Johannes Graumann wrote:
ronggui wrote:
I think you should use labels = c(= 0.66, = 0.33, = -0.33,
= -0.66) .
Note the quote in each element of the vector.
No. Am using that at the moment. Does not give me the nice
plotmath 'greater-than' etc. signs ...
Then please read ?plotmath
Guan-Hua Huang wrote:
Dear All,
My R version: 2.0.1 ; OS using: Linux. I install the package cluster by
using install.packages(cluster). After install it, it runs fine for
function clara, but it does not work for function fanny. I did the following
things:
library(cluster)
Farrel Buchinsky wrote:
I am using dgc.genetics to perform TDT analysis on SNP data from a cohort of
trios.
I now have a file with about 6008 variables. The first few variables related
to the pedigree data such as the pedigree ID the person ID etc. Thereafter
each variable is a specific
mingyu shi wrote:
Hello,
I have an ordinary linear regression model.
for example y~x1+x2+x3;
I need to compute the percentage of variance explained by each covariate.
can anyone tell me how I can compute it in R?
Please try to solve your homework yourself or by reading a good textbook
-
Then please read ?plotmath and use it:
labels = expression( = 0.66, == 0.33, = -0.33, = -0.66)
pie(1:3,lab=)
text(locator(3),lab=expression(=1,=2,=3))
Error: attempt to use zero-length variable name
text(locator(3),lab=expression(NULL=1,NULL=2,NULL=3))
What 's the problem come from?
Quin Wills wrote:
I want use a validation set for my classification tree rather than the
default 10-fold validation in rpart() but can't see which arguments to use
to get this right. Advice appreciated thanks. I assume that this is
possible!
You cannot for the internal algorithm that
ronggui wrote:
Then please read ?plotmath and use it:
labels = expression( = 0.66, == 0.33, = -0.33, = -0.66)
pie(1:3,lab=)
text(locator(3),lab=expression(=1,=2,=3))
Error: attempt to use zero-length variable name
text(locator(3),lab=expression(NULL=1,NULL=2,NULL=3))
What
Dear Spencer
Thank you for your answer. You are correct. The adjustSigma
argument is only used for the ML method. In the code there is:
stdFixed - sqrt(diag(as.matrix(object$varFix)))
if (object$method == ML adjustSigma == TRUE) {
stdFixed - sqrt(object$dims$N/(object$dims$N -
I fit a proportional odds model
with the polr-function of the MASS package from
Venables and Ripley
Applying the confint method to calculate confidence intervals for the
parameters I get
the following error message
Waiting for profiling to be done...
Re-fitting to get Hessian
Error in X[,
On Tue, 2 May 2006, Brian O'Gorman wrote:
I can't make R on a Solaris 9 box. Does anyone have any suggestions? Thanks
in advance.
Looks like the effect of some header you are including from Solaris9. Try
adding
#undef open
near the top of src/main/connections.c, after fcntl.h is included.
Many thanks. I'm using it for pruning and was hoping that rpart allows use
of a validation set rather than cross-validation for generating a CP/error
table.
-Original Message-
From: Uwe Ligges [mailto:[EMAIL PROTECTED]
Sent: 03 May 2006 07:53
To: Quin Wills
Cc: r-help@stat.math.ethz.ch
Hello Anne,
you will find the necessary details in the following paper, its
publication in 'Biometrika' is cited in ?ur.pp.
http://cowles.econ.yale.edu/P/cp/p07a/p0706.pdf
Best,
Bernhard
Hello,
Could someone give me a hint about what might be the
difference between running urppTest
with
Dear Jiang,
We do a lot of plots of that sort. At first I read the time values with
as.Date(timevariable, format = %d/%m/%Y). Then You can plot along like
Plot(as.Date-variable, your target variable, type = l)
If You set axes = F than you can customize the xAxis with axis(1, ...)
