[R] update.lm question
Hello,
at the Rgui command line I can easily remove a term from a fitted lm
object, like
fit - lm(y~x1+x2+x3, data=myData)
update(fit, .~.-x1)
However, I would like to do this in a function with term given as string, like
removeTerm - function(linModel, termName) { ??? }
removeTerm(fit, x1)
but I can not fill the ???. I already tried
removeTerm - function(linModel, termName) { update(linModel, .~. - termName },
removeTerm - function(linModel, termName) { update(linModel, .~. -
as.name(termName) },
removeTerm - function(linModel, termName) { update(linModel, .~. -
eval(termName) },
removeTerm - function(linModel, termName) { update(linModel, .~. -
eval.parent(termName) },
removeTerm - function(linModel, termName) { update(linModel, .~. -
get(termName) },
but these attempts produce error messages.
Can you advise me here?
Kind regards,
Karsten
__
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.
Re: [R] update.lm question
You need to review:
?update
?update.formula
?as.formula
The way should be clear at that point. If not, then include a
reproducible data example to work with.
--
David
On Nov 15, 2009, at 9:23 AM, Karsten Weinert wrote:
Hello,
at the Rgui command line I can easily remove a term from a fitted lm
object, like
fit - lm(y~x1+x2+x3, data=myData)
update(fit, .~.-x1)
However, I would like to do this in a function with term given as
string, like
removeTerm - function(linModel, termName) { ??? }
removeTerm(fit, x1)
but I can not fill the ???. I already tried
removeTerm - function(linModel, termName) { update(linModel, .~. -
termName },
removeTerm - function(linModel, termName) { update(linModel, .~. -
as.name(termName) },
removeTerm - function(linModel, termName) { update(linModel, .~. -
eval(termName) },
removeTerm - function(linModel, termName) { update(linModel, .~. -
eval.parent(termName) },
removeTerm - function(linModel, termName) { update(linModel, .~. -
get(termName) },
but these attempts produce error messages.
Can you advise me here?
Kind regards,
Karsten
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
__
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.
Re: [R] update.lm question
Hello David,
of course I read those help pages and searched the r-help archive.
I agree that the way should be clear, but I am stuck and looking for
help. Here is reproducible code
removeTerm1 - function(linModel, termName) { update(linModel, .~. - termName) }
removeTerm2 - function(linModel, termName) { update(linModel, .~. -
as.name(termName)) }
removeTerm3 - function(linModel, termName) { update(linModel, .~. -
eval(termName)) }
removeTerm4 - function(linModel, termName) { update(linModel, .~. -
eval.parent(termName)) }
removeTerm5 - function(linModel, termName) { update(linModel, .~. -
get(termName)) }
myData - data.frame(x1=rnorm(10), x2=rnorm(10), x3=rnorm(10), y=rnorm(10))
fit - lm(y~x1+x2+x3, data=myData)
# all this does not work, as I am expecting the function to return a
lm object with formula y~x2+x3
removeTerm1(fit, x1)
removeTerm2(fit, x1)
removeTerm3(fit, x1)
removeTerm4(fit, x1)
removeTerm5(fit, x1)
Any help appreciated,
kind regards,
Karsten Weinert
2009/11/15 David Winsemius dwinsem...@comcast.net:
You need to review:
?update
?update.formula
?as.formula
The way should be clear at that point. If not, then include a reproducible
data example to work with.
--
David
__
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.
Re: [R] update.lm question
On 15/11/2009 9:23 AM, Karsten Weinert wrote:
Hello,
at the Rgui command line I can easily remove a term from a fitted lm
object, like
fit - lm(y~x1+x2+x3, data=myData)
update(fit, .~.-x1)
However, I would like to do this in a function with term given as string, like
removeTerm - function(linModel, termName) { ??? }
removeTerm(fit, x1)
but I can not fill the ???. I already tried
removeTerm - function(linModel, termName) { update(linModel, .~. - termName },
removeTerm - function(linModel, termName) { update(linModel, .~. -
as.name(termName) },
removeTerm - function(linModel, termName) { update(linModel, .~. -
eval(termName) },
removeTerm - function(linModel, termName) { update(linModel, .~. -
eval.parent(termName) },
removeTerm - function(linModel, termName) { update(linModel, .~. -
get(termName) },
but these attempts produce error messages.
Can you advise me here?
There are two problems:
1. .~. is different from . ~ ..
2. You need to construct the formula . ~ . - x1, and none of your
expressions do that. You need to use substitute() or bquote() to edit a
formula. For example, I think both of these should work:
removeTerm - function(linModel, termName)
update(linModel, bquote(. ~ . - .(as.name(termName
removeTerm - function(linModel, termName)
update(linModel, substitute(. ~ . - x, list(x=as.name(termName
Duncan Murdoch
__
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.
Re: [R] update.lm question
On Nov 15, 2009, at 11:15 AM, Karsten Weinert wrote:
Hello David,
of course I read those help pages and searched the r-help archive.
