2)), list(e=expression(a+b expression(a + b) eval(substitute(substitute(e,list(b=2)), list(e=quote(expression(a+b) expression(a + 2) and the root cause of this is that expression(a+b) expression(a + b) quote(expression(a+b)) expression(a + b) There's a group of cases where deparse and parse are not exact inverses, because some objects are not directly representable in the language, so you get instead a function call that *evaluates* to a similar object; the simplest case is vectors of length 1: dput(rnorm(1)) -0.214713202300464 dput(rnorm(2)) c(0.92727004398074, 0.437854016853309) -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] substitute question
[This appears to have bounced so I am sending it again. Apologies
if it gets there twice.]
Thanks. Thus it seems that there are two types of expressions and calls:
1. fully expanded
2. partially expanded
and that fully expanded ones are a prerequisite for substitution.
body() and quote() produce such fully expanded expressions.
Using a small utility function we can investigate this:
recurse - function( x, idx = NULL )
if ( length( x ) 0 ) {
for( i in seq( along = x ) )
if (length(x[[i]])1)
Recall( x[[i]], c(idx, i))
else {
if (length(idx)) cat(idx,)
cat( i, class(x[[i]]), : )
cat( rep(\t,length(idx) + 2) )
print( x[[i]] )
}
}
f - function(){a+1}
eb - body(f)
class(eb)
recurse(eb)
eq - quote(function(){a+1})
class(eq)
recurse(eq)
ep - parse(text=deparse(f))
class(ep)
recurse(ep)
The output that the above is shown below. It shows that
body() and quote() produce fully expanded expression style objects
although body's is of class { and quote is of class call.
However, parse(text=deparse(f)) also produces a fully expanded
expression style object of class expression yet substitution
does not occur with that. Thus full vs. partial expansion is likely
a necessary but not a sufficient condition. There is something
else but I don't know what it is.
f - function(){a+1}
eb - body(f)
class(eb)
[1] {
recurse(eb)
1 name : `{`
2 1 name : `+`
2 2 name : a
2 3 numeric : [1] 1
eq - quote(function(){a+1})
class(eq)
[1] call
recurse(eq) # lines begin with list indices and class name
1 name : `function`
2 NULL : NULL
3 1 name : `{`
3 2 1 name : `+`
3 2 2 name : a
3 2 3 numeric : [1] 1
4 NULL : NULL
ep - parse(text=deparse(f))
class(ep)
[1] expression
recurse(ep)
1 1 name : `function`
1 2 NULL : NULL
1 3 1 name : `{`
1 3 2 1 name : `+`
1 3 2 2 name : a
1 3 2 3 numeric : [1] 1
1 4 NULL : NULL
Date: Thu, 18 Mar 2004 17:27:20 -0800 (PST)
From: Thomas Lumley [EMAIL PROTECTED]
To: Gabor Grothendieck [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED], [EMAIL PROTECTED]
Subject: Re: [R] substitute question
On Thu, 18 Mar 2004, Gabor Grothendieck wrote:
I don't think I expressed myself very well on that.
Looking at what we get from the example:
z - substitute(substitute(expression(f),list(a=quote(b))),list(f=f))
z
substitute(expression(function ()
{
a + 1
}), list(a = quote(b)))
class(z);mode(z);typeof(z)
[1] call
[1] call
[1] language
we see that the function seems to be expanded correctly and
the statement does produce a call object. However,
applying eval one, two or three times does not give what
you would think if you looked at z above.
Maybe we didn't express ourselves well enough.
Looking at z above isn't enough. z is a call to substitute().
Its first operand is an expression. The expression contains a single term,
which is a function.
If you typed
notz- quote(substitute(expression(function ()
{
a + 1
}), list(a = quote(b
you would obtain something that deparsed the same as z, and so looked the
same, but was actually different. In notz the first operand of substitute
is an expression containing multiple terms, which if evaluated would
return a function.
substitute() goes though this expression and checks each term to see if it
is `a`. In z there is only one term and it isn't `a`. In notz there is
(after sufficient recursion) an `a` and it gets replaced.
So
z[[2]][[2]]
function ()
{
a + 1
}
notz[[2]][[2]]
function() {
a + 1
}
are the respective operands, and they still look the same. But
mode(z[[2]][[2]])
[1] function
mode(notz[[2]][[2]])
[1] call
length(z[[2]][[2]])
[1] 1
length(notz[[2]][[2]])
[1] 4
and if we try to find the actual `a` in there
notz[[2]][[2]][[3]][[2]][[2]]
a
z[[2]][[2]][[3]][[2]][[2]]
Error in z[[2]][[2]][[3]] : object is not subsettable
-thomas
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Re: [R] substitute question
Great detective work, Tony! Thanks.
Date: Fri, 19 Mar 2004 17:51:27 -0700
From: Tony Plate [EMAIL PROTECTED]
To: [EMAIL PROTECTED], [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: Re: [R] substitute question
I don't think it is correct that there are two types of expressions and
calls. There is only one type of these things. I believe the relevant
distinction here is between 'call' objects (which can be informally thought
of as the parse trees of unevaluated R code, and which formally have mode
'call' in R) and other things like objects of mode 'function', etc.
However, this is all does pretty confusing when using substitute(),
because, substitute() does go inside objects that have mode 'call', but
doesn't go inside of objects that have mode 'function' or
'expression'. What makes it more confusing is that sometimes 'call'
objects can be wrapped up in expression objects. Note that parse(text=)
returns a 'call' object wrapped in an 'expression' object, whereas quote()
returns a 'call' object -- I believe that in general it is true that
parse(text=XXX)[[1]] === quote(XXX).
Earlier in this discussion, Peter Dalgard stated you can only do
substitutions on language objects and then used the function is.language()
in an example, which I took at that time to imply that substitute() would
go inside objects for which is.language() returned true. However, from
experimenting, it seems that is.call() rather than is.language() is the
appropriate test.