The
Dear Rexperts,
I have recently build R-2.3.0 from source on a Linux system and have
encountered the following problem (perhaps not exactly a problem, but a
minor display flaw):
demo()
+here is how the output looks in less+
Demos in package E28098baseE28099:
is.things
Hi,
I've a matrix in 20*11 order. There are 11 variables,
i.e 11 columns and each variable have 20 row data. Now
i want to calculate covariance between any variable
with others taking 4 rows at a time, so that there
will be 5 blocks. How can i do this using any
R-function? If i want to do it in
lapply(split(mat, gl(nrow(mat)/4, 4, nrow(mat)), cov)
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
On Wed, 3 May 2006, Quin Wills wrote:
Many thanks. I'm using it for pruning and was hoping that rpart allows use
of a validation set rather than cross-validation for generating a CP/error
table.
Since it is not documented how to, why do you expect to? Indeed, why do
you think it would be a
You are almost certainly using a UTF-8 locale and have the character set
for less and/or your terminal set incorrectly. This works exactly as
intended on a properly configured Linux system in single-byte and UTF-8
locales.
This would be quite inappropriate as a bug report on R.
On Wed, 3 May
Is it not true that cross-validation can sometimes over estimate
classification error - versus bringing in an external validation data set
and checking its classification error? I was trying to test this out, but
from what I see either way seems to be much of muchness.
-Original Message-
I tried your code, but it's giving the following
error..
Error in match.fun(FUN) : argument FUN is missing,
with no default
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
Hi,All,
I am trying to fit a GJR-GARCH model in R:
r_t = mu + e_t
h_t = alp_0 + alp_1 * e_(t-1)^2 + alp_2 * s_(t-1) * e_(t-1)^2 + beta *
h_(t-1)
where r_t = return (on day t), h_t = conditional volatility on day t,
and s_(t-1) = 1 if e_(t-1) 0 (zero otherwise).
I have downloaded the
sorry (need a data.frame) :
lapply(split(as.data.frame(mat), gl(nrow(mat)/4, 4, nrow(mat))), cov)
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Dear R users,
Suppose I have a 4*5 data frame like this:
Date x1 x2 x3 x4
1/1/2004100 22 12 34
2/1/200433-22 33 12
3/1/20041212 -115
4/1/20041 223 22 1
Now I want to get a correlation matrix
Dear R users,
For those of you who like to use the vim editor on a Unix
platform, I put together a little homepage for my vim plugin
which is also known as vim script No. 1048, featuring a
screenshot:
http://www.uft.uni-bremen.de/chemie/ranke/index.php?page=vim_R_linux
Have a good day!
On 5/3/06, Uwe Ligges [EMAIL PROTECTED] wrote:
Looks like you have to be much more specific:
tdt() is a function within dgc.genetics.
dgc.genetics is a package written by David Clayton and available at
http://www-gene.cimr.cam.ac.uk/clayton/software/
It consists of extensions to the genetics
Arun Kumar Saha wrote:
Dear R users,
Suppose I have a 4*5 data frame like this:
Date x1 x2 x3 x4
1/1/2004100 22 12 34
2/1/200433-22 33 12
3/1/20041212 -115
4/1/20041 223 22 1
Now I
Hi R-users,
I have a data set with a particular type of interval-censored data: right
and left censored data, but no other type of interval censoring. I need to
fit a model with covariates to evaluate the influence of them on the
survival time.
First: I use Surv of library(survival) to create
Farrel Buchinsky wrote:
On 5/3/06, Uwe Ligges [EMAIL PROTECTED] wrote:
Looks like you have to be much more specific:
tdt() is a function within dgc.genetics.
dgc.genetics is a package written by David Clayton and available at
http://www-gene.cimr.cam.ac.uk/clayton/software/
It
Prof Brian Ripley wrote:
You are almost certainly using a UTF-8 locale and have the character set
for less and/or your terminal set incorrectly. This works exactly as
intended on a properly configured Linux system in single-byte and UTF-8
locales.