But did you look at the as.formula page?
I agree that the way should be clear, but I am stuck and looking for
help. Here is reproducible code
And here is what seemed like the obvious extension of your example but
using as.formula:
myData - data.frame(x1=rnorm(10), x2=rnorm(10), x3=rnorm(10),
y=rnorm(10))
fit - lm(y~x1+x2+x3, data=myData)
remvterm - function(ft, termname) update(ft, as.formula(paste(.~.-,
termname, sep=)))
remvterm(fit, x3)
Call:
lm(formula = y ~ x1 + x2, data = myData)
Coefficients:
(Intercept) x1 x2
-0.2598 -0.0290 -0.2645
--
David
removeTerm1 - function(linModel, termName) { update(linModel, .~. -
termName) }
removeTerm2 - function(linModel, termName) { update(linModel, .~. -
as.name(termName)) }
removeTerm3 - function(linModel, termName) { update(linModel, .~. -
eval(termName)) }
removeTerm4 - function(linModel, termName) { update(linModel, .~. -
eval.parent(termName)) }
removeTerm5 - function(linModel, termName) { update(linModel, .~. -
get(termName)) }
myData - data.frame(x1=rnorm(10), x2=rnorm(10), x3=rnorm(10),
y=rnorm(10))
fit - lm(y~x1+x2+x3, data=myData)
# all this does not work, as I am expecting the function to return a
lm object with formula y~x2+x3
removeTerm1(fit, x1)
removeTerm2(fit, x1)
removeTerm3(fit, x1)
removeTerm4(fit, x1)
removeTerm5(fit, x1)
Any help appreciated,
kind regards,
Karsten Weinert
2009/11/15 David Winsemius dwinsem...@comcast.net:
You need to review:
?update
?update.formula
?as.formula
The way should be clear at that point. If not, then include a
reproducible
data example to work with.
--
David
__
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.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
__
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.
Re: [R] update.lm question
On 15/11/2009 11:28 AM, Duncan Murdoch wrote:
On 15/11/2009 9:23 AM, Karsten Weinert wrote:
Hello,
at the Rgui command line I can easily remove a term from a fitted lm
object, like
fit - lm(y~x1+x2+x3, data=myData)
update(fit, .~.-x1)
However, I would like to do this in a function with term given as string, like
removeTerm - function(linModel, termName) { ??? }
removeTerm(fit, x1)
but I can not fill the ???. I already tried
removeTerm - function(linModel, termName) { update(linModel, .~. - termName },
removeTerm - function(linModel, termName) { update(linModel, .~. -
as.name(termName) },
removeTerm - function(linModel, termName) { update(linModel, .~. -
eval(termName) },
removeTerm - function(linModel, termName) { update(linModel, .~. -
eval.parent(termName) },
removeTerm - function(linModel, termName) { update(linModel, .~. -
get(termName) },
but these attempts produce error messages.
Can you advise me here?
There are two problems:
1. .~. is different from . ~ ..
Oops, wrong. Those are the same. Sorry...
Duncan Murdoch
2. You need to construct the formula . ~ . - x1, and none of your
expressions do that. You need to use substitute() or bquote() to edit a
formula. For example, I think both of these should work:
removeTerm - function(linModel, termName)
update(linModel, bquote(. ~ . - .(as.name(termName
removeTerm - function(linModel, termName)
update(linModel, substitute(. ~ . - x, list(x=as.name(termName
Duncan Murdoch
__
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.
__
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.
Re: [R] update.lm question
Thanks Duncan and David
for opening my eyes :-). It took quite a while but I think I learned a
lot about lm today. I used your advice to produce added variable
plots as mentioned here [1], [2]. I would bet someone did it in R
already (may leverage.plot in car) but it was worth doing it myself.
Kind regards,
Karsten.
[1] http://www.minitab.com/support/documentation/answers/AVPlots.pdf
[2]
http://www.mathworks.com/access/helpdesk/help/toolbox/stats/addedvarplot.html
plotAddedVar.lm - function(
linModel,
termName,
main=,
xlab=paste(termName, | andere),
ylab=paste(colnames(linModel$model)[1], | andere),
cex=0.7, ...) {
oldpar - par(no.readonly = TRUE); on.exit(par(oldpar))
par(mar=c(3,4,0.4,0)+0.1, las=1, cex=cex)
yData = residuals(update(linModel, substitute(. ~ . - x,
list(x=as.name(termName)
xData = residuals(update(linModel, substitute(x ~ . - x,
list(x=as.name(termName)
plot(xData, yData, main=main, xlab=, ylab=)
mtext(side=2, text=ylab, line=3, las=0, cex=cex)
mtext(side=1, text=xlab, line=2, las=0, cex=cex)
abline(h=0)
abline(a=0, b=coefficients(linModel)[termName], col=blue)
}
plotAddedVar - function(linModel,...) UseMethod(plotAddedVar)
__
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.