Here are some simple examples.
esub - function(expr, sublist) do.call(substitute, list(expr, sublist))
e1 - parse(text=a + 1)
e2 - quote(a + 1)
e1
expression(a + 1)
e2
a + 1
mode(e1)
[1] expression
mode(e2)
[1] call
identical(e1[[1]], e2)
[1] TRUE
# substitute() doesn't go inside e1, even though is.language(e1) is TRUE
c(is.language(e1), is.call(e1))
[1] TRUE FALSE
esub(e1, list(a=as.name('b')))
expression(a + 1)
c(is.language(e2), is.call(e2))
[1] TRUE TRUE
esub(e2, list(a=as.name('b')))
b + 1
c(is.language(e1[[1]]), is.call(e1[[1]]))
[1] TRUE TRUE
esub(e1[[1]], list(a=as.name('b')))
b + 1
identical(e2, e1[[1]])
[1] TRUE
ef - Quote(function() a + 1)
f - function() a + 1
c(is.language(ef), is.call(ef))
[1] TRUE TRUE
esub(ef, list(a=as.name('b')))
function() b + 1
c(is.language(f), is.call(f))
[1] FALSE FALSE
esub(f, list(a=as.name('b')))
function ()
a + 1
c(is.language(body(f)), is.call(body(f)))
[1] TRUE TRUE
esub(body(f), list(a=as.name('b')))
b + 1
I also see that in S-plus 6.2, substitute() behaves differently -- it does
go inside objects of mode 'call' and 'expression' and substitutes 'b' for
'a' in every case above. To run the above code in S-plus, first do:
body - function(f) f[[1]]
quote - Quote
Although there isn't much to guide one in the documentation ?substitute,
the R Language manual does have some discussion of substitute() and
'expression' objects.
-- Tony Plate
At Thursday 10:58 PM 3/18/2004, Gabor Grothendieck wrote:
Thanks. Thus it seems that there are two types of expressions and calls:
1. fully expanded
2. partially expanded
and that fully expanded ones are a prerequisite for substitution.
body() and quote() produce such fully expanded expressions.
Using a small utility function we can investigate this:
recurse - function( x, idx = NULL )
if ( length( x ) 0 ) {
for( i in seq( along = x ) )
if (length(x[[i]])1)
Recall( x[[i]], c(idx, i))
else {
if (length(idx)) cat(idx,)
cat( i, class(x[[i]]), : )
cat( rep(\t,length(idx) + 2) )
print( x[[i]] )
}
}
f - function(){a+1}
eb - body(f)
class(eb)
recurse(eb)
eq - quote(function(){a+1})
class(eq)
recurse(eq)
ep - parse(text=deparse(f))
class(ep)
recurse(ep)
The output that the above is shown below. It shows that
body() and quote() produce fully expanded expression style objects
although body's is of class { and quote is of class call.
However, parse(text=deparse(f)) also produces a fully expanded
expression style object of class expression yet substitution
does not occur with that. Thus full vs. partial expansion is likely
a necessary but not a sufficient condition. There is something
else but I don't know what it is.
f - function(){a+1}
eb - body(f)
class(eb)
[1] {
recurse(eb)
1 name : `{`
2 1 name : `+`
2 2 name : a
2 3 numeric : [1] 1
eq - quote(function(){a+1})
class(eq)
[1] call
recurse(eq) # lines begin with list indices and class name
1 name : `function`
2 NULL : NULL
3 1 name : `{`
3 2 1 name : `+`
3 2 2 name : a
3 2 3 numeric : [1] 1
4 NULL : NULL
ep - parse(text=deparse(f))
class(ep)
[1] expression
recurse(ep)
1 1 name : `function`
1 2 NULL : NULL
1 3 1 name : `{`
1 3 2 1 name : `+`
1 3 2 2 name : a
1 3 2 3 numeric : [1] 1
1 4 NULL : NULL
Date: Thu, 18 Mar 2004 17:27:20 -0800 (PST)
From: Thomas Lumley [EMAIL PROTECTED]
To: Gabor Grothendieck [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED], [EMAIL PROTECTED]
Subject: Re: [R] substitute question
On Thu, 18 Mar 2004, Gabor
RE: [R] substitute question
On Wed, 17 Mar 2004, Gabor Grothendieck wrote:
I left out the brackets in my last email but the problem
(a reappears after have been substituted out) still remains:
z - substitute( function(){a+1}, list(a=quote(b)) )
z
function() {
b + 1
}
eval(z)
function(){a+1}
Interesting.
Appearances are misleading, however:
z - substitute( function(){a+1}, list(a=quote(b)) )
z
function() {
b + 1
}
f-eval(z)
f()
Error in f() : Object b not found
f
function(){a+1}
attr(f,source)-NULL
f
function ()
{
b + 1
}
So it isn't that eval(z) has a+1 inside, it just has a source attribute
with a+1.
Looking more carefully at z
as.list(z)
[[1]]
`function`
[[2]]
NULL
[[3]]
{
b + 1
}
[[4]]
[1] function(){a+1}
so the original construction of z has kept the source (not, however, as a
source attribute).
There is method to our madness here. It is impossible (or at least too
complicated) to keep comments in the right place as a function is
parsed and deparsed. In the old days, comments would occasionally move
around, sometimes in very misleading ways (IIRC with if(){}else{} cases)
Now we keep a copy of the source code with functions created interactively
or with source(), and drop the comments on parsing. This is controlled by
options(keep.source).
If you do a lot of substitute()-style programming you may want
options(keep.source=FALSE).
-thomas
PS: There are, of course, interesting possibilities for creative abuse of
the source attribute
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RE: [R] substitute question
This is because of the saved attribute source on z (that doesn't get
printed out before evaluating z, because z is not yet then a function).
