Katie2009 wrote:
I'm trying to analyse some excel data in R. The problem is that when i
input the data with the first column as absolute values, everything works
fine, can analyse as normal. When I leave the first column unchanged to
import negative numbers as well I get:
Error in
I'm trying to analyse some excel data in R. The problem is that when i input
the data with the first column as absolute values, everything works fine,
can analyse as normal. When I leave the first column unchanged to import
negative numbers as well I get:
Error in storage.mode(y) - double :
Good Morning,
I have a graph with groups of variables. I have include the group
names as variables so that I can have them positioned correctly.
Unfortunately this means that the group names have to follow all of
the same rules as the variables within the groups. I would rather
have those
Katie2009 wrote:
hi dieter,
the method i'm using is in excel, copying the data, then in r
w-read.delim(clipboard)
w-as.data.frame(w)
i've been doing a bit more fiddling, and have identified the 'class' of
the column that i'm having trouble with, is classified as 'factor' whilst
At 19:18 09/05/2009, U.H wrote:
Dear R gurus,
I have data for which I want to estimate the markov transition matrix
that generated the sequence, and preferably obtain some measure of
confidence for that estimation.
RSiteSearch(markov, restrict = function)
shows that there are numerous
hi dieter,
the method i'm using is in excel, copying the data, then in r
w-read.delim(clipboard)
w-as.data.frame(w)
i've been doing a bit more fiddling, and have identified the 'class' of the
column that i'm having trouble with, is classified as 'factor' whilst the
rest are numeric.
if i
I have a graph with groups of variables. I have include the group
names as variables so that I can have them positioned correctly.
Unfortunately this means that the group names have to follow all of
the same rules as the variables within the groups. I would rather
have those group names
Hi all,
I am unable to install R package on RHEL version 5.1. While running
* ./configure* command following is printed on to the screen and no *
makefile* is created. Can anyone help me out ??? Thanks in advance.
./configure
checking build system type... x86_64-unknown-linux-gnu
Dear John,
In RLQ (similarly to other methods in ade4), inertia are given by
eigenvalues. Thus, the percentage of inertia are obtained simply by:
rlq1$eig/sum(rlq1$eig)*100
For more details on outputs or RLQ, you can have a look at
Hi,
I am trying to plot a 3d scatter plot using the rgl package but am having
trouble.
I have a matrix containing x, y and z co-ordinates and a fourth element related
to a count variable.
I would like to plot the points using plot3d with the sizes of the points
related to the fourth element in
Has anyone else had experience with CentOS and the latest version of R?
$uname -r
2.6.18-92.1.22.el5
On occasion, I need to have the LD_LIBRARY_PATH environment variable
which I set in ~/.bash_profile. Until R-2.9.0 that has worked fine,
but now when I check with Sys.getenv(LD_LIBRARY_PATH), I
Thank you for the suggestions, I'm sure irr is what I'm looking for. Thanks
again.
spencerg wrote:
To look up observer agreement, you might consider the
RSiteSearch package. The RSiteSearch.function looks only for
matches in help pages of contributed packages.
Having a problem running the ace command in ape.
After reading my table into R and then putting the names of the rows in the
table in same order as the tree I can't get ace to run and get the message
Erreur dans as.matrix(x) :
dims [produit 25] ne correspond pas à la longueur de l'objet [0]
Hi Zoé,
try
plot3d(swadt[,1],swadt[,2],swadt[,3],radius=swadt[,4],type=s)
Hope this helps. regards. Olivier
Zoe Hoare wrote:
Hi,
I am trying to plot a 3d scatter plot using the rgl package but am having
trouble.
I have a matrix containing x, y and z co-ordinates and a fourth element
Hello,
With the help of seek I can start readBin from any byte offset within my
file that I deem appropriate.
What I would like to do is to be able to define the endpoint of that read as
well. Is there any solution to that already out there?
Thanks for any hints, Joh
Phillip Porter wrote:
Good Morning,
I have a graph with groups of variables. I have include the group
names as variables so that I can have them positioned correctly.
