Try
?update.packages
Ross Darnell
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Wensui Liu
Sent: Tuesday, 11 September 2007 11:45 AM
To: r-help@stat.math.ethz.ch
Subject: [R] install packages automatically
Dear Listers,
I am a little tired of
Thanks Frede
I didn't know about the "r" type.
Ross Darnell
-Original Message-
From: Frede Aakmann Tøgersen [mailto:[EMAIL PROTECTED]
Sent: Mon 10-Sep-07 4:45 PM
To: Ross Darnell; r-help@stat.math.ethz.ch
Subject: SV: [R] lattice panel.lmline problem
Why not use
the last line without the "if"
statement which obviously means it's ignored.
Of course I may be taking the wrong tack completely to get the result I
need. Any advice would be appreciated
Ross Darnell
[[alternative HTML version deleted]]
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The answer to your first question is
curve(ifelse((x>-1)&(x<1),x^2+1,x+2),from=-10,to=10)
Sorry I cannot understand the next part of your query.
Ross Darnell
On Sun, 2007-09-02 at 00:31 -0700, JHawk3 wrote:
> Hi all,
>
> I'm a relatively new user to the R interface,
Then you can regenerate the factor by
common.without.Method4 <- subset(common, Method!=4)
common.without.Method4 <- transform(common.without.Method4,
Method = factor(Method))
Ross Darnell
-Original Message-
From:
Does this help
common.without.Method4 <- subset(common, Method!=4)
xyplot(MeanBxg ~ PercentVarExplained | bdg.f * bdx.f,
data=common.without.Method4,
groups=Method.f, type="l", auto.key=T)
Ross Darnell
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]
osest to 12pm?
Perhaps for condition 2 for a animal which.min(time-1200) (replace 1200 with
properly defined timestamp representing 12:00pm)
Ross Darnell
-Original Message-
From: [EMAIL PROTECTED] on behalf of Tim Sippel
Sent: Wed 08-Aug-07 9:37 AM
To: r-help@stat.math.ethz.ch
Su
You could read the file and test each line (try readline()) but you need some
rule to distinguish between text lines and data lines. You could then reread
the file with the number of skip lines defined.
Ross Darnell
-Original Message-
From: [EMAIL PROTECTED] on behalf of Tom Cohen
Manuel
Jim's may be what you want-- a list of numerics with names P, T and Q or
a list of character strings?
> str <- "P = 0.0, T = 0.0, Q = 0.0"
> str(as.vector(unlist(strsplit(str,",")),mode="list"))
List of 3
$ : chr "P = 0.0"
$ : chr " T = 0.0"
$ : chr " Q = 0.0"
-Original Message-
Is this what your want?
as.vector(unlist(strsplit(str,",")),mode="list")
Ross Darnell
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Manuel Morales
Sent: Friday, 27 July 2007 10:39 AM
To: r-help
Subject: [R] Convert string to lis
Ana
You are estimating a random coefficient model on 5 individuals (mean
and variance). Are you sure this is wise?
Ross Darnell
- Original Message -
From: Ana Conesa <[EMAIL PROTECTED]>
Date: Thursday, July 5, 2007 1:21 am
Subject: [R] questions on lme function
>
> Dear
Alternatively generate the log-likelihood using the sum(dpois(y,
fitted(model), log = TRUE))
Regards
Ross Darnell
Doxastic wrote:
>
> You're right. I do need to learn more. I never learned null/residual
> deviance. I know the deviance is equivalent to an anova decompostio
I would appreciate help trying
to set different plot types in a lattice plot with multiple responses e.g.
xyplot(y+fval~x)
but have "y" as points and "fval" as a line.
I have tried
xyplot(y+fval~x,type=c("p","l"))
but this results in both plots
I cannot reproduce Fig. 13.2 in MASS4.
> plot(gehan.surv,fun='cloglog')
Warning message:
2 x values <= 0 omitted from logarithmic plot in: xy.coords(x, y,
xlabel, ylabel, log)
and the x-axis is badly scaled.
I was wondering if someone can help
Rega
Hi Vincent
This would seem logical but in this case doesn't work.
It doesn't seem to be a Sweave problem (feature) at all but within R as
Hadley stated.
Within R try
> quote(women[1])
women[1]
now try
> quote("["(women,1))
women[1]
So it's parsed and "normalised" (there's a familiar term);
I am wanting to use the "[" operator in an S-chunk, e.g.
<<>>=
str(women)
women$height
women[,1]
"["(women,1)
@
to show the equivalence of three methods of extracting an element from
a data.frame.
However Sweave returns the last of these as
women[1]
in the S input chunk
How can I force it
described.