To complete your example:
z - substitute( function(){a+1}, list(a=quote(b)) )
z
function() {
b + 1
}
eval(z)
function(){a+1}
ze - eval(z)
attributes(ze)
$source
[1] function(){a+1}
attr(ze, source) - NULL
ze
function ()
{
b + 1
}
I previously wrote on this topic:
Date: Fri, 24 Oct 2003 09:42:55 -0600
To: [EMAIL PROTECTED], Peter Dalgaard [EMAIL PROTECTED]
From: Tony Plate [EMAIL PROTECTED]
Subject: Re: [R] what's going on here with substitute() ?
Mime-Version: 1.0
Content-Type: text/plain; charset=us-ascii; format=flowed
Peter, thank you for the explanation. This is indeed what is happening.
Might I suggest the following passage for inclusion in the help page for
function, and possibly also body, in the DETAILS or WARNING section:
Note that the text of the original function definition is saved as an
attribute source on the function, and this is printed out when the
function is printed. Hence, if the function body is changed in some way
other than by assigning a value via body() (which removes the source
attribute), the printed form of the function may not be the same as the
actual function body.
Something along these lines could also go in the help for eval, though if
it were only there it might be very difficult to find if one were trying to
look up puzzling behavior of a function.
Here is a transcript that shows what is happening, with another suggestion
following it.
eval(substitute(this.is.R - function() X,
list(X=!is.null(options(CRAN)[[1]]
this.is.R
function() X
body(this.is.R)
[1] TRUE
attributes(this.is.R)
$source
[1] function() X
attributes(this.is.R) - NULL
this.is.R
function ()
TRUE
# the source attribute comes from function definition:
attributes(function() X)
$source
[1] function() X
# and seems to be added by eval:
attr(eval(parse(text=function() TRUE)[[1]]), source)
[1] function() TRUE
# we can assign bogus source
attr(this.is.R, source) - a totally bogus function body
this.is.R
a totally bogus function body
# assigning to body() removes source
body(this.is.R) - list(666)
this.is.R
function ()
666
attr(this.is.R, source)
NULL
An even better approach might be something that gave a warning on printing
if the parsed source attribute was not identical to the language object
being printed. This would probably belong in the code for case LANGSXP:
in the function PrintValueRec in main/print.c (if it were written in R, I
could contribute a patch, but right now I don't have time to try to
understand the C there.) R code to do the test could be something like this:
f - this.is.R
identical(f, eval(parse(text=attr(f, source))[[1]]))
[1] FALSE
f - function() TRUE
identical(f, eval(parse(text=attr(f, source))[[1]]))
[1] TRUE
-- Tony Plate
At Wednesday 08:09 PM 3/17/2004, Gabor Grothendieck wrote:
I left out the brackets in my last email but the problem
(a reappears after have been substituted out) still remains:
z - substitute( function(){a+1}, list(a=quote(b)) )
z
function() {
b + 1
}
eval(z)
function(){a+1}
---
Date: Wed, 17 Mar 2004 20:10:43 -0500 (EST)
From: Gabor Grothendieck [EMAIL PROTECTED]
[ Add to Address Book | Block Address | Report as Spam ]
To: [EMAIL PROTECTED]
Subject: [R] substitute question
Consider the following example:
# substitute a with b in the indicated function. Seems to work.
z - substitute( function()a+1, list(a=quote(b)) )
z
function() b + 1
# z is an object of class call so use eval
# to turn it into an object of class expression; however,
# when z is evaluated, the variable a returns.
eval(z)
function()a+1
Why did a suddenly reappear again after it had already been replaced?
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RE: [R] substitute question
Tony, Thomas. Thanks for your help. Your comments were very
useful.
Unfortunately, my next step gives me a new round of problems.
The following is the same as the last example except that instead
of hard coding the function into the expression I wanted to
pass it to the expression. It seems like one has to do a double
substitute to get this effect but I am having problems getting this
to work.
In the code below we first show that the keep.source option has been
set to FALSE so that the source attribute does not mislead us.
Then we do a double substitute. The outer substitute just creates
the substitute that was in my previous question. This outer
substitute produces z, a call object. So far its as expected.
We then do ze - eval(z) but we get a function object right away. I was
expecting that we get an expression object. Even worse, the function does not
have a replaced with b -- even though this did work in the previous example
that I posted.
What's wrong?
options()$keep.source
[1] FALSE
f - function(){a+1}
z - substitute(substitute(f,list(a=quote(b))),list(f=f))
class(z)
[1] call
as.list(z)
[[1]]
substitute
[[2]]
function ()
{
a + 1
}
[[3]]
list(a = quote(b))
ze - eval(z)
class(ze)
[1] function
ze
function ()
{
a + 1
}
attr(ze,source)
NULL
---
Date: Thu, 18 Mar 2004 09:00:02 -0800 (PST)
From: Thomas Lumley [EMAIL PROTECTED]
To: Gabor Grothendieck [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: RE: [R] substitute question
On Wed, 17 Mar 2004, Gabor Grothendieck wrote:
I left out the brackets in my last email but the problem
(a reappears after have been substituted out) still remains:
z - substitute( function(){a+1}, list(a=quote(b)) )
z
function() {
b + 1
}
eval(z)
function(){a+1}
Interesting.
Appearances are misleading, however:
z - substitute( function(){a+1}, list(a=quote(b)) )
z
function() {
b + 1
}
f-eval(z)
f()
Error in f() : Object b not found
f
function(){a+1}
attr(f,source)-NULL
f
function ()
{
b + 1
}
So it isn't that eval(z) has a+1 inside, it just has a source attribute
with a+1.
Looking more carefully at z
as.list(z)
[[1]]
`function`
[[2]]
NULL
[[3]]
{
b + 1
}
[[4]]
[1] function(){a+1}
so the original construction of z has kept the source (not, however, as a
source attribute).