Unfortunately this means that the group names have to follow all of
the same rules as the variables within the groups. I
Hi Spencer,
Thanks for suggesting the genD function. In attempting it, I have
rearranged my function from phat1 ~ ... to ... - 1, it apparently doesn't
like the first one :) But when I run it, it tells me the partials are all
zero. I'm trying out a simple MNL equation before I expand it to
Hi Sir,
I want to install the package cluster in R language.
I could be installing, but I cant load that package..
I have faced the following warning in installing time..
Warning: cannot remove prior installation of package 'cluster'
So, I tried to update the package, but again I faced the
Hello,
I have been struggling for quite some time to find a solution for the following
problem. I have a data frame which is organized by block and trial. Each trial
is represented across several rows in this data frame. I'd like to extract the
first x rows per trial and block.
For example
LS,
We would like to inform you about the new course Pharmacokinetic
and pharmacodynamic modeling and simulation by Dr. Jan Freijer
in Amsterdam, November 28th.
The course is meant for users of R or S-PLUS in the
bio-pharmaceutical sciences who would like to apply it to
clinical trial
try this,
library(plyr)
ddply(d, .(block, trial), function(.d) .d[1:2, ])
block trial x y
1 1 1 605 150
2 1 1 603 148
3 1 2 607 148
4 1 2 605 152
HTH,
baptiste
On 11 May 2009, at 13:49, Jens Bölte wrote:
Hello,
I have been struggling for quite
Try this:
do.call(rbind, by(DF, DF[1:2], head, 2))
On Mon, May 11, 2009 at 7:49 AM, Jens Bölte boe...@psy.uni-muenster.de wrote:
Hello,
I have been struggling for quite some time to find a solution for the
following problem. I have a data frame which is organized by block and
trial. Each
Here is a way of doing it:
x
block trial x y
1 1 1 605 150
2 1 1 603 148
3 1 1 604 140
4 1 1 600 140
5 1 1 590 135
6 1 1 580 135
7 1 2 607 148
8 1 2 605 152
10 1 2 600 158
do.call(rbind, lapply(split(x,
Can you be more specific on how you want to define the endpoint of that
read. What is the criteria you want to use? Can you read in a block and
then search of the pattern?
On Mon, May 11, 2009 at 7:05 AM, Johannes Graumann johannes_graum...@web.de
wrote:
Hello,
With the help of seek I can
good point, i forgot about head (!),
library(plyr)
ddply(d, .(block, trial), head, 2)
block trial x y
1 1 1 605 150
2 1 1 603 148
3 1 2 607 148
4 1 2 605 152
On 11 May 2009, at 14:04, Gabor Grothendieck wrote:
Try this:
do.call(rbind, by(DF, DF[1:2],
On May 10, 2009, at 7:54 PM, David Winsemius wrote:
?subset
?%in%
(I have gotten tired of converting dataframes that are presented in
a non-executable form, such as is supported by the dput function.
So, ... you should read those help pages and take the obvious path
to success.)
You might try again, assuming you are a Windows user, which was
suggested by your choice of repositories but not actually stated by
you. The repository may have been temporarily unavailable. It seems to
be open for service now with a version of cluster that is 1.11.13.
--
David Winsemius
The anova process is only statistically valid for nested models, i.e., where
one includes the other as a proper subset. A model with continuous age and one
with discrete age are not nested; your p-value will be meaningless.
Terry Therneau
__
On 11/05/2009 7:05 AM, Johannes Graumann wrote:
Hello,
With the help of seek I can start readBin from any byte offset within my
file that I deem appropriate.
What I would like to do is to be able to define the endpoint of that read as
well. Is there any solution to that already out there?
