Douglas/R-help,
Thanks for your reply. I did try the solution but the result is not
what I expect and I also get the following warning message:
---
Warning message:
number of columns of result
is not a multiple of vector length (arg 1) in: rbind(1, c(6, 9, 10,
12,
And be careful - R is case-sensitive. You have "surv(...)" instead of
"Surv(...)" in your code, that will probably give an error.
The coding is as you have it - 1=failure, 0=censored.
Petr
Christos Hatzis napsal(a):
> The Surv object contains the information on the type of censoring.
> Look
The Surv object contains the information on the type of censoring.
Look at ?Surv
for an explanation of how censored events are represented.
-Christos
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Lu, Jiang
> Sent: Saturday, April 28, 2007 11:
Dear r-helpers,
This is my first time to run survival analysis. Currently, I have a
data set which contains two variables, the variable of time to event
(or time to censoring) and the variable of censor indicator. For the
indicator variable, it was coded as 0 and 1. 0 represents right
censor, 1 me
Just out of curiosity, I took the default "iris" example in the RF
helpfile...
but seeing the admonition against using the formula interface for large data
sets, I wanted to play around a bit to see how the various options affected
the output. Found something interesting I couldn't find documentati
I don't think you can mix plotmath and Hershey but you could do this:
plot(1, xlab = quote(epsilon))
z <- list(x = 1.00823, y = 0.4475955)
text(z$x, z$y, "\\pp", vfont = c("serif", "plain"), xpd = TRUE, cex = .7)
You can set z via:
z <- locator()
On 4/28/07, Matthew Neilson <[EMAIL PROTECTED]
IIRC you have a yes/no smoking variable scored 1/2 ?
It is possibly being read in as a factor not as an
integer.
try
class(df$smoking.variable)
to see .
--- Natalie O'Toole <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I'm getting an error message:
>
> Error in df[, 1:4] * df[, 5] : non-numeri
Thanks for your response, Gabor.
That works quite nicely. The documentation states that it is not possible to
mix and match Hershey fonts with plotmath symbols. My *ideal* scenario would be
to write the
perpendicular symbol as a subscript (specifically, I would like to have "
\epsilon_{\perp}
Hi,
I'm getting an error message:
Error in df[, 1:4] * df[, 5] : non-numeric argument to binary operator
In addition: Warning message:
Incompatible methods ("Ops.data.frame", "Ops.factor") for "*"
here is my code:
##reading in the file
happyguys<-read.table("c:/test4.dat", header=TRUE, row.na
Sorry folks,
With some further checking, it turns out that this sampling
scheme does not conform to the relevant null.
:-(
Chuck
On Sat, 28 Apr 2007, Charles C. Berry wrote:
> Nick Cutler sms.ed.ac.uk> writes:
>
>>
>> I would like to be able to randomise presence-absence (i.e. binary)
>> ma
Thanks Charles, it seems to be working fine now, But I had to
uninstall/reinstall both R and WinEdt.
2007/4/28, Uwe Ligges <[EMAIL PROTECTED]>:
>
>
> Charles Annis, P.E. wrote:
> > Dimitri:
> >
> > Several of us early Vista users have encountered difficulties that
> > ultimately were related to Vi
Hello,
I have two "related" questions, one about MCMClogit and the other about
BRUGS:
Given the data on nausea due to diuretic and nsaid below:
nsaid diureticyes no
0 0 185 6527
0 1 53 1444
1 0 42 1293
1
Reply to self:
set border=NA, stupid.
Paul Artes wrote:
>
> Dear all,
>
> hist ( ) plots a horizontal line at y=0 when the respective bin is empty.
> I can deal with this by modifying the hist object before plotting it
> (x$density[x$density == 0] <- NA), but I'm sure I've seen a more elegant
Dear "stat",
Interesting claim to a name!
In any case, var(X) where X is the data matrix with n rows of 5-variables
should do the trick.
Btw, please read the posting guide: your question is legitimate, hiding your
identity ("stat stat") is not.