There is method to our madness here. It is impossible (or at least too
complicated) to keep comments in the right place as a function is
parsed and deparsed. In the old days, comments would occasionally move
around, sometimes in very misleading ways (IIRC with if(){}else{} cases)
Now we keep a copy of the source code with functions created interactively
or with source(), and drop the comments on parsing. This is controlled by
options(keep.source).
If you do a lot of substitute()-style programming you may want
options(keep.source=FALSE).
-thomas
PS: There are, of course, interesting possibilities for creative abuse of
the source attribute
---
Date: Thu, 18 Mar 2004 08:41:54 -0700
From: Tony Plate [EMAIL PROTECTED]
To: [EMAIL PROTECTED], [EMAIL PROTECTED]
Subject: RE: [R] substitute question
This is because of the saved attribute source on z (that doesn't get
printed out before evaluating z, because z is not yet then a function).
To complete your example:
z - substitute( function(){a+1}, list(a=quote(b)) )
z
function() {
b + 1
}
eval(z)
function(){a+1}
ze - eval(z)
attributes(ze)
$source
[1] function(){a+1}
attr(ze, source) - NULL
ze
function ()
{
b + 1
}
I previously wrote on this topic:
Date: Fri, 24 Oct 2003 09:42:55 -0600
To: [EMAIL PROTECTED], Peter Dalgaard [EMAIL PROTECTED]
From: Tony Plate [EMAIL PROTECTED]
Subject: Re: [R] what's going on here with substitute() ?
Mime-Version: 1.0
Content-Type: text/plain; charset=us-ascii; format=flowed
Peter, thank you for the explanation. This is indeed what is happening.
Might I suggest the following passage for inclusion in the help page for
function, and possibly also body, in the DETAILS or WARNING section:
Note that the text of the original function definition is saved as an
attribute source on the function, and this is printed out when the
function is printed. Hence, if the function body is changed in some way
other than by assigning a value via body() (which removes the source
attribute), the printed form of the function may not be the same as the
actual function body.
Something along these lines could also go in the help for eval, though if
it were only there it might be very difficult to find if one were trying to
look up puzzling behavior of a function.
Here is a transcript that shows what is happening, with another suggestion
following it.
eval(substitute(this.is.R - function() X,
list(X=!is.null(options(CRAN)[[1]]
this.is.R
function() X
body(this.is.R)
[1] TRUE
attributes(this.is.R)
$source
[1] function() X
attributes(this.is.R) - NULL
this.is.R
function ()
TRUE
# the source attribute comes from
RE: [R] substitute question
Gabor, I suspect you are overlooking the fact that substitute() does not
evaluate its first argument. So, in the previous example you were giving
an unevaluated expression as the first argument of substitute(). In your
most recent example you are supplying a function object as the first
argument of substitute() (by supplying it as the value to be substituted
for f). I think you can get what you want by first doing f -
Quote(function(){a+1}):
f - Quote(function(){a+1})
f
function() {
a + 1
}
z - substitute(substitute(f,list(a=quote(b))),list(f=f))
z
substitute(function() {
a + 1
}, list(a = quote(b)))
eval(z)
function() {
b + 1
}
eval(eval(z)) # this displays the source attribute
function(){a+1}
body(eval(eval(z)))
{
b + 1
}
hope this helps,
Tony Plate
At Thursday 12:00 PM 3/18/2004, you wrote:
Tony, Thomas. Thanks for your help. Your comments were very
useful.
Unfortunately, my next step gives me a new round of problems.
The following is the same as the last example except that instead
of hard coding the function into the expression I wanted to
pass it to the expression. It seems like one has to do a double
substitute to get this effect but I am having problems getting this
to work.
In the code below we first show that the keep.source option has been
set to FALSE so that the source attribute does not mislead us.
Then we do a double substitute. The outer substitute just creates
the substitute that was in my previous question. This outer
substitute produces z, a call object. So far its as expected.
We then do ze - eval(z) but we get a function object right away. I was
expecting that we get an expression object. Even worse, the function does not
have a replaced with b -- even though this did work in the previous
example
that I posted.
What's wrong?
options()$keep.source
[1] FALSE
f - function(){a+1}
z - substitute(substitute(f,list(a=quote(b))),list(f=f))
class(z)
[1] call
as.list(z)
[[1]]
substitute
[[2]]
function ()
{
a + 1
}
[[3]]
list(a = quote(b))
ze - eval(z)
class(ze)
[1] function
ze
function ()
{
a + 1
}
attr(ze,source)
NULL
---
Date: Thu, 18 Mar 2004 09:00:02 -0800 (PST)
From: Thomas Lumley [EMAIL PROTECTED]
To: Gabor Grothendieck [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: RE: [R] substitute question
On Wed, 17 Mar 2004, Gabor Grothendieck wrote:
I left out the brackets in my last email but the problem
(a reappears after have been substituted out) still remains:
z - substitute( function(){a+1}, list(a=quote(b)) )
z
function() {
b + 1
}
eval(z)
function(){a+1}
Interesting.
Appearances are misleading, however:
z - substitute( function(){a+1}, list(a=quote(b)) )
z
function() {
b + 1
}
f-eval(z)
f()
Error in f() : Object b not found
f
function(){a+1}
attr(f,source)-NULL
f
function ()
{
b + 1
}
So it isn't that eval(z) has a+1 inside, it just has a source attribute
with a+1.
Looking more carefully at z
as.list(z)
[[1]]
`function`
[[2]]
NULL
[[3]]
{
b + 1
}
[[4]]
[1] function(){a+1}
so the original construction of z has kept the source (not, however, as a
source attribute).
There is method to our madness here. It is impossible (or at least too
complicated) to keep comments in the right place as a function is
parsed and deparsed. In the old days, comments would occasionally move
around, sometimes in very misleading ways (IIRC with if(){}else{} cases)
Now we keep a copy of the source code with functions created interactively
or with source(), and drop the comments on parsing. This is controlled by
options(keep.source).