I have a large series of Visual FoxPro (.dbf) files that I want to
convert to a plain text (.csv) file. I wrote a quick script to do all of
this for me. With a little tweaking it seemed to work like a charm. But,
when I looked at my newly created text files carefully, I discovered a
problem. Every
Hi R users,
I'm trying to run a SVM - regression using e1071 package but the function svm()
all the time apply a classification method rather than a regression.
svm.m1 - svm(st ~ ., data = train, cost = 1000, gamma = 1e-03)
Parameters:
SVM-Type: C-classification
SVM-Kernel:
beetle2 wrote:
For a homework question.
I was wondering if rcmdr has a function to plot a graph of a bivariate
function of X and Y.
I have a function with joint pdf
fX,Y(x,y) = x+y for 0x1 , 0y1
I've tried
x - seq(0,1,.001)
y - seq(0,1,.001)
r = x+y
plot(r)
but it seems to
It would be nice if each package went through a peer-review and had a
related publication (either in R-news or J Stat Soft). This publication can
then be used as the official citation for the package. However, this still
would not address updates and versions of the package.
Ravi.
Hello, I was just wondering whether with the ace command in ape it was
possible to incorporate polymorphic characters in ancestral character
estimation. I am currently working with distibutions and often
distributions are not restricted to a single geographic area. Thanks for
any help.
Dear adegenet users,
I am trying to calculate geographical barriers using monmonier.
However, I consistently get the error dimnames do not match array
extent.
- import of the distance matrix: read.table
- conversion 1: as.dist(data, diag=TRUE, upper =TRUE)
- conversion 2: as.matrix(data)
-
Hi all, is there any function to find some words in a character-string? For
example suppose the string is : gdfsa-sdhchc-88, now I want to find
whether this string contains sdhch. Is there any R function to do that?
Regards,
--
View this message in context:
Andrew RODRIGUES wrote:
Having a problem running the ace command in ape.
After reading my table into R and then putting the names of the rows in
the
table in same order as the tree I can't get ace to run and get the message
Erreur dans as.matrix(x) :
dims [produit 25] ne
At 12:37 AM -0700 5/11/09, Katie2009 wrote:
hi dieter,
the method i'm using is in excel, copying the data, then in r
w-read.delim(clipboard)
w-as.data.frame(w)
i've been doing a bit more fiddling, and have identified the 'class' of the
column that i'm having trouble with, is classified
On 5/11/2009 10:07 AM, RON70 wrote:
Hi all, is there any function to find some words in a character-string? For
example suppose the string is : gdfsa-sdhchc-88, now I want to find
whether this string contains sdhch. Is there any R function to do that?
See ?grep. There are several functions
The following is TRUE if the indicated string is found:
regexpr(sdhch, gdfsa-sdhchc-88) 0
See ?regexpr
Either of the next two will break up a string into words. The
first needs to know the delimiters whereas the second
needs to know the contents:
strsplit(gdfsa-sdhchc-88, \\W+)[[1]]
[1]
Hi Ron,
Look up the grep() function.
Cheers,
--
*Luc Villandré*
/Biostatistician
McGill University Health Center -
Montreal Children's Hospital Research Institute/
RON70 wrote:
Hi all, is there any function to find some words in a character-string? For
example suppose the string is :
In stats::D, I was wondering why variables are represented as symbols
in expressions, but as strings in lists of variables:
D(quote(x^2),x) = 2*x
D(quote(x^2),quote(x)) = error Variable must be a character string
Strings are not allowed in the expression to denote variables:
I tried to install Rcompression (for reading Matlab files)
options(CRAN = c(getOption(CRAN), http://www.omegahat.org/R;))
install.packages(Rcompression)
Warning: unable to access index for repository
http://www.omegahat.org/R/bin/windows/contrib/2.9
Warning message:
In getDependencies(pkgs,
I have have a package that I wrote called StreamMetabolism; which I
use to calculate single station stream metabolism from diurnal oxygen
curves. I would love to publish something about it (I am also an
entering PhD student and need publications); however, I am not sure
the applicability out side
Dear List,
Consider the following example
x=c(1,2,3,4,5)
y=c(2,4,6,8,10)
linearmodel=lm(y~x)
To predict a y-value if you know the corresponding x value is very easy
with the command predict.
predict(linearmodel, newdata=(x=1.5))
The other way around, to predict an x-value with a
2009/5/6 Emmanuel Charpentier charp...@bacbuc.dyndns.org:
Le mercredi 06 mai 2009 à 00:22 -0400, Farrel Buchinsky a écrit :
Is R an appropriate tool for data manipulation and data reshaping and data
organizing?
