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
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
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
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 the
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
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
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 work:
Matt #
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
of a
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 4/28/2007 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 it
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
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
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
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 enough (||),
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
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
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,
Thomas
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
axis, it
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 2 factor
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
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
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 quite surprising
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_07 -NULL
Nick Cutler s0455078 at 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
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
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
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 mail client?
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 variables? or
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 16:36:55
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
way.
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
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 Vista's
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 s0455078 at sms.ed.ac.uk writes:
I would like to be able to randomise presence-absence (i.e. binary)
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.names=1)
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}
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$smoking.variable)
to see .
--- Natalie O'Toole [EMAIL PROTECTED] wrote:
Hi,
I'm getting an error message:
Error in df[, 1:4] * df[, 5] : non-numeric
I don't think you can mix plotmath and Hershey but you could do this:
plot(1, xlab = quote(epsilon))
z - list(x = 1.00823, y = 0.4475955)
text(z$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] wrote:
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
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
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 11:10
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 at
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