If you do a lot of substitute()-style programming you may want
options(keep.source=FALSE).
-thomas
PS: There are, of course, interesting possibilities for creative abuse of
the source attribute
---
Date: Thu, 18 Mar 2004 08:41:54 -0700
From: Tony Plate [EMAIL PROTECTED]
To: [EMAIL PROTECTED], [EMAIL PROTECTED]
Subject: RE: [R] substitute question
This is because of the saved attribute source on z (that doesn't get
printed out before evaluating z, because z is not yet then a function).
To complete your example:
z - substitute( function(){a+1}, list(a=quote(b)) )
z
function() {
b + 1
}
eval(z)
function(){a+1}
ze - eval(z)
attributes(ze)
$source
[1] function(){a+1}
attr(ze, source) - NULL
ze
function ()
{
b + 1
}
I previously wrote on this topic:
Date: Fri, 24 Oct 2003 09:42:55 -0600
To: [EMAIL PROTECTED], Peter Dalgaard [EMAIL PROTECTED]
From: Tony Plate [EMAIL PROTECTED]
Subject: Re: [R] what's going on here with substitute() ?
Mime-Version: 1.0
Content-Type: text/plain; charset=us-ascii; format=flowed
Peter, thank you for the explanation. This is indeed what is happening.
Might I suggest the following passage for inclusion in the help page for
function, and possibly also body, in the DETAILS or WARNING section:
Note that the text of the original function definition is saved as an
attribute source on the function, and this is printed out when
RE: [R] substitute question
On Thu, 18 Mar 2004, Gabor Grothendieck wrote:
Tony, Thomas. Thanks for your help. Your comments were very
useful.
Unfortunately, my next step gives me a new round of problems.
The following is the same as the last example except that instead
of hard coding the function into the expression I wanted to
pass it to the expression. It seems like one has to do a double
substitute to get this effect but I am having problems getting this
to work.
The problem is that f is a function, not an expression. You need to work
with body(f)
Either
f-function(){a+1}
body(f)-do.call(substitute,list(body(f),list(a=quote(b f
function ()
{
b + 1
}
or
f-function(){a+1}
body(f)-eval(substitute(substitute(expr,list(a=quote(b))),list(expr=body(f
f
function ()
{
b + 1
}
-thomas
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RE: [R] substitute question
Anyone Knows How Shoud I Build Rasch Models (Partial Credit Models) In R? []s Leonard Assis EstatÃstico - CONFE 7439 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] substitute question
Tony, Thomas, Thanks, again.
If the problem with my last example was just that I was passing a function
rather than an unevaluated expression, then why don't the following
return f with b in place of a? In both cases, a is still there in the
final output.
f - function() { a + 1 }
z - substitute(substitute(f=f,list(a=quote(b))),list(f=parse(text=deparse(f
eval(eval(z))
or
f - function() { a + 1 }
z - substitute(substitute(expression(f),list(a=quote(b))),list(f=f))
eval(eval(eval(z)))
---
Date: Thu, 18 Mar 2004 11:29:39 -0800 (PST)
From: Thomas Lumley [EMAIL PROTECTED]
To: Gabor Grothendieck [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED], [EMAIL PROTECTED]
Subject: RE: [R] substitute question
On Thu, 18 Mar 2004, Gabor Grothendieck wrote:
Tony, Thomas. Thanks for your help. Your comments were very
useful.
Unfortunately, my next step gives me a new round of problems.
The following is the same as the last example except that instead
of hard coding the function into the expression I wanted to
pass it to the expression. It seems like one has to do a double
substitute to get this effect but I am having problems getting this
to work.
The problem is that f is a function, not an expression. You need to work
with body(f)
Either
f-function(){a+1}
body(f)-do.call(substitute,list(body(f),list(a=quote(b f
function ()
{
b + 1
}
or
f-function(){a+1}
body(f)-eval(substitute(substitute(expr,list(a=quote(b))),list(expr=body(f
f
function ()
{
b + 1
}
-thomas
---
Date: Thu, 18 Mar 2004 12:15:45 -0700
From: Tony Plate [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: RE: [R] substitute question
Gabor, I suspect you are overlooking the fact that substitute() does not
evaluate its first argument. So, in the previous example you were giving
an unevaluated expression as the first argument of substitute(). In your
most recent example you are supplying a function object as the first
argument of substitute() (by supplying it as the value to be substituted
for f). I think you can get what you want by first doing f -
Quote(function(){a+1}):
f - Quote(function(){a+1})
f
function() {
a + 1
}
z - substitute(substitute(f,list(a=quote(b))),list(f=f))
z
substitute(function() {
a + 1
}, list(a = quote(b)))
eval(z)
function() {
b + 1
}
eval(eval(z)) # this displays the source attribute
function(){a+1}
body(eval(eval(z)))
{
b + 1
}
hope this helps,
Tony Plate
At Thursday 12:00 PM 3/18/2004, you wrote:
Tony, Thomas. Thanks for your help. Your comments were very
useful.
Unfortunately, my next step gives me a new round of problems.
The following is the same as the last example except that instead
of hard coding the function into the expression I wanted to
pass it to the expression. It seems like one has to do a double
substitute to get this effect but I am having problems getting this
to work.
In the code below we first show that the keep.source option has been
set to FALSE so that the source attribute does not mislead us.
Then we do a double substitute. The outer substitute just creates
the substitute that was in my previous question. This outer
substitute produces z, a call object. So far its as expected.
We then do ze - eval(z) but we get a function object right away. I was
expecting that we get an expression object. Even worse, the function does not
have a replaced with b -- even though this did work in the previous
example
that I posted.