Thank you very much for the hint! I'll be
On Wed, 3 May 2006, Ulrich Halekoh wrote:
I fit a proportional odds model
with the polr-function of the MASS package from
Venables and Ripley
Applying the confint method to calculate confidence intervals for the
parameters I get
the following error message
Waiting for profiling to be
On Wed, 3 May 2006, Berta wrote:
My question is: How could I fit a survival model that include right-left
censoring and time-dependent covariates?
Not with anything in the survival package, unfortunately.
-thomas
Thomas Lumley Assoc. Professor, Biostatistics
[EMAIL
Are there any online courses for learning R?
__
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
Hello Scott,
see:
http://www.statistics.com/content/courses/R/index.html
http://www.statistics.com/content/courses/graphicsR/index.html
http://www.statistics.com/content/courses/modelingR/index.html
there has been a recent posting on this list, if I recall it correctly.
Best,
B.
How does one create a vector whose contents is the list of variables in a
dataframe pertaining to a particular pattern?
This is so simple but I cannot find a straightforward answer.
I want to be able to pass the contents of that list to a for loop.
So let us assume that one has a dataframe whose
Dear list,
how can I compute the inverse of the X'WX matrix (inverse of the weighted sum
of squares and crossproducts matrix) from an object of class lm from a
weigthed linear regression?
Thanks, Sven
__
R-help@stat.math.ethz.ch mailing list
I noticed that:
http://www.r-project.org/useR-2006/
seems to be inexistent (page not found).
Best,
Bernhard
Dr. Bernhard Pfaff
Global Structured Products Group
(Europe)
Invesco Asset Management Deutschland GmbH
Bleichstrasse 60-62
D-60313 Frankfurt am Main
Tel: +49(0)69 298 07230
Fax:
On Wed, 2006-05-03 at 10:46 -0400, Farrel Buchinsky wrote:
How does one create a vector whose contents is the list of variables in a
dataframe pertaining to a particular pattern?
This is so simple but I cannot find a straightforward answer.
I want to be able to pass the contents of that list
Farrel Buchinsky wrote:
How does one create a vector whose contents is the list of variables in a
dataframe pertaining to a particular pattern?
This is so simple but I cannot find a straightforward answer.
I want to be able to pass the contents of that list to a for loop.
So let us assume
Column names in iris that contain the string Sepal:
grep(Sepal, names(iris), value = TRUE)
On 5/3/06, Farrel Buchinsky [EMAIL PROTECTED] wrote:
How does one create a vector whose contents is the list of variables in a
dataframe pertaining to a particular pattern?
This is so simple but I
Here's an example.
dfr - data.frame(A1=1:10,A2=21:30,B1=31:40,B2=41:50)
vars - colnames(dfr)
for (v in vars[grep(B,vars)]) print(mean(dfr[,v]))
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Farrel
Buchinsky
Sent: Wednesday, May 03, 2006 10:46 AM
Pfaff, Bernhard Dr. wrote:
I noticed that:
http://www.r-project.org/useR-2006/
seems to be inexistent (page not found).
Probably lost during the move of CRAN master. For the meantime use
http://www.ci.tuwien.ac.at/Conferences/useR-2006/
Uwe Ligges
Best,
Bernhard
Dr. Bernhard
I think you need
summary(lm.obj)$cov.unscaled
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web:
Dear list,
I am new here. often times, I have a question about how to evaluate
logistic model fit the data or not, do you guys can help me with some
guidence? thanks.
Youme
--
Yuezhou Jing
Center for Human Growth Development University of Michigan
___ ___
exactly!
May thanks!
-Original Message-
From: Dimitris Rizopoulos [mailto:[EMAIL PROTECTED]
Sent: Wed 5/3/2006 5:40 PM
To: Garbade, Sven
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Inverse X'WX matrix from weighted linear regression
I think you need
summary(lm.obj)$cov.unscaled
I
Uwe Ligges [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
vnames - paste(Height, 1:20, sep=.)