[ Large Snip ! ... ]
Depends on what you have to do.
I've done what can be more
Dear R users,
I have what I think it is a very simple question concerning plots in R. If
you could help me I would be very grateful.
How can I include a legend in my plot? Below I give an example of my data
and the code I am using.
What I wish to accomplish is a legend saying that the black
Dear Mafalda,
?legend is what you are looking for. Take a look at the help page. Here is
one way to do what you asked for:
legend('topleft', c('A','B','C'),lty=1, col=c(1,2,4))
HTH,
Jorge
On Mon, May 11, 2009 at 11:26 AM, Mafalda Viana via...@tcd.ie wrote:
Dear R users,
I have what I
Hi !
From an Ensembl annotation like ENSG /// ENSGy /// ENSG, I am
trying to keep only the first part: ENSG. I wasn't able to find any
helpful information about how to do it. Could you help me with that please ?
Is the use of the equivalent to the Excel * (any text) a good way
Hi,
I have been working with R for the last year and using the UKFSST package to
look at satellite tag track data and SST information. Fpr those not familiar,
the package uses the positions estimated by the satellite tags themselves and
the associated SST data from servers (in this case, from
Hi useR's
I have created a simple map of the world using the following code:
m - map(xlim=c(-180,180), ylim=c(-90,90))
map.axes()
I then create a grid of dimension 36x72 using the code:
map.grid(m, nx=72, ny=36, labels=FALSE, col=black)
This gives 2592 grid cells. In a separate data set of
Good afternoon,
My name is Thiago. I'm a graduate student and affiliated to an Anuran
Bioacustic and Systematics Laboratory at UFU (Federal University of
Uberlândia) in Brazil. The professor in charge of the lab (Dr. Ariovaldo
Antonio Giaretta) and I have recently been making use of R,
For a homework question.
I was wondering if rcmdr has a function to plot a graph of a bivariate
function of X and Y.
I have a function with joint pdf
fX,Y(x,y) = x+y for 0x1 , 0y1
I've tried
x - seq(0,1,.001)
y - seq(0,1,.001)
r = x+y
plot(r)
but it seems to just add them together say
At 9:25 PM +1000 5/11/09, Jim Lemon wrote:
Hi Phillip,
I'm not exactly sure how you are positioning the labels. Is it
possible to give us an example with some data that will produce a
plot to show what you want? It shouldn't be too hard to do.
Jim
Data is something along the lines of:
Can anyone tell me what is skip=2, skip =7
From ?read.csv:
skip: integer: the number of lines of the data file to skip before
beginning to read data.
and %in% mean here?
%in% matches values; see ?'%in%', and example('%in%')
Regards,
Richie.
Mathematical Sciences Unit
HSL
You may also want to look at the rootogram function in the vcd package.
You could also use the updateusr function in the TeachingDemos package to reset
the y-scale before plotting the curve (works well if you just want the shape of
the curve, probably more work than the other suggestions if you
Rob Steele wrote:
I'm finding that readLines() and read.fwf() take nearly two hours to
work through a 3.5 GB file, even when reading in large (100 MB) chunks.
The unix command wc by contrast processes the same file in three
minutes. Is there a faster way to read files in R?
Thanks!
Colleagues,
I have encountered a problem in version 2.9 of R, running in both
Vista and OS X. My code is quite lengthy but the critical line is:
if (file.exists(FILENAME)) readLines(FILENAME))
This triggers the error:
Error in file(con, r) : cannot open the connection
Thank you Stephanie. This is perfect!