What's wrong?
options()$keep.source
[1] FALSE
f - function(){a+1}
z - substitute(substitute(f,list(a=quote(b))),list(f=f))
class(z)
[1] call
as.list(z)
[[1]]
substitute
[[2]]
function ()
{
a + 1
}
[[3]]
list(a = quote(b))
ze - eval(z)
class(ze)
[1] function
ze
function ()
{
a + 1
}
attr(ze,source)
NULL
---
Date: Thu, 18 Mar 2004 09:00:02 -0800 (PST)
From: Thomas Lumley [EMAIL PROTECTED]
To: Gabor Grothendieck [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: RE: [R] substitute question
On Wed, 17 Mar 2004, Gabor Grothendieck wrote:
I left out the brackets in my last email but the problem
(a reappears after have been substituted out) still remains:
z - substitute( function(){a+1}, list(a=quote(b)) )
z
function() {
b + 1
}
eval(z)
function(){a+1}
Interesting.
Appearances are misleading, however:
z - substitute( function(){a+1}, list(a=quote(b)) )
z
function() {
b + 1
}
f-eval(z)
f()
Error in f() : Object b not found
f
function(){a+1}
attr(f,source)-NULL
f
function ()
{
b + 1
}
So it isn't that eval(z) has a+1 inside, it just has a source attribute
with a+1.
Looking more carefully at z
as.list(z)
[[1]]
`function`
[[2]]
NULL
[[3]]
{
b + 1
}
[[4]]
[1] function(){a+1}
so the original construction of z has kept the source (not, however, as a
source attribute).
There is method to our madness here. It is impossible (or at least too
complicated) to keep comments in the right place as a function is
parsed
Re: [R] substitute question
Gabor Grothendieck [EMAIL PROTECTED] writes:
Tony, Thomas, Thanks, again.
If the problem with my last example was just that I was passing a function
rather than an unevaluated expression, then why don't the following
return f with b in place of a? In both cases, a is still there in the
final output.
f - function() { a + 1 }
z - substitute(substitute(f=f,list(a=quote(b))),list(f=parse(text=deparse(f
eval(eval(z))
or
f - function() { a + 1 }
z - substitute(substitute(expression(f),list(a=quote(b))),list(f=f))
eval(eval(eval(z)))
As far as I can see you don't quote *the expression that creates f* in
either of the above
f - quote(function() { a + 1 })
attr(f,source) - NULL
g - eval(substitute(substitute(f, list(a=quote(b))),list(f=f)))
g
g() # oops, g is not a function but a call to function
g - eval(g)
b - 500
g()
If you have only the function to begin with, try something along these
lines
f - function() { a + 1 }
g - f
attr(g,source) - NULL
body(g) - eval(substitute(substitute(f, list(a=quote(b))),list(f=body(f
g
g()
(The real pain in these examples is that substitute autoquotes its
expr argument. Therefore, when you want to modify an expression that
is already stored in a variable, you need an extra outer layer of
eval(substitute(...)) to poke the content of the variable into the
inner substitute. An esub function with standard evaluation
semantics would make this much easier.)
--
O__ Peter Dalgaard Blegdamsvej 3
c/ /'_ --- Dept. of Biostatistics 2200 Cph. N
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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Re: [R] substitute question
From: Peter Dalgaard [EMAIL PROTECTED] (The real pain in these examples is that substitute autoquotes its expr argument. Therefore, when you want to modify an expression that is already stored in a variable, you need an extra outer layer of eval(substitute(...)) to poke the content of the variable into the inner substitute. An esub function with standard evaluation semantics would make this much easier.) That is one of the frustrations of using substitute. The other is that even if you do perform two levels of substitute, as I have been trying, you still can't count on it working for an arbitrary unevaluated expression, as my examples show. Even putting aside the source attribute which is super confusing until you know about it, all the solutions that I can see to the problem I presented are ugly. (1) One can either pick apart the function using body, or (2) I assume one could convert the function to text and paste together the correct substitute command with the text of the function inserted in its argument. The quote mechanism works but is not applicable if all you have is the function itself (as you point out). This is sooo frustrating. Thanks to you, Tony and Thomas for all your help. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] substitute question
Gabor Grothendieck [EMAIL PROTECTED] writes: From: Peter Dalgaard [EMAIL PROTECTED] (The real pain in these examples is that substitute autoquotes its expr argument. Therefore, when you want to modify an expression that is already stored in a variable, you need an extra outer layer of eval(substitute(...)) to poke the content of the variable into the inner substitute. An esub function with standard evaluation semantics would make this much easier.) That is one of the frustrations of using substitute. The other is that even if you do perform two levels of substitute, as I have been trying, you still can't count on it working for an arbitrary unevaluated expression, as my examples show. Er, I don't think so. All I have seen is a couple of cases where you tried to pass something that was not a language object (e.g. a function as opposed to an expression or call generating a function.) Even putting aside the source attribute which is super confusing until you know about it, all the solutions that I can see to the problem I presented are ugly. (1) One can either pick apart the function using body, or (2) I assume one could convert the function to text and paste together the correct substitute command with the text of the function inserted in its argument. The quote mechanism works but is not applicable if all you have is the function itself (as you point out). This is sooo frustrating. Well, the certain road to frustration is to try to do the impossible. A function is not a language object and you can only do substitutions on language objects. f - function()x g - quote(function()x) f function()x g function() x mode(f) [1] function mode(g) [1] call is.language(f) [1] FALSE is.language(g) [1] TRUE However, a function is basically a triplet consisting of an argument list, a body, and an environment, the middle of which *is* a language object. So I don't think your (1) is ugly, it's the logical way to proceed. -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] substitute question