Interesting but not suitable. It creates a name even if such a variable does
not exist. I have 6000 variables. The numeric component of the variable name
goes from 0 to about 1000.
On Tuesday 02 May 2006 23:33, Uwe Ligges wrote:
Then please read ?plotmath and use it:
labels = expression( = 0.66, == 0.33, = -0.33, = -0.66)
Error in lab != : comparison is not allowed for expressions
In addition: Warning message:
is.na() applied to non-(list or vector) in: is.na(lab -
Johannes Graumann wrote:
On Tuesday 02 May 2006 23:33, Uwe Ligges wrote:
Then please read ?plotmath and use it:
labels = expression( = 0.66, == 0.33, = -0.33, = -0.66)
Error in lab != : comparison is not allowed for expressions
In addition: Warning message:
is.na() applied to
Yes, this has been reported to the webmasters (not really much point in
telling a list).
cran.r-project.org and www.r-project.org now point at machines at
wu-wien.ac.at (the DNS was changed in the last 24 hours). I have also
found that rsync is not being allowed through (e.g. for
On 5/3/06, Uwe Ligges [EMAIL PROTECTED] wrote:
Johannes Graumann wrote:
On Tuesday 02 May 2006 23:33, Uwe Ligges wrote:
Then please read ?plotmath and use it:
labels = expression( = 0.66, == 0.33, = -0.33, = -0.66)
Error in lab != : comparison is not allowed for expressions
Gabor Grothendieck wrote:
On 5/3/06, Uwe Ligges [EMAIL PROTECTED] wrote:
Johannes Graumann wrote:
On Tuesday 02 May 2006 23:33, Uwe Ligges wrote:
Then please read ?plotmath and use it:
labels = expression( = 0.66, == 0.33, = -0.33, = -0.66)
Error in lab != : comparison is
Uwe Ligges wrote:
Johannes Graumann wrote:
On Tuesday 02 May 2006 23:33, Uwe Ligges wrote:
Then please read ?plotmath and use it:
labels = expression( = 0.66, == 0.33, = -0.33, = -0.66)
Error in lab != : comparison is not allowed for expressions
In addition: Warning message:
On Wednesday 03 May 2006 09:05, Uwe Ligges wrote:
Ah, I see, this happens in pie()'s line:
if (!is.na(lab - labels[i]) lab != ) {
where lab is one element of the expression.
I'd like to propose to change that line to
if (!is.na(lab - labels[i]) nchar(lab) 0) {
What's the
Maybe because ?pie says
labels: a vector of character strings giving names for the slices.
For empty or NA labels, no pointing line is drawn either.
Note, nothing about expressions
pie() contains
if (!is.na(lab - labels[i]) lab != ) {
and so it would be easy to
Dear Erin:
The documentation for my currently installed version of
garchFit{fSeries} indicates that the default include.mean = TRUE.
How do you know it didn't work? With the default include.mean =
TRUE, the following took just under a minute on my IBM T-30 notebook:
How would one go about getting sprintf to use the
values of a vector without having to specify each
argument individually?
v - c(1, 2, -1.197114, 0.1596687)
iv - c(3, 1, 2, 4)
sprintf(%9.2f\t%d\t%d\t%8.3f, v[3], v[1], v[2], v[4])
[1] -1.20\t1\t2\t 0.160
Essentially, desired effect would
On Wed, 3 May 2006, Johannes Graumann wrote:
On Wednesday 03 May 2006 09:05, Uwe Ligges wrote:
Ah, I see, this happens in pie()'s line:
if (!is.na(lab - labels[i]) lab != ) {
where lab is one element of the expression.
I'd like to propose to change that line to
if (!is.na(lab -
Dear R users,
I face to a nested pattern and despite the numerous examples in the help I am
still confused.
I sampled bugs in different habitats within sites which were within rivers
themselves within different regions.