On May 10, 12:34 pm, Stephanie Kovalchik sko...@ucla.edu wrote:
JC,
If each row are the counts for a 2 x 2 contingency table - so for the
ith contingency table you have counts for row 1 c(Y08[i],Z08[i]) and
row 2 (Y09[i],Z09[i]) then you could use
A large chunk of the function below could be replaced with a call to the Reduce
function. I don't know if it would be faster, slower, or depend on the
situation, but it might make it a little more readable.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
Is this what you want (using regular expressions):
x - ENSG /// ENSGy /// ENSG
sub(^([[:alpha:]]+).*, \\1 file://0.0.0.1/, x)
[1] ENSG
On Mon, May 11, 2009 at 9:01 AM, Amélie Baud ameli...@yahoo.fr wrote:
Hi !
From an Ensembl annotation like ENSG /// ENSGy ///
It seems that the structure of your data is that you have two groups
(Real Bad Stuff and Other Crazy Things) which are then subdivided into
further categories. I'd be tempted to set your data up like this:
dfr - data.frame(
score=c(23, 14, 17, 8, 43, 13),
group=rep(c(Real Bad Stuff,
Is the variable st character or a factor? What does str(train$st) show?
--
Max
__
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
On Mon, 11 May 2009, Dennis Fisher wrote:
Colleagues,
I have encountered a problem in version 2.9 of R, running in both Vista and
OS X. My code is quite lengthy but the critical line is:
if (file.exists(FILENAME)) readLines(FILENAME))
This triggers the error:
Error in
Why not use the x=T and y=T paramemters and see if :
linearmodel=lm(y~x, x=T, y=T)linearmodel$model
(linearmodel$model)
y x
1 2 1
2 4 2
3 6 3
4 8 4
5 10 5
give you useful results. If you want to interpolate or
extrapolate, then search for threads regarding those tasks.
--
David
Dear all,
I have P value of a list of markers. Now I need the chi square value
with degrees of freedom 2.
I noticed there are several Chisquare functions (dchisq, pchisq,
qchisq, and rchisq), but it seems all are not for my purpose.
In microsoft excel, there is a function CHINV to do
On May 11, 2009, at 9:46 AM, Ben Bolker wrote:
beetle2 wrote:
For a homework question.
I was wondering if rcmdr has a function to plot a graph of a
bivariate
function of X and Y.
I have a function with joint pdf
fX,Y(x,y) = x+y for 0x1 , 0y1
I've tried
x - seq(0,1,.001)
y -
On 5/11/2009 3:36 PM, Anyuan Guo wrote:
Dear all,
I have P value of a list of markers. Now I need the chi square value
with degrees of freedom 2.
I noticed there are several Chisquare functions (dchisq, pchisq,
qchisq, and rchisq), but it seems all are not for my purpose.
In
-- Forwarded message --
From: r-help-boun...@r-project.org
Date: Mon, May 11, 2009 at 10:24 PM
Subject: The results of your email commands
To: das.moumita.onl...@gmail.com
The results of your email command are provided below. Attached is your
original message.
- Results:
On 5/11/2009 10:32 AM, dxc13 wrote:
Hi useR's
I have created a simple map of the world using the following code:
m - map(xlim=c(-180,180), ylim=c(-90,90))
map.axes()
I then create a grid of dimension 36x72 using the code:
map.grid(m, nx=72, ny=36, labels=FALSE, col=black)
This gives 2592 grid
On Mon, 2009-05-11 at 12:36 -0700, Anyuan Guo wrote:
Dear all,
I have P value of a list of markers. Now I need the chi square value
with degrees of freedom 2.
I noticed there are several Chisquare functions (dchisq, pchisq,
qchisq, and rchisq), but it seems all are not for my
Thank you very much Mr Onkelinx, I appreciated your answer.
I'm sorry about my lack of knowledge on this subject, but I don't understand
why residual variance is fixed at 1 with the binomial family and I didn't
manage to find any exhaustive explanation for that.