Date: 18 Mar 2004 23:52:47 +0100 From: Peter Dalgaard [EMAIL PROTECTED] To: [EMAIL PROTECTED] Cc: [EMAIL PROTECTED], [EMAIL PROTECTED], [EMAIL PROTECTED], [EMAIL PROTECTED] Subject: Re: [R] substitute question Gabor Grothendieck [EMAIL PROTECTED] writes: From: Peter Dalgaard [EMAIL PROTECTED] (The real pain in these examples is that substitute autoquotes its expr argument. Therefore, when you want to modify an expression that is already stored in a variable, you need an extra outer layer of eval(substitute(...)) to poke the content of the variable into the inner substitute. An esub function with standard evaluation semantics would make this much easier.) That is one of the frustrations of using substitute. The other is that even if you do perform two levels of substitute, as I have been trying, you still can't count on it working for an arbitrary unevaluated expression, as my examples show. Er, I don't think so. All I have seen is a couple of cases where you tried to pass something that was not a language object (e.g. a function as opposed to an expression or call generating a function.) The parse/deparse one definitely is an expression: z - parse(text=deparse(f)) class(z);mode(z);typeof(z) [1] expression [1] expression [1] expression In the other one, it is not an expression in the inner substitute but should be by the time it gets to the outer one since I explicitly added expression(...). It is expanded which shows it did do that part right. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] substitute question
Gabor, you might have less frustration if you work with Peter's esub() suggestion (use it instead of substitute()): esub - function(expr, subst.list) do.call(substitute, list(expr, subst.list)) When you say In the other one, it is not an expression in the inner substitute but should be by the time it gets to the outer one since I explicitly added expression(...). It is expanded which shows it did do that part right. something to bear in mind is the natural order of evaluation is reversed with nested substitute()'s (because the inner substitute() is quoted by the outer substitute()) (but I haven't carefully reasoned through this being an explanation for your particular frustration here.) hope this might help, Tony Plate At Thursday 04:06 PM 3/18/2004, Gabor Grothendieck wrote: Date: 18 Mar 2004 23:52:47 +0100 From: Peter Dalgaard [EMAIL PROTECTED] To: [EMAIL PROTECTED] Cc: [EMAIL PROTECTED], [EMAIL PROTECTED], [EMAIL PROTECTED], [EMAIL PROTECTED] Subject: Re: [R] substitute question Gabor Grothendieck [EMAIL PROTECTED] writes: From: Peter Dalgaard [EMAIL PROTECTED] (The real pain in these examples is that substitute autoquotes its expr argument. Therefore, when you want to modify an expression that is already stored in a variable, you need an extra outer layer of eval(substitute(...)) to poke the content of the variable into the inner substitute. An esub function with standard evaluation semantics would make this much easier.) That is one of the frustrations of using substitute. The other is that even if you do perform two levels of substitute, as I have been trying, you still can't count on it working for an arbitrary unevaluated expression, as my examples show. Er, I don't think so. All I have seen is a couple of cases where you tried to pass something that was not a language object (e.g. a function as opposed to an expression or call generating a function.) The parse/deparse one definitely is an expression: z - parse(text=deparse(f)) class(z);mode(z);typeof(z) [1] expression [1] expression [1] expression In the other one, it is not an expression in the inner substitute but should be by the time it gets to the outer one since I explicitly added expression(...). It is expanded which shows it did do that part right. ___ No banners. No pop-ups. No kidding. Introducing My Way - http://www.myway.com __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] substitute question
On Thu, 18 Mar 2004, Gabor Grothendieck wrote:
From: Peter Dalgaard [EMAIL PROTECTED]
(The real pain in these examples is that substitute autoquotes its
expr argument. Therefore, when you want to modify an expression that
is already stored in a variable, you need an extra outer layer of
eval(substitute(...)) to poke the content of the variable into the
inner substitute. An esub function with standard evaluation
semantics would make this much easier.)
That is one of the frustrations of using substitute.
The other is that even if you do perform two levels of substitute,
as I have been trying, you still can't count on it working for
an arbitrary unevaluated expression, as my examples show.
You can, but a function isn't an arbitrary unevaluated expression
You still aren't passing an expression containg the symbol `a`. You are
passing an expression containing the function f, whose body is an
expression containing the symbol `a`. Functions aren't expressions.
f - function() { a + 1 }
z - substitute(substitute(f=f,list(a=quote(b))),list(f=parse(text=deparse(f)$
eval(eval(z))
or
f - function() { a + 1 }
z - substitute(substitute(expression(f),list(a=quote(b))),list(f=f))
eval(eval(eval(z)))
(1) One can either pick apart the function using body, or
Actually, I think this is fairly natural -- it is only body(f) that is an
expression. Certainly substitute() could have been written to operate on
functions as well, but it wasn't.
-thomas
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Re: [R] substitute question
I don't think I expressed myself very well on that.
Looking at what we get from the example:
z - substitute(substitute(expression(f),list(a=quote(b))),list(f=f))
z
substitute(expression(function ()
{
a + 1
}), list(a = quote(b)))
class(z);mode(z);typeof(z)
[1] call
[1] call
[1] language
we see that the function seems to be expanded correctly and
the statement does produce a call object. However,
applying eval one, two or three times does not give what
you would think if you looked at z above. In all cases,
a is still there. Some posters have claimed that the
problem is that list(f=f) cannot hold a function and must be
an expression but that's not what the help page says, the
examples are not all of class expression or call and
even if that were true then my parse/deparse example should
work since it is unquestionably an expression.