The habitat correspond to different substrata (not systematically
Try this:
do.call(sprintf, c(%9.2f\t%d\t%d\t%8.3f, as.list(v[iv])))
On 5/3/06, Paul Roebuck [EMAIL PROTECTED] wrote:
How would one go about getting sprintf to use the
values of a vector without having to specify each
argument individually?
v - c(1, 2, -1.197114, 0.1596687)
iv - c(3, 1,
Dear Erin:
I just realized I had misread your question. Have you tried sending
your question directly to the 'tseries' maintainer, listed with
help(package=tseries)?
While I misread your question earlier, it looks to me like you could
get what you want by first
Finally fixed for the next release - and will also include the bugsLog()
stuff as well.
Best,
Uwe
Gregor Gorjanc wrote:
Hello Paul,
thank you very much for this report. You caught a bug in R2WinBUGS that
was introduced by me. I added support for winepath in 1.1-1 version.
Since I switch
You could finagle things with do.call (maybe right your own function
that does this if you will use it often). Here is your example:
v - c(1, 2, -1.197114, 0.1596687)
iv - c(3, 1, 2, 4)
tmp - c(list(%9.2f\t%d\t%d\t%8.3f),as.list(v[iv]))
do.call('sprintf',tmp)
[1] -1.20\t1\t2\t 0.160
On 5/3/2006 12:31 PM, Paul Roebuck wrote:
How would one go about getting sprintf to use the
values of a vector without having to specify each
argument individually?
v - c(1, 2, -1.197114, 0.1596687)
iv - c(3, 1, 2, 4)
sprintf(%9.2f\t%d\t%d\t%8.3f, v[3], v[1], v[2], v[4])
[1]
I have got factor from read.xls:
is(factor_value)
[1] factor oldClass
[288] -0.32 0.180.180.18-0.32 0.180.68
[295] 0.680.18
43 Levels: -0.05 -0.13 -0.15 -0.18 -0.20 -0.26 ... 1.33
If I am using the funciton as.real(factor_value)
I get
[271] 17
I have a data set with ~10 variables (i.e. columns).
I wrote this little function to replace missing values
with zero.
sz - function(x) { ifelse(is.na(x)==F,x,0) }
Can anyone help with a function that replaces missing
values with the median of the non-missing values?
sz - function(x) ifelse(is.na(x), median(x, na.rm=TRUE), x)
Gabor
On Wed, May 03, 2006 at 10:06:40AM -0700, r user wrote:
I have a data set with ~10 variables (i.e. columns).
I wrote this little function to replace missing values
with zero.
sz - function(x) { ifelse(is.na(x)==F,x,0)
You can use as.is = TRUE arg to read.xls to get character
data rather than factors.
On 5/3/06, Knut Krueger [EMAIL PROTECTED] wrote:
I have got factor from read.xls:
is(factor_value)
[1] factor oldClass
[288] -0.32 0.180.180.18-0.32 0.180.68
[295] 0.680.18
Prof Brian Ripley ripley at stats.ox.ac.uk writes:
On Wed, 3 May 2006, Johannes Graumann wrote:
What's the canonical way of patching something like this in R? Redefining
the
function at the start of your script?
There are namespace issues, so the canonical way is to change the
I thank you for your answer. It was really helpful.
I purchased Pinheiro and Bates last year for the reasons you mentionned.
I checked Sec. 5.2 and think I might use the following :
model.var - update(model2,weights=varIdent(form=~1|Limb))
which gives :
Variance function:
Structure: Different
Dear R users,
I am have a question on SAM analysis - two class unpaired. I am not
sure how exactly SAM calculates the fold change for logged2 transformed
data. The output produced by SAM -
rbind(siggenes.table$genes.up,siggenes.table$genes.lo) - has numerator,
denominator, d-statistic, fold
As a workaround you could use pie3D in the plotrix package
with height=0 and theta=pi, e.g.