May you kindly suggest me some
Thank you very much Mr Onkelinx, I appreciated your answer.
I'm sorry about my lack of knowledge on this subject, but I don't understand
why residual variance is fixed at 1 with the binomial family and I didn't
manage to find any exhaustive explanation for that.
May you kindly suggest me some
On 11-May-09 19:36:00, Anyuan Guo wrote:
Dear all,
I have P value of a list of markers. Now I need the chi square
value with degrees of freedom 2.
I noticed there are several Chisquare functions (dchisq, pchisq,
qchisq, and rchisq), but it seems all are not for my purpose.
In
Hi there,
I have been trying to fit an NLME to my data. My dataset has two category
levels - one is a fixed effect (level1) and one is a random effect (level2),
though so far I have only experimented with the highest level grouping (fixed,
level1), with the following code:
mod1 - nlme(H ~
This is really odd,
I've installed the binary of robustbase through synaptic on an ubuntu
8.04 machine and get:
library(robustbase)
Error in dyn.load(file, DLLpath = DLLpath, ...) :
unable to load shared library
'/usr/lib/R/site-library/robustbase/libs/robustbase.so':
libRlapack.so:
Hi all.
Recently i have been used maxlik package for optimizing a function with 5
parameters but i couldn't define gradient argument in function maxLik;
How i can define a command to receive my goal? Whether I can change the core of
this package?
Thanks for your attention and reply
A.
I am not a statistician and not a computer scientist by education. I
consider myself an R novice and came to R - thanks to my boss - from
an SPSS background. I work for a market research company and the most
typical data files we deal with are not huge - up to several thousand
rows and up to a
*Dear R Users;*
* *
*I´m writing to ask you how can I do Survivals Curves using Time-dependent
covariates? Which packages I need to Install?*
*Thanks for your help, I look foward you reply.*
**
*Sincerelly.*
**
*Nancy Tejerina.*
[[alternative HTML version deleted]]
Hi DD,
if you have a .shp file (which in principal should exist of the kind
of map your describing) you can load into R using this packages:
http://r-spatial.sourceforge.net/
In addition, there is also the R-Geo list which is much more suited
for this kind of questions:
Lindsay Banin wrote:
Hi there,
I have been trying to fit an NLME to my data. My dataset has two category
levels - one is a fixed effect (level1) and one is a random effect
(level2), though so far I have only experimented with the highest level
grouping (fixed, level1), with the
Le lundi 11 mai 2009 à 23:20 +0800, ronggui a écrit :
[ Snip... ]
But, at least in my trade, the ability to handle Excel files is a must
(this is considered as a standard for data entry. Sigh ...).
[ Re-snip... ]
I don't think Excel is a standard tool for data entry. Epidata entry
is
windows xp
R 2.8.1
I am trying to plot the -log(log(survival)) to visually test the proportional
hazards assumption of a Cox regression. The plot, which should give two lines
(one for each treatment) gives only one line and a warning message. I would
appreciate help getting two lines, and an
for an undergraduate course in working with R I would like to explain
the R-environment and the associated file system. But I am not sure if
I have understand all of that correctly myself.
Is there a document or a website out there which explains that in an
understandable way? Things like;
?Startup
should help.
Bert Gunter
Nonclinical Biostatistics
467-7374
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Martin Batholdy
Sent: Monday, May 11, 2009 12:53 PM
To: r-help@r-project.org
Subject: [R] explanation - what
Dear R-users,
I am applying professor Koenker's code for fixed effect quantile regression.
However, I need to bootstrap and cluster the standard errors in my fitted
model.
Assuming that I need to bootstrap then cluster the standard errors by s (the
strata indicator in Prof. Koenker's code),
Hi R-helpers,
I would like to read into R all the .csv files that are in my working
directory, without having to use a read.csv statement for each file.
Each .csv would be read into a separate dataframe which would acquire
the filename of the .csv.