Best wishes,
Ranjan
On Sat, 28 Apr 2007 16:36:
stat stat wrote:
> Dear all R users,
>
> I wanted to calculated a sample Variance covariance matrix of a five-variate
> normal distribution. However I stuck to calculate each element of that
> matrix. My question is should I calculate ordinary variance and covariances,
> taking pairwise variable
Hi Sam,
> "SamMc" == Sam McClatchie <[EMAIL PROTECTED]>
> on Fri, 27 Apr 2007 11:21:56 -0700 writes:
SamMc> System:Linux kernel 2.6.15 Ubuntu dapper
...
SamMc> Has anyone figured out how to make the R-help digest
SamMc> more easily readable in the Thunderbird mai
> "tom" == tom soyer <[EMAIL PROTECTED]>
> on Sat, 28 Apr 2007 08:15:39 -0500 writes:
tom> I wanted to understand exactly how acf and pacf works,
tom> so I tried to calculate ac and pac manually. For ac, I
tom> used the standard acf formula: acf(k) =
tom> sum(X(t)-Xbar)
Thanks for your reply Frank.
I realize that the cex values are relative. The problem is that for
high resolutions, cex has to be high (e.g., around 5) for the tick
and curve labels to be legible. The curve and tick labels were tiny
when I used cex=1 and set the other cex values to 1/5. But when
Nick Cutler sms.ed.ac.uk> writes:
>
> I would like to be able to randomise presence-absence (i.e. binary)
> matrices whilst keeping both the row and column totals constant. Is
> there a function in R that would allow me to do this?
>
> I'm working with vegetation presence-absence matrices bas
Natalie O'Toole napsal(a):
> Does anyone know why it is giving me this error? Any help would be greatly
> appreciated!!
>
> Thanks,
>
> Nat
>
>
>
> myfile<-("c:/test2.txt")
> mysubset<-myfile
> mysubset$Y_Q02 <-mysubset$DVSELF <-NULL
> mysubset2<-mysubset
> mysubset2$Y_Q10B <-mysubset2$GP2_
On Sat, 28 Apr 2007, Eric Thompson wrote:
> Professor Ripley,
>
> Thank you for your comments.
>
> On 4/28/07, Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
>> How big is abs(lam - lam2[i])/Meps ?
>
> Here is the result on my system:
>
>> abs(lam - lam2[i])/Meps
> [1] 60 18 17 0 12
>
> which is qu
Professor Ripley,
Thank you for your comments.
On 4/28/07, Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
> How big is abs(lam - lam2[i])/Meps ?
Here is the result on my system:
> abs(lam - lam2[i])/Meps
[1] 60 18 17 0 12
which is quite surprising to me that the maximum value exactly equals
the
Dear all R users,
I wanted to calculated a sample Variance covariance matrix of a five-variate
normal distribution. However I stuck to calculate each element of that matrix.
My question is should I calculate ordinary variance and covariances, taking
pairwise variables? or I should take partial
Pedro Sobral wrote:
> Dear R super-users,
>
> I am quite new in using R and I am not managing to edit factors.
>
> Lest suppose that one has the following data:
>
> Factor A
> Factor B
> Factor C
> Claims
>
> Factor A has 3 factors (1,2 and 3). To simplify the glm model I only want to
> have
> "Arun" == Arun Kumar Saha <[EMAIL PROTECTED]>
> on Thu, 26 Apr 2007 23:44:03 +0530 writes:
Arun> Dear all R-users,
Arun> I would like to draw a tangent of a given function for a particular
(given)
Arun> point. However the straight line representing it should not cut any
Thomas Funke wrote:
> Hi all,
>
> I tried to do normalization of affymetrix data with bioconductor on a
> Linux server. When I read in the cel files all seemed ok. But the next
> step caused an error. With Win XP all works fine. Did anyone experience
> similar problems?