library(plotrix)
pie3D(1:3, height = 0, theta = pi,
labels = expression( = 1, == 2, = 3))
On 5/3/06, Johannes Graumann [EMAIL PROTECTED] wrote:
On Wednesday 03 May 2006 09:05, Uwe Ligges wrote:
R-users,
I'm seeking any suggestions on optimizing some code for speed. Here's
the setup: the code below is part of a larger chunk that is calculating
Fst values across loci and alleles. This chunk is designed to calculate
the proportion ('p.a') of an allele ('a') at a locus in each
sum() looks faster to me
tmp - as.logical(rbinom(100,1,.5))
system.time(for (i in 1:1) length(tmp[tmp]))
[1] 0.09 0.00 0.09 NA NA
system.time(for (i in 1:1) sum(tmp))
[1] 0.03 0.00 0.03 NA NA
I am running on Windows XP
Sys.getenv()[20]
Dear R users,
I am trying to perform a hypothesis test on a marked point pattern. I
would like to calculate the mean of the absolute value of the
difference of marks between nearest neigbours, randomize the marks
among points, then calculate this mean again. Ideally, I would test
whether random
I am trying to install R-2.3.0 on a 64bit linux box and encounter
several error during the make step. I'd appreciate any help. Error
messages follow:
[EMAIL PROTECTED] R-2.3.0]# make
make[4]: Entering directory
`/state/partition1/apps/packages/R-2.3.0/src/modules/lapack'
gcc -shared
Richard,
Thanks! I never would've thought of using sum() to count the TRUEs in a
logical vector, but it makes perfect sense.
Cheers,
e.
Richard M. Heiberger wrote:
sum() looks faster to me
tmp - as.logical(rbinom(100,1,.5))
system.time(for (i in 1:1) length(tmp[tmp]))
[1]
Hello,
I have a data set with a grouping variable (TRIPID) and several other
variables. TRIPID is repeated in some areas and I would like to use a
function like aggregate to sum the variable UNITS according to TRIPID.
However I would also like to retain the other variables as they are in
the
As it says, your ATLAS library has not compiled for use in a shared
object.
This is discussed in the R-admin manual which you were asked to read (in
the INSTALL file) if you have any problems.
You don't even tell us what Linux distro this is. Try configuring with
--without-blas.
On Wed, 3
Suppose we want to sum C over levels of A and that B is constant
within levels of A. Then:
DF - data.frame(A = gl(2,2), B = gl(2,2), C = 1:4) # test data
do.call(rbind, by(DF, DF$A, function(x) replace(x[1,], C, sum(x$C
On 5/3/06, Guenther, Cameron [EMAIL PROTECTED] wrote:
Hello,
I
Is it possible to make disappear the id numbers from scatter.dudi (mc
analysis) ?
a - as.factor(c(1, 2, 3, 2, 1))
b - as.factor(c(3, 2, 3, 1, 1))
x - as.factor(c(1, 2, 2, 1, 3))
y - as.factor(c(2, 2, 3, 1, 1))
dat - data.frame(a=a, b=b,x=x,y=y)
summary(dat)
dat
require(ade4)
dat.acm -
Hi List,
I want to know how to test whether an image or generally a random
field, is stationary. I am sure there are a lot of intuitive ways to do this
but right now I need a formal test which can gives the p-value. And any of
those tests already implemented in R? Thanks a lot!
Yuefeng
It would be nice to have something like stata's outreg that lets regression
output go into a form like
Specification (1) Specification (2)
Var 1 coef(1,1) coef(1,2)
se(1,1) se(1,2)
Var 2 coef(2,1) coef(2,2)
se(2,1)
Hi
After a little work i found a way of using near 8bit color output in
GDD. I edited the GDD/src/GDDtalk.c file...