I would appreciate any suggestions
Ross Darnell
University of Queensland
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
,"X","Xcopy")
model <- lm(y~X+Xcopy,data=Xnew)
coef(model) # reports NA for Xcopy estimate
summary(model) # ditto
coef(summary(model)) # drops NA for Xcopy estimate
Is this intentional? Is there a way of extracting the coefficient table
from the summary output with the aliased
es 0 when it should (IMHO and a few others perhaps) be
1 - RSS/TSS(corrected for the mean only)
RSS = sum(resid(my.model)^2)
TSS = sum((y-mean(y))^2)
Have I missed this somewhere in the FAQ's or elsewhere?
Regards
Ross Darnell
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Sweavers
As the title suggests I would appreciate any help to include S code in
ordinary paragraph mode. I can use the textsl font but is isn't the
same.
Thanks
Ross Darnell
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I would appreciate help translating the following lme model to an lmer
function.
lme(lognrms ~ Group*Rotation*muscle*side*support*arms,
random=~1|Subject/Stratum2/rep, data=Data)
Many thanks
Ross Darnell
[EMAIL PROTECTED]
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R
R-colleagues
I have adapted the anova.lmlist function to use the model object name as
the first column in the output instead of the string "Model n".
If there is general agreement can the change be implemented into the
stats package?
Regards
Ross Darnell
--
University of
necessarily the same as the
original data frame. You need a bit more work to get the right model.matrix
that correspond to the newdata. It's not clear to me whether you want to
return model matrix or model frame, but in either case it's not sufficient
to just use `newdata'.
Andy
Fro
fit <- napredict(na.act, fit)
se.fit <- napredict(na.act, se.fit)
}
predictors <- if (missing(newdata)) model.matrix(object) else newdata
pred <- list(predictors=predictors,
fit = fit, se.fit = se.fit,
residual.scale = residual.scale)
}
Is it possible to produce a histogram directly using the hist()
function with the common borders removed?
It can be done by plotting the histogram object using type 's'teps.
my.hist <- hist(x,plot=FALSE)
plot(my.hist$breaks,c(0,my.hist$counts),type='s')
I would apprec
#x27;EMG/Graphics'))
try(winMenuDel("EMG"))
winMenuAdd("EMG")
winMenuAdd("EMG/Graphics")
winMenuAddItem("EMG/Graphics","Plot trace","plot.trace()")
}
"plot.trace" <- function(){
x <- eval(parse(text=trace.dialog(
ng if anyone else is experiencing this problem
Regards
Ross Darnell
--
University of Queensland, Brisbane QLD 4067 AUSTRALIA
Email: <[EMAIL PROTECTED]>
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Frank Harrell wrote
> On Wed, 05 Nov 2003 09:22:34 +1000
> Ross Darnell <[EMAIL PROTECTED]> wrote:
>
> > I have fitted a logistic regression model
> >
> > > failed.lr2$call
> > lrm(formula = failed ~ Age + task2 + Age:task2, data = time.long,
>
variable Age does not have limits defined by datadist
>
I get an error. The NA values in ddist seem to be the problem but I
don't understand the datadist information.
Can anyone help explain why this is failing.
Thanks
Ross Darnell
--
University of Queensland, Brisbane QLD
interested in attending can find out more by looking at
http://www.sci.usq.edu.au/staff/dunn/qstatconf/workshops.html
--
Ross Darnell
Organising committee
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-subject variance.
Many thanks
Ross
Douglas Bates <[EMAIL PROTECTED]> writes:
> Ross Darnell <[EMAIL PROTECTED]> writes:
>
>> I would like some advice on how if possible, to test the following
>>
>> I have subjects each measured several times. The subjects a
lme0 <- lme(y~group,random=~1|subject)
I did think that the model that defined a specific between-subject
variance for each group was
update(lme0,.~., weights=varIdent(form=~1|group))
but I am not sure.
I would appreciate any help.
Ross Darnell
qnorm(beta/2))^2
s = standard deviation
delta = clinically important difference
alpha = Type I error
beta = Type II error
--
Ross Darnell
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her random
>effects.
>Looking forward to your help.
It may be that you want to use the "weights=varIdent(form=~1|groups)"
argument which will fit a variance for each level of the groups
factor.
--
Ross Darnell
School of Health and Rehabilitation Sciences
University of Queensla
The norm package of Joe Schafer can be used. It has a function called
'prelim' (or very similar) that does this.
Regards
Ross Darnell
Email: <[EMAIL PROTECTED]>
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?
2. Why does the toplevel frame shrink if though I define the width of
the menubar frame and set the fill="x"?
Thanks
Ross Darnell
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require(tcltk) || stop("tcltk support is absent")
# Generate the main frame
.fr <- tktoplevel(widt
I cannot find the glmmNQ function in the MASS package (or anywhere else I have tried)
mentioned on page 296 of MASS4.
I would appreciate directions.
Thanks
--
Ross Darnell
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