As an example:
Mark-read.csv(Mark.csv)
I really dont know hot to use it - but have a look at ?source
On Mon, May 11, 2009 at 10:37 PM, Mark Na mtb...@gmail.com wrote:
Hi R-helpers,
I would like to read into R all the .csv files that are in my working
directory, without having to use a read.csv statement for each file.
Each .csv
I'm working with genotype data in a frequency table:
a=matrix(1:16, nrow=4)
rownames(a)=c(AA,AT,TA,TT)
a
[,1] [,2] [,3] [,4]
AA159 13
AT26 10 14
TA37 11 15
TT48 12 16
'AT' and 'TA' are essentially the same, and I'd like to combine (add) the
Hi,
Something like this perhaps:
files - dir( pattern = \\.csv$ )
for( x in files){
assign( sub( \\.csv$, , x ) , read.csv(x), envir = .GlobalEnv )
}
Romain
Mark Na wrote:
Hi R-helpers,
I would like to read into R all the .csv files that are in my working
directory, without having to
Woo, global jinx!
On Mon, May 11, 2009 at 1:55 PM, Romain Francois romain.franc...@dbmail.com
wrote:
Hi,
Something like this perhaps:
files - dir( pattern = \\.csv$ )
for( x in files){
assign( sub( \\.csv$, , x ) , read.csv(x), envir = .GlobalEnv )
}
Romain
Mark Na wrote:
Hi
On Sat, May 9, 2009 at 12:17 PM, Berwin A Turlach
ber...@maths.uwa.edu.au wrote:
log(H) = log(n) - log( 1/x_1 + 1/x_2 + ... + 1/x_n)
...But we need to calculate the logarithm of a sum from the logarithms of the
individual terms.
...The way to calculate log(x+y) from lx=log(x) and ly=log(y) ...
Hi,
I'm using stepAIC with a CoxPH model on some data. As
we expect, there are a bunch of predictors, and stepAIC
throws some out and then converges on others.
Anyway, when I run cox.zph on the final model, I get
some evidence of Beta(t) time dependence. Should I
be concerned about the impact of
Hi,
I once made this function (essentially the same as Romain),
assignFiles -
function (pattern = csv, strip = (_|.csv|-| ), ...) # strip is
any pattern you want to remove from the filenames
{
listFiles - list.files(pattern = pattern, all.files = FALSE,
full.names = FALSE,
On 12/05/2009, at 8:41 AM, Andreas Christoffersen wrote:
I really dont know hot to use it - but have a look at ?source
No; ``source'' is irrelevant here.
On Mon, May 11, 2009 at 10:37 PM, Mark Na mtb...@gmail.com wrote:
Hi R-helpers,
I would like to read into R all the .csv
Hi
I'm a relative newbie to R and had a query concerning plotting. I have
generated a par(mfrow = c(2, 2)) graphic with 10 rose diagrams using
circular. What I wanted to add to each individual plot was n = x for the
number of data observations in each dataset. How might I go about doing
this??
Some time ago, I posted a note about what I considered to be a bug in
axis.POSIXt() for R 2.8.x, relating to whether timezones in the data are
obeyed on the axes. A link to that note, and to a quick and helpful
response, is at the following URL
Hi R-helpers,
I must use WinXP at work, and I'm missing a particular feature that's
available in R on my Mac at home...
In WinXP I would like to launch R by dragging and releasing a text
file on top of the R shortcut on my desktop. Then, I would like the
working directory to be automatically set
On Mon, 2009-05-11 at 09:54 -0300, Joyce, Warren wrote:
Hi,
I have been working with R for the last year and using the UKFSST package to
look at satellite tag track data and SST information. Fpr those not familiar,
the package uses the positions estimated by the satellite tags themselves
There are a several possibilities using batchfiles whose
home page is at:
http://batchfiles.googlecode.com
When on that home page note that there are different versions of
these utilities for XP and Vista.
There are two utilities that could be used. Description is
for Vista so for XP you may
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