There was this loop:
if (!strcmp(it, png8)) {
if (nl3 strcmp(fn+nl-4,.png)) strcat(fn, .png);
out = fopen (fn, wb);
if
I think this is what you are asking for. It would be relatively easy
to package this into a function that takes a list of lm objects as
its argument.
tmp - data.frame(matrix(rnorm(100), 25, 4))
names(tmp) - c(y,x1,x2,x3)
t1.lm - lm(y ~ x1, data=tmp)
t2.lm - lm(y ~ x2, data=tmp)
t3.lm - lm(y ~
Hi all
I'm trying to install R 2.3.0 under Suse 10.0. As I'm using SSH to login
into the SUSE server, I can't use YAST2, so I have to use rpm -i in the
shell. The system tells me that I need some other packages such as
xorg-x11-fonts-100dpi, blas, libgfortran.so.0(). Is there some website
[zhihua li]
I'm trying to install R 2.3.0 under Suse 10.0. As I'm using SSH to
login into the SUSE server, I can't use YAST2,
I presume this is because you cannot remotely mount the CD's or DVD's?
The next time you visit your server, if possible, copy your distribution
media to your hard
Thanks for your tips. But I don't understand why coping the distribution
media to the server's hard disks would enable me to use yast2 remotely with
ssh? Actually I can launch yast2 remotely now. After connecting to the
remote server by ssh and then typing yast2, a yast interface will appear
Youme,
The question has been asked before, you can have a look
at the R archives for that issue using:
RSiteSearch(logistic goodness of fit).
Last time I did this, the command found some 53 entries!
Hope it helps,
Augusto
Augusto Sanabria. MSc,
On Wednesday 03 May 2006 20:28, zhihua li wrote:
Hi all
I'm trying to install R 2.3.0 under Suse 10.0. As I'm using SSH to login
into the SUSE server, I can't use YAST2, so I have to use rpm -i in the
shell. The system tells me that I need some other packages such as
Uwe Ligges ligges at statistik.uni-dortmund.de writes:
The same way. lapply() and sapply() should work for almost all functions
given, if nothing strange happens with environemnts, which is the case here:
The problem is tdt() itself. Note that it has its argument data set to
Larry
I tried using yast in my shell ( i have the root authority). The yast
ncurses did appear in my shell, but i can't control the panel. For example,
it says press F1 for help, but pressing my F1 just resulted in decrease
of my screen light (the default function of the F1 key in mac). So
Zhihua,
In Linux you use the tab key to move from field to field, or you can go alt-O
for example to hit the OK button. Windows is similar. You need the suse
installation source to be configured so that it can load the dependent
packages (libgfortran, etc.).
Larry
On Wednesday 03 May 2006
Hello,
Does anybody know how to calculate local goodness of fit at the different
quantiles of a censored quantile regression? I suppose I am looking to see
if there is a specific command in R for this...
Thanks for your help,
Tanya
__
I am using dgc.genetics to perform TDT analysis on SNP data from a cohort of
trios.
I now have a file with about 6008 variables. The first few variables related
to the pedigree data such as the pedigree ID the person ID etc. Thereafter
each variable is a specific locus or marker. The
I followed the examples of previous posts about R2HTML to practice
exporting a data to a clipboard, but the result is not as smooth as I
had expected:
library(R2HTML)
data(iris)
HTML(iris, file(clipboard,w), append=FALSE)
I got an error message:
HTML(iris, file(clipboard,w), append=FALSE)
Dear Community,
This is largely a repost with some new information.
I'm interested in developing a package that could ease the
command-line learning curve for new users. It would provide more
detailed syntax checking and commentary as feedback. It would try to
anticipate common new-user
The following should work
sz - function(x) { ifelse(is.na(x) == F, x, median(x, na.rm=TRUE)) }
best, isaia.
Original Message
Subject: [R] function to replace missing values with median value?
Date: Wed, 3 May 2006 10:06:40 -0700 (PDT)
From: r user [EMAIL PROTECTED]
To: rhelp
there is also a replace function
-Messaggio originale-
Da: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] conto di
[EMAIL PROTECTED]
Inviato: Thursday, May 04, 2006 07:31 AM
A: [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Oggetto: [R] [Re:] function to replace
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