Date: Thu, 18 Mar 2004 16:41:31 -0700
From: Tony Plate [EMAIL PROTECTED]
To: [EMAIL PROTECTED], [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED], [EMAIL PROTECTED]
Subject: Re: [R] substitute question
Gabor, you might have less frustration if you work with Peter's esub()
suggestion (use it instead of substitute()):
esub - function(expr, subst.list) do.call(substitute, list(expr,
subst.list))
When you say
In the other one, it is not an expression in the inner substitute
but should be by the time it gets to the outer one since I explicitly added
expression(...). It is expanded which shows it did do that part right.
something to bear in mind is the natural order of evaluation is reversed
with nested substitute()'s (because the inner substitute() is quoted by the
outer substitute()) (but I haven't carefully reasoned through this being
an explanation for your particular frustration here.)
hope this might help,
Tony Plate
At Thursday 04:06 PM 3/18/2004, Gabor Grothendieck wrote:
Date: 18 Mar 2004 23:52:47 +0100
From: Peter Dalgaard [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED], [EMAIL PROTECTED],
[EMAIL PROTECTED], [EMAIL PROTECTED]
Subject: Re: [R] substitute question
Gabor Grothendieck [EMAIL PROTECTED] writes:
From: Peter Dalgaard [EMAIL PROTECTED]
(The real pain in these examples is that substitute autoquotes its
expr argument. Therefore, when you want to modify an expression that
is already stored in a variable, you need an extra outer layer of
eval(substitute(...)) to poke the content of the variable into the
inner substitute. An esub function with standard evaluation
semantics would make this much easier.)
That is one of the frustrations of using substitute.
The other is that even if you do perform two levels of substitute,
as I have been trying, you still can't count on it working for
an arbitrary unevaluated expression, as my examples show.
Er, I don't think so. All I have seen is a couple of cases where you
tried to pass something that was not a language object (e.g. a
function as opposed to an expression or call generating a function.)
The parse/deparse one definitely is an expression:
z - parse(text=deparse(f))
class(z);mode(z);typeof(z)
[1] expression
[1] expression
[1] expression
In the other one, it is not an expression in the inner substitute
but should be by the time it gets to the outer one since I
explicitly added expression(...). It is expanded which shows
it did do that part right.
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Re: [R] substitute question
Date: Thu, 18 Mar 2004 15:56:59 -0800 (PST) From: Thomas Lumley [EMAIL PROTECTED] On Thu, 18 Mar 2004, Gabor Grothendieck wrote: (1) One can either pick apart the function using body, or Actually, I think this is fairly natural -- it is only body(f) that is an expression. Certainly substitute() could have been written to operate on functions as well, but it wasn't. In the context of R, natural is performing operations on whole objects at once. Its the same difference as indexing vs. vector operations. I am not sure where this came from about not operating on functions. Are you sure? I know just about everyone is saying this but the help page does not refer to that and the example I gave where we do use list(f=f) for f a function does seem to expand properly (see my last posting) although the subsitution never appears to take effect. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] substitute question
On Thu, 18 Mar 2004, Gabor Grothendieck wrote:
I don't think I expressed myself very well on that.
Looking at what we get from the example:
z - substitute(substitute(expression(f),list(a=quote(b))),list(f=f))
z
substitute(expression(function ()
{
a + 1
}), list(a = quote(b)))
class(z);mode(z);typeof(z)
[1] call
[1] call
[1] language
we see that the function seems to be expanded correctly and
the statement does produce a call object. However,
applying eval one, two or three times does not give what
you would think if you looked at z above.
Maybe we didn't express ourselves well enough.
Looking at z above isn't enough. z is a call to substitute().
Its first operand is an expression. The expression contains a single term,
which is a function.
If you typed
notz- quote(substitute(expression(function ()
{
a + 1
}), list(a = quote(b
you would obtain something that deparsed the same as z, and so looked the
same, but was actually different. In notz the first operand of substitute
is an expression containing multiple terms, which if evaluated would
return a function.
substitute() goes though this expression and checks each term to see if it
is `a`. In z there is only one term and it isn't `a`. In notz there is
(after sufficient recursion) an `a` and it gets replaced.
So
z[[2]][[2]]
function ()
{
a + 1
}
notz[[2]][[2]]
function() {
a + 1
}
are the respective operands, and they still look the same. But
mode(z[[2]][[2]])
[1] function
mode(notz[[2]][[2]])
[1] call
length(z[[2]][[2]])
[1] 1
length(notz[[2]][[2]])
[1] 4
and if we try to find the actual `a` in there
notz[[2]][[2]][[3]][[2]][[2]]
a
z[[2]][[2]][[3]][[2]][[2]]
Error in z[[2]][[2]][[3]] : object is not subsettable
-thomas
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Re: [R] substitute question
On Thu, 18 Mar 2004, Gabor Grothendieck wrote: I am not sure where this came from about not operating on functions. Are you sure? I know just about everyone is saying this but the help page does not refer to that and the example I gave where we do use list(f=f) for f a function does seem to expand properly (see my last posting) although the subsitution never appears to take effect. Yes, I am sure. It's not that it is invalid to have a function as the first argument of substitute, it's that substitute() does not recurse into either the body or the formal arguments of the function (or into the environment, though no-one would expect that). -thomas __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] substitute question
Consider the following example: # substitute a with b in the indicated function. Seems to work. z - substitute( function()a+1, list(a=quote(b)) ) z function() b + 1 # z is an object of class call so use eval # to turn it into an object of class expression; however, # when z is evaluated, the variable a returns. eval(z) function()a+1 Why did a suddenly reappear again after it had already been replaced? __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] substitute question
I left out the brackets in my last email but the problem
(a reappears after have been substituted out) still remains:
z - substitute( function(){a+1}, list(a=quote(b)) )
z
function() {
b + 1
}
eval(z)
function(){a+1}
---
Date: Wed, 17 Mar 2004 20:10:43 -0500 (EST)
From: Gabor Grothendieck [EMAIL PROTECTED]
[ Add to Address Book | Block Address | Report as Spam ]
To: [EMAIL PROTECTED]
Subject: [R] substitute question
Consider the following example:
# substitute a with b in the indicated function. Seems to work.
z - substitute( function()a+1, list(a=quote(b)) )
z
function() b + 1
# z is an object of class call so use eval
# to turn it into an object of class expression; however,
# when z is evaluated, the variable a returns.
eval(z)
function()a+1
Why did a suddenly reappear again after it had already been replaced?
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