Here I am in a simulation study where I want to find different values
of x and y such that f(x,y)=c (some known constant) w.r.t. x, y >0,
y<=x and x<=c1 (another known constant). Can anyone please tell me how
to do it efficiently in R. One way I thought that I will draw
different random numbers fro
Erin Hodgess wrote:
Your data set has 5 variables; res$V1, res$V2etc
When you plot, you need to do:
hist(res$V1)
Yes, but it seems that there is only one row of data? If the intention
was just to read 5 numbers into a vector, then use scan(), not
read.table(). (A histogram is pretty us
Does your res.dat only have one row and you want this to one variable?
For read.table each new case has to start on a new line.
If your data are in a "just one row for a variable" format,
res <- scan("C:/Documents and Settings/db/Desktop/res.dat")
will probably do what you want.
Debasish Roy wr
Alex Baugh wrote:
Hi Folks,
I have repeated measures for data on association time (under 2
acoustic condtions) in male and female frogs as they grow to adulthood
(6 timepoints). Thus, two within-subject variables (Acoustic
Condition: 2 levels, Timepoint: 6 levels) and one between-subject
variabl
Sorry, but Thomas needs a format something else can read, e.g. 'a flat
file'.
write.table() is better than write(), and you may want to write in
sections of, say, 1million rows.
The comparison you quote is for ca 7-14 secs to impoort 1m rows of 4
integers. 6 secs is what write.table(DF, "fo
On Mon, 28 Apr 2008, Bin Yue wrote:
Dear friends:
I am a new R user, and not very good at statistics and maths
I find it difficult to find the minimum of this item:
((p*(b-1)-1)*(p+1)^(b-1)+1)/p^2
It seems that the method optim can find the minimum. I tried several
times , but R
mad_bassie <[EMAIL PROTECTED]> wrote in
news:[EMAIL PROTECTED]:
>
> I simulated data for the ANOVA-test where the condition of equal
> variances was not accomplished.
> I have three groups:
> X<-rnorm(50,30,5)
> Y<-rnorm(50,30,10)
> Z<-rnorm(50,30,5)
> (this is just an examplethe variables
Dear friends:
I am a new R user, and not very good at statistics and maths
I find it difficult to find the minimum of this item:
((p*(b-1)-1)*(p+1)^(b-1)+1)/p^2
It seems that the method optim can find the minimum. I tried several
times , but R constanly tells that " failure to
Thx
I verified by typing in plot(1,1) at the R prompt -- sure enough an X window
popped up on my desktop.
Off to investigate Linux GUI front ends!
Thx again
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: Monday, April 28, 2008 5:20 PM
To: Dumblauskas, Jerr
Your data set has 5 variables; res$V1, res$V2etc
When you plot, you need to do:
hist(res$V1)
On Mon, Apr 28, 2008 at 8:47 PM, Debasish Roy <[EMAIL PROTECTED]> wrote:
> I tried to read a dataset and then draw a histogram with a freqeuncy density
> curve fitted on the top of this.
> I used t
I tried to read a dataset and then draw a histogram with a freqeuncy density
curve fitted on the top of this.
I used the command
res <- read.table("C:/Documents and Settings/db/Desktop/res.dat")
and found
> res[1:5]
V1 V2 V3 V4 V5
1 -0.4806 0.5075 0.0491 -0.7985 -0.666
T
I tried to read a dataset and then draw a histogram with a freqeuncy density
curve fitted on the top of this.
I used the command
res <- read.table("C:/Documents and Settings/Desktop/res.dat")
and found
> res[1:5]
??? V1?? V2?? V3? V4? ?V5
1? -0.4806? 0.5075? 0.049
?save
This will write the object to a file in the fastest manner. Here is an example:
> x <- runif(12500/8)
> object.size(x)
[1] 12524
> system.time(save(x, file='/tempxx.Rdata'))
user system elapsed
56.840.86 87.97
>
This wound up to be 79MB on disk after compression.
With
Prof,
Thanks for your generous assistance.
I'm unsure, but an thinking that to utilize one of MS SQL Server's bulk import
utilities, I'll need to export my dataframe to a "flat-file".
Any tips on the best approach for exporting such a large dataframe to a
flat-file? Is write() or write.table()
I am try to create a plot using xyplot(). I created a function panel.fun() wich
generate segment, but I need use two columns from dataset2.
library(lattice)
panel.fun <- function(x, y, minx, maxx, miny, maxy)
{
panel.xyplot(x,y)
# since I don't know how to call dataset2 inside panel
I simulated data for the ANOVA-test where the condition of equal variances
was not accomplished.
I have three groups:
X<-rnorm(50,30,5)
Y<-rnorm(50,30,10)
Z<-rnorm(50,30,5)
(this is just an examplethe variables might still change depending on
how clear the results are)
Now I want to construc
Excellent points Steve,
I am ever expanding my understanding in the area and power is an interesting
one. I do a lot of choice modelling myself, and I am confounded (grin) by
the optimal way to develop designs with conditional levels (deliberate
confounds) etc.
Thanks for that.
Regards P
Hi R users
Here is an example of my data (only a small part)
Year Season A_Time R_Time O_Time All_Time AandR_Time
A_number R_numberO_numberAandR_numberall_number
1999 Winter 9.590741 35.312963 42.759524 23.228431 18.164815
18 9
> Hi R users
> Here is an example of my data (only a small part)
>
> Year Season A_Time R_Time O_Time All_Time AandR_Time
> A_number R_numberO_numberAandR_numberall_number
> 1999 Winter 9.590741 35.312963 42.759524 23.228431 18.164815
> 189
On 28-Apr-08 21:42:19, Dumblauskas, Jerry wrote:
> I think this helps.
>
> Are you saying the RGUI I see when I launch the windows version of R
> will not show in Linux, just any commands that display graphics?
> I'll do a quick test.
> Thx
That is correct. There is an RGUI for Linux, but it is a
On Mon, 28 Apr 2008, Hutchinson,David [PYR] wrote:
I am getting an unanticipated result attempting to format a datetime
string
date1 <- strptime('2005 297 1030 35', '%Y %j %H%M %S') # returns
"2005-10-24 10:12:35"
date2 <- strptime('2005 297 912 35', '%Y %j %H%M %S') # returns NA
Are
Hi Folks,
I have repeated measures for data on association time (under 2
acoustic condtions) in male and female frogs as they grow to adulthood
(6 timepoints). Thus, two within-subject variables (Acoustic
Condition: 2 levels, Timepoint: 6 levels) and one between-subject
variable (Sex:male or femal
I think this helps.
Are you saying the RGUI I see when I launch the windows version of R will not
show in Linux, just any commands that display graphics?
I'll do a quick test.
Thx
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: Monday, April 28, 2008 4:37
Hi, thank you all for those who helped me on prediction of newdata for
linear model,
it is my new question on the prediction of coxph for newdata, for example,
i have the model:
coxout<-coxph( Surv(time, status) ~ x
predict(coxout) will give the fitted values
I have tried predict(coxout, newda
On 28-Apr-08 21:22:51, Dumblauskas, Jerry wrote:
> I just installed R on my 64 bit SUSE Linux system
> -- I compiled with the default x windows support on..
>
> This may be a newbie question (apologies in advance)--
> but how does this show up in X?
>
> I have SSH'd in to my box and set my displa
I am getting an unanticipated result attempting to format a datetime
string
date1 <- strptime('2005 297 1030 35', '%Y %j %H%M %S') # returns
"2005-10-24 10:12:35"
date2 <- strptime('2005 297 912 35', '%Y %j %H%M %S') # returns NA
Are there any ways to adjust the format argument (e.g.,
I just installed R on my 64 bit SUSE Linux system -- I compiled with the
default x windows support on..
This may be a newbie question (apologies in advance)-- but how does this show
up in X?
I have SSH'd in to my box and set my display -- I can run xcalc OK -- but when
I hit the R binary it ju
The POSIXct/lt functions do this (provided your OS is up to it).
x <- c("2007-03-01 12:00:00", "2007-03-15 12:00:00",
"2007-04-01 12:00:00")
xx <- as.POSIXct(x, tz="EST5EDT")
format(xx, tz="Europe/London")
[1] "2007-03-01 17:00:00" "2007-03-15 16:00:00" "2007-04-01 17:00:00"
format
Well, I'll go ahead and (partially) answer my own question.
This seems to do the trick for Britain:
> as.POSIXlt(as.POSIXct("2007-04-01 12:00:00"), tz="GB")
[1] "2007-04-01 17:00:00 BST"
(And it also points out that I got my conversions wrong in my toy
example!)
I had originally tried this by se
I'm interested in restricting the pairwise comparisons of interaction
effects in a multi-way factorial ANOVA, because I find comparisons of
interactions between all different variables different to interpret.
For example (supposing a p<0.10 cutoff just to be able to use this
example):
> summary(fm
Hello,
I'm trying to convert times in the EST/EDT (New York) format to times in
the GMT/BST (London) and UTC+9 format (Tokyo). That is, if I know what
time it is in New York, what is local time in London and Tokyo?
Ex:
Here's the conversion from New York EST/EDT time to London's GMT/BST
time zone
Thank you very much for your replies , but what i don't understand from jim's
reply is this command
dates <- seq(tsp(x)[1]+5/12, tsp(x)[2], .5)
Removing the +5/12 gives me the perfect solution, while keeping it makes the
axis run from the second value plotted and the first value from my ts
ts() only knows about monthly and quarterly series, not biannual ones.
You need to look into methods for irregular series (zoo, its, tseries
...), or add dates yourself (?plot.POSIXct). E.g.
x<-sample (10:100, 17)
x<-ts(x, start=c(1998,1), frequency=2)
xtime <- seq(as.Date("1998-05-01"), by="
I think the short answer is that RODBC is not designed for that, because
ODBC is not. There seems to be an ODBC extension specific to SQL Server
to do so (somewhere said 'SQL Server version 7.0 or later', which may not
apply to you).
I'm pretty unlikely to add support for just one database, e
Hi, thank you all for those who helped me on prediction of newdata for
linear model,
it is my new question on the prediction of coxph for newdata, for example,
i have the model:
coxout<-coxph( Surv(time, status) ~ x
predict(coxout) will give the fitted values
I have tried predict(coxout, newda
Hi List,
I have the following time series and i want to be able to plot it while having
the x-axis running from 1998 to 2006 but in a bi-annual format
So here is my example:
x<-sample (10:100, 17)
x<-ts(x, start=c(1998,1), frequency=2)
plot(x)
When I plot the ts it goes from 1998 to 2006 by
HUGE help, thanks...
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Mike Prager
Sent: Monday, April 28, 2008 12:05 PM
To: [EMAIL PROTECTED]
Subject: Re: [R] Use of recordPlot
"Beck, Kenneth (STP)" <[EMAIL PROTECTED]> wrote:
To be able to page through p
On Mon, 28 Apr 2008, Prof Brian Ripley wrote:
Another possible fix is to copy $prefix/R/lib64/bin/Rscript to $prefix/bin,
Sorry, sent in the middle of editing: that should be
$prefix/lib64/R/bin/Rscript (omit 64 on a 32-bit system).
for those who no longer have the sources around.
I've adde
Another possible fix is to copy $prefix/R/lib64/bin/Rscript to
$prefix/bin, for those who no longer have the sources around.
I've added an extra dependence in R-patched which will I hope prevent
those two copies being different (and often they are not for me).
On Mon, 28 Apr 2008, Gregg Phill
I am using R2.6.0 on Windows Small Business Server 2003. I apologize if the
answer to my question is available
I have searched but have not found anything
that I thought helped me.
I have a dataframe that contains ~4.5 million rows and 5 columns. (see
memory and df details below). I a
Yes,
Re-running make install at the src/unix level seems to have fixed it. I
originally ran make install at the top level after running ./configure
--prefix=/usr/local/r-2.7.0.
Thank You for your assistance,
Gregg.
Prof Brian Ripley wrote:
I have recently built R version 2.7.0 on a CentOS
I have recently built R version 2.7.0 on a CentOS release 4.6, 64-bit
system using the following command:
./configure --prefix=/usr/local/r-2.7.0
Everything built fine and works properly except for Rscript which is
looking for files in the original build directory instead of the
specified in
Gregg,
On 28 April 2008 at 10:42, Gregg Phillips wrote:
| Good Morning,
|
| I have recently built R version 2.7.0 on a CentOS release 4.6, 64-bit
| system using the following command:
|
| ./configure --prefix=/usr/local/r-2.7.0
|
| Everything built fine and works properly except for Rscript w
Can you please post the following offer to the R listserv members?
Chapman & Hall/CRC Press is delighted to offer you a 20% off Discount on our
latest and bestselling R books. Please order online at www.crcpress.com.
Enter promotion code 783EM to apply discount.
Recently Published!
"Beck, Kenneth (STP)" <[EMAIL PROTECTED]> wrote:
To be able to page through plots, open the graphics device with
recording turned on before plotting. Then you can use PageUp and
PageDown keys to see the plots in sequence. The plots are stored
into .SavedPlots. (AFAIK they cannot be accessed with
Thanks, the reference to the wiki and ways to submit patches is a help.
-Original Message-
From: Kingsford Jones [mailto:[EMAIL PROTECTED]
Sent: Thursday, April 24, 2008 10:56 PM
To: Duncan Murdoch
Cc: Martin Maechler; r-help@r-project.org; Beck, Kenneth (STP); Bert
Gunter
Subject: Re: [
Good Morning,
I have recently built R version 2.7.0 on a CentOS release 4.6, 64-bit
system using the following command:
./configure --prefix=/usr/local/r-2.7.0
Everything built fine and works properly except for Rscript which is
looking for files in the original build directory instead of th
Try this:
x <- sample(0:1, rep = TRUE, 1e6, prob=c(0.4, 0.6))
sample(which(x == 1), 1)
On Mon, Apr 28, 2008 at 1:12 PM, <[EMAIL PROTECTED]> wrote:
>
> I have a vector consisting of zeroes and ones, e.g. 101001.
>
> I would like to pick a random element from this vector equal to one. In
> the
> e
I have a vector consisting of zeroes and ones, e.g. 101001.
I would like to pick a random element from this vector equal to one. In the
example above, that means that only elements 1, 3, or 6 can be chosen.
Is there a simple and fast way to do that? (my vector of zeroes and ones is
very large)
Paul;
You asked
>using ...
> optFederov(~.,dat,6)
>... does the job with good efficiency.
>
>I would be interested to know what your objection to this is S
I have no issue with AlgDesign in principle, but the question was
specifically about _fractional_ factorials, so I answered that.
As t
DAVID ARTETA GARCIA wrote:
Dear list,
after fitting an lrm with the Design package (stored as "mymodel") I try
running a summary, but I get the following error:
dim(mydata)
[1] 235 9
names(mydata)
[1] "id" "VAR1" "VAR2" "VAR3" "VAR4" "VAR5" "VAR6" "VAR7" "VAR8"
summary(mymodel)
What
I've used the following to get a few different line types. However, I'd be interested to
hear from someone with expertise in creating line types that are "maximally
distinguishable" by human eyes.
ltys = c("22", "44", "13", "1343", "73", "2262", "12223242", "F282", "F4448444",
"224282F2", "F1"
More than likely.
Take a look at the 'workspace' argument in ?fisher.test and review the
second paragraph in Details:
"For 2 by 2 cases, p-values are obtained directly using the (central or
non-central) hypergeometric distribution. Otherwise, computations are
based on a C version of the FOR
I posted this last night but I think I figured out the problem.
I checked on the underlying equation for the Fisher Exact Test for tables
greater than 2X2. It looks like I have an insane amount of factorials to be
multiplied.
Is that problem? Is it an overflow issue?
Jeff
L
On Mon, Apr 28, 2008 at 03:33:31PM +0100, John C Frain wrote:
> Thanks for the step by step guide. It installs a base R and some
> libraries very well. I tries to load some of the Rmetrics libraries,
> using install.packages() nu ran into problems. I used apt-get install
> to install make and gc
> > a<-as.data.frame(matrix(rnorm(100),nrow=10,ncol=10))
> > b<-which(a$V1>0.8)
> > a_indexb<-a[b,]
> > a_notIndexB<-a[!b,]
> > nrow(a_notIndexB)
> [1] 0
> a_notIndexB <- a[-b,]
> nrow(a_notIndexB)
[1] 9
cu
Philipp
--
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik
Tech
Caio,
using algdesign code below (this produces a full factorial 2*3*3 full
design)
gen.factorial(c(2,3,3))
X1 X2 X3
1 -1 -1 -1
2 1 -1 -1
3 -1 0 -1
4 1 0 -1
5 -1 1 -1
6 1 1 -1
7 -1 -1 0
8 1 -1 0
9 -1 0 0
10 1 0 0
11 -1 1 0
12 1 1 0
13 -1 -1 1
14 1 -1 1
15 -
On Apr 28, 2008, at 10:40 AM, Georg Ehret wrote:
Dear R Community, A simple problem (for some of you): I wish to
index a
data.frame by all elements NOT in my index
E.g.:
a<-as.data.frame(matrix(rnorm(100),nrow=10,ncol=10))
b<-which(a$V1>0.8)
b
[1] 1 4 6 10
a_indexb<-a[b,]
a_notIndex
Thanks for the clarification, this helps a lot, expecially your comment
about "source is real". I will stop my effort to save the graphs
permanently, the scripts don't take long to generate.
But the end of your response hits on the key issue: when I create
several graphs, I would like to be able
On 28.04.2008, at 16:40, Georg Ehret wrote:
E.g.:
a<-as.data.frame(matrix(rnorm(100),nrow=10,ncol=10))
b<-which(a$V1>0.8)
b
[1] 1 4 6 10
a_indexb<-a[b,]
a_notIndexB<-a[!b,]
nrow(a_notIndexB)
[1] 0
Indexing a on b is not a problem (a_indexb), but how can do get
only the
elements left
Thanks for the step by step guide. It installs a base R and some
libraries very well. I tries to load some of the Rmetrics libraries,
using install.packages() nu ran into problems. I used apt-get install
to install make and gcc from the Xandros repository. gcc also included
binutils, gcc-4.1 and
Dear R Community, A simple problem (for some of you): I wish to index a
data.frame by all elements NOT in my index
E.g.:
> a<-as.data.frame(matrix(rnorm(100),nrow=10,ncol=10))
> b<-which(a$V1>0.8)
> b
[1] 1 4 6 10
> a_indexb<-a[b,]
> a_notIndexB<-a[!b,]
> nrow(a_notIndexB)
[1] 0
Indexing a
Thanks Matthias and Gabor. You're both right. Matthias, I think that *was*
what I read although now I realize that Gabor's suggestion is more along the
lines of what I want to do. Though I have found one situation where this
wouldn't really work properly:
par(mfrow=c(2,2))
plot(rnorm(10))
plot(
Hi Jake,
are you looking for argument "panel.first" of "plot.default"?
?plot.default
panel.first: an expression to be evaluated after the plot axes are set
up but before any plotting takes place. This can be useful
for drawing background grids or scatterplot smooths.
hth,
Mat
alg-design doesn;t include fractional factorials; it includes optimal
designs.
The BHH2 does include fractional factorials for 2-level designs.
conf.design is perhaps even better; conf.design generates specified
confounded fractional factorials for multi-level designs provided that
the number of
See:
?frame
e.g.
setHook("plot.new", function(...) cat("Starting plot\n"))
On Mon, Apr 28, 2008 at 9:22 AM, Jake Michaelson
<[EMAIL PROTECTED]> wrote:
> Hi all,
>
> I would like to be able to call a custom function automatically before
> plot.new() is called (more specifically, before a new pl
> I want to process a maximum likelihood estimation for a parametric
> regression survival time model with multiple events per subject.
Data sets with multiple records per subjects are used for several things, you
need to tell me what it is that you want to accomplish. Multiple records is a
m
Hi all,
I would like to be able to call a custom function automatically before
plot.new() is called (more specifically, before a new plot is created on the
current graphics device). Recently I've been poking around in the help
files of some of the low(er) level plotting functions, and I seem to
r
alg-design will do the trick
regards paul
- Original Message -
From: "Caio Azevedo" <[EMAIL PROTECTED]>
To: "R - discussion list" <[EMAIL PROTECTED]>
Sent: Monday, April 28, 2008 11:11 PM
Subject: [R] Fractional Factorial Design
Hi all,
Does anybody know if it is possible to build
Hi all,
Does anybody know if it is possible to build a fractional factorial design
in R? That is, suppose that we want do design an experiment with 3 factors
with 2, 3 and 3 levels, respectivly. However we want to consider, let's say,
only 6 from all possible level combinations. Does R design such
dear list,
my question concerns the use of `eval' in defining the model formula
for `nls' (version 2.6.2.).
consider the following simple example, where the same model and data
are used to perform unweighted and weighted fits. I intentionally
used very uneven weights to guarantee large difference
Hi ,
I am using the deal package (for learning and comparing the bayesian
network).
I am getting the negative score for the heuristics search , i am not able to
interpret the result .
more negative(smaller) is better or larger value is better.Please Help me
out .I will be thankful to you.
Manish
options(warnPartialMatchArgs=TRUE) is available to those who do regrets,
e.g.
options(warnPartialMatchArgs=TRUE)
fooA("hi",m=1)
foo.default: m is 1
Warning message:
In fooA("hi", m = 1) : partial argument match of 'm' to 'model'
On Mon, 28 Apr 2008, Peter Dalgaard wrote:
Aaron Rendahl wrot
Dear list,
after fitting an lrm with the Design package (stored as "mymodel") I
try running a summary, but I get the following error:
dim(mydata)
[1] 235 9
names(mydata)
[1] "id" "VAR1" "VAR2" "VAR3" "VAR4" "VAR5" "VAR6" "VAR7" "VAR8"
summary(mymodel)
Error in `contrasts<-`(`*tmp*`, v
Hi,
I was just wondering if there are any books on R that have applications
using Hidden Markov models
or anyone with tutorials/exercises that can be shared.
Ideally, these would have a bio/ecological slant though not neccesarily.
Any suggestions would be appreciated.
Thanks!
[[alternat
Aaron Rendahl wrote:
> I'm seeing some funny behavior when using methods (the older S3 type)
> and having variables that start with the same letter. I have a vague
> recollection of reading something about this once but now can't seem
> to find anything in the documentation. Any explanation, or a
I'm seeing some funny behavior when using methods (the older S3 type)
and having variables that start with the same letter. I have a vague
recollection of reading something about this once but now can't seem
to find anything in the documentation. Any explanation, or a link to
the proper d
Werner Wernersen wrote:
Hi,
I am plotting several lines into one plot and would
like them to be distinguishable in print later on as
well. Thus, my question is: Is there a larger set of
such line types available like the sets available for
colors?
Maybe somebody has already put in the work to d
Hi,
I'll take care of it.
Thank you for the report.
Detlef
On Mon, 28 Apr 2008 12:31:37 +0200
Peter Dalgaard <[EMAIL PROTECTED]> wrote:
> Ralf Goertz wrote:
> > Peter Dalgaard, Montag, 28. April 2008:
> >
> >> Peter Dalgaard wrote:
> >>
> >
> >
> >> The 10.2 version that I just ins
Hi,
I am plotting several lines into one plot and would
like them to be distinguishable in print later on as
well. Thus, my question is: Is there a larger set of
such line types available like the sets available for
colors?
Maybe somebody has already put in the work to define
some additional good
Ralf Goertz wrote:
> Peter Dalgaard, Montag, 28. April 2008:
>
>> Peter Dalgaard wrote:
>>
>
>
>> The 10.2 version that I just installed does seem to work though:
>> viggo:~/>R
>>
>> R version 2.7.0 (2008-04-22)
>> Copyright (C) 2008 The R Foundation for Statistical Computing
>> ISBN 3-9
With the help of some reproducible code from Tania I traked this down. She
started with all=NULL as the first argument, and merge() was failing when
there were no common columns and no rows in one of the inputs (as
expand.grid failed). I've fixed that in R-patched.
Using all for your object w
Peter Dalgaard, Montag, 28. April 2008:
> Peter Dalgaard wrote:
> The 10.2 version that I just installed does seem to work though:
> viggo:~/>R
>
> R version 2.7.0 (2008-04-22)
> Copyright (C) 2008 The R Foundation for Statistical Computing
> ISBN 3-900051-07-0
> ...
> > air
> airmilesairqual
Peter Dalgaard wrote:
> Ralf Goertz wrote:
>
>> Hi,
>>
>> after updating to 2.7.0, command line completion doesn't work anymore. I
>> understand that the package rcompgen is now part of utils. I hadn't used
>> rcompgen before but completion worked without it (I double checked it
>> using an 2.6.
It works for many other people in 2.7.0, and nothing in R has been changed
(this was part of R in 2.6.x, not part of rcompgen).
Assuming you have not set R_COMPLETION, it is a SuSE-specfic problem.
If you installed a binary, try building from the sources.
On Mon, 28 Apr 2008, Ralf Goertz wrote:
> x1<-paste("A", 1:6, sep = "")
> x2<- round(rgamma(6,2,1))
> x3<-paste("B", 1:6, sep = "")
> x4<- round(rgamma(6,2,1))
> data1 <- data.frame(x1,x2,x3,x4)
> I would like to get
> data2 <- c(A1=4, A2=1, A3=0,...)
> Is there any standard for such a case?
I presume that 4, 2, 0 are the first few va
Ralf Goertz wrote:
> Hi,
>
> after updating to 2.7.0, command line completion doesn't work anymore. I
> understand that the package rcompgen is now part of utils. I hadn't used
> rcompgen before but completion worked without it (I double checked it
> using an 2.6.2 version on another machine). Now,
On Mon, 28 Apr 2008, Kenn Konstabel wrote:
On Sun, Apr 27, 2008 at 4:43 AM, Greg Snow <[EMAIL PROTECTED]> wrote:
What if mylist <- list( 1:10, 101:110 , some.other.things) so the first 2
elements are vectors of length 10. then mylist[1:2] makes sense as still
being a list with the 2 vectors.
Zarrar Shehzad wrote:
Hi
I would like to make a color bar with a gradient of colors that represent
values between 0-1. I used the gradient.rect function in plotrix to make
the actually color bar but I don't know how to put a range of values (0-1)
next to the color bar. How do I do this?
Hi Z
Hi all,
I have the following
x1<-paste("A", 1:6, sep = "")
x2<- round(rgamma(6,2,1))
x3<-paste("B", 1:6, sep = "")
x4<- round(rgamma(6,2,1))
data1 <- data.frame(x1,x2,x3,x4)
I would like to get
data2 <- c(A1=4, A2=1, A3=0,...)
Is there any standard for such a case?
Thank you very much in advance
Dear R users!
I want to process a maximum likelihood estimation for a parametric
regression survival time model with multiple events per subject.
the STATA command for this survival regression is:
use survreg
stset failure(exercise), id(optionid)
local regressors itm posret negret
streg `regress
Hi,
after updating to 2.7.0, command line completion doesn't work anymore. I
understand that the package rcompgen is now part of utils. I hadn't used
rcompgen before but completion worked without it (I double checked it
using an 2.6.2 version on another machine). Now, it doesn't work, even
after s
On Sun, Apr 27, 2008 at 4:43 AM, Greg Snow <[EMAIL PROTECTED]> wrote:
> What if mylist <- list( 1:10, 101:110 , some.other.things) so the first 2
> elements are vectors of length 10. then mylist[1:2] makes sense as still
> being a list with the 2 vectors. What should mylist[[1:2]] return in this
marcelha mukim wrote:
Hi Sarah:
Thank you very much, but my problem remains. What I want is not duplicate
the sample, but be able to divide each frequency for a factor(specific for
each value).For example:
Values 1 2 34 5
Frequency 10 34 56 67 98
Factor
Dear All,
A while back, the question was raised whether R would run on an Eee PC:
http://tolstoy.newcastle.edu.au/R/e3/help/07/12/6564.html
There were some positive responses and some suggestions for getting this to
work:
http://tolstoy.newcastle.edu.au/R/e3/help/07/12/7244.html
For those who
_
An Advanced Course on the S Language
Amsterdam, 6 June 2008
By Longhow Lam
_
For whom?
Existing users of R and S-PLUS who wou
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