>
> Thanks,
>
> Tho
Charles Annis, P.E. wrote:
> Dimitri:
>
> Several of us early Vista users have encountered difficulties that
> ultimately were related to Vista's treating only the Administrator as having
> permission to make some changes. Even if you are the only user, you still
> do not have administrative pr
Dimitri:
Several of us early Vista users have encountered difficulties that
ultimately were related to Vista's treating only the Administrator as having
permission to make some changes. Even if you are the only user, you still
do not have administrative privileges by default. (I think this is a
Its available in the Hershey fonts:
plot(0, 0, type = "n")
text(0, 0, "A \\pp B", vfont = c("serif", "plain"))
On 4/28/07, Matthew Neilson <[EMAIL PROTECTED]> wrote:
> Hey,
>
> Does anyone know of an equivalent to the LaTeX \perp (perpendicular)
> symbol for adding to R plots? Parallel is easy e
Dear R super-users,
I am quite new in using R and I am not managing to edit factors.
Lest suppose that one has the following data:
Factor A
Factor B
Factor C
Claims
Factor A has 3 factors (1,2 and 3). To simplify the glm model I only want to
have 2 factor (let's say 1 and 3).
I should I do
Hey,
Does anyone know of an equivalent to the LaTeX \perp (perpendicular)
symbol for adding to R plots? Parallel is easy enough ("||"), but I
haven't been
able to find a way of adding perpendicular. The plotmath documentation
doesn't mention how to do it, so I'm inclined to think that it doesn't
Hi,
I have a new computer with Windows Vista and I am trying to use
RWinEdt, which I have always used. I am using R version 2.5.
The installation of the RWinEdt library is funny. First, it didn't
install at all. Then, I uninstalled/reinstalled both R and WinEdt,
downloaded the package again from
On 4/28/2007 6:20 AM, AJ Rossini wrote:
>
> I agree entirely with Gabor. My advice would be to just ignore the people
> who
> think differently
That's fairly bad advice, in that many of the people who actually
provide helpful advice are old-fashioned, and like to know who they're
providing
Hello!
I need to compare 2 datasets whether they come from the same distribution. I
use function ks.boot{Matching}. And what is the confidence level of the
p-value, returned by ks.boot function?
The code is:
set=read.table("http://stella.sai.msu.ru:8080/~gala/data/testsets.csv";,
On 4/28/07, Ajit Pawar <[EMAIL PROTECTED]> wrote:
> Greetings,
> This might be something very simple but a nice solution eludes me!!
>
>I have a function that I call within sapply that generates data frame
> in each call. Now when sapply returns me back the result - it's in the form
>
> "Matt" == Matthew Neilson <[EMAIL PROTECTED]>
> on Fri, 27 Apr 2007 15:54:20 +0100 writes:
Matt> Hey Felix,
Matt> So basically what you want is a figure containing a block of four
plots, with a main title for the figure? If that's the case then something like
this should wo
Hi,
I wanted to understand exactly how acf and pacf works, so I tried to
calculate ac and pac manually. For ac, I used the standard acf formula:
acf(k) = sum(X(t)-Xbar)(X(t-k)-Xbar))/sum(X(t)-Xbar)^2. But for pac, I could
not figure out how to calculate it by hand. I understand that in both R and
Hi all,
I tried to do normalization of affymetrix data with bioconductor on a
Linux server. When I read in the cel files all seemed ok. But the next
step caused an error. With Win XP all works fine. Did anyone experience
similar problems?
Thanks,
Thomas
> PI <- ReadAffy()
> PI
AffyBatch
Brian O'Connor wrote:
> R-Masters,
>
> I need to produce high resolution line plots and place labels on the
> curves. It seems that cex must be high relative to the other cex
> values in order to produce sufficiently large & legible tick labels
> at high resolutions. But high cex values cause t
R-Masters,
I need to produce high resolution line plots and place labels on the
curves. It seems that cex must be high relative to the other cex
values in order to produce sufficiently large & legible tick labels
at high resolutions. But high cex values cause the curve labels to
become giganti
Greetings,
This might be something very simple but a nice solution eludes me!!
I have a function that I call within sapply that generates data frame
in each call. Now when sapply returns me back the result - it's in the form
of a "list of data frames". so in order to extract the infor
I agree entirely with Gabor. My advice would be to just ignore the people who
think differently -- however, if you want those particular folks to respond,
you'll have to play by their rules. (and if you don't play by their rules,
you'll just have to ignore the consequences -- this _IS_ the i
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