Hello,
When I try to to obtain the expected risk for a new dataset using coxph in the
survival package I get an error. Using the example from ?coxph:
> test1 <- list(time= c(4, 3,1,1,2,2,3),+
> status=c(1,NA,1,0,1,1,0),+ x= c(0, 2,1,1,1,0,0),+
>
When I use a model fit with LME, I get an error if I try to use "predict" with
a dataset consisting of a single line.
For example, using this data:
> simpledata
Y t D ID
1 -1.464740870 1 0 1
2 1.222911373 2 0 1
3 -0.605996798 3 0 1
4 0.155692707 4 0 1
5 3.849619772 1 0
From the DESCRIPTION for Matrix:
SystemRequirements: GNU make
Presumably you have a BSD make on your FreeBSD system. This has come up
before, and FreeBSD users have succeeded with GNU make.
On Sat, 7 Jun 2008, ronggui wrote:
[EMAIL PROTECTED] MAKE=gmake
[EMAIL PROTECTED] R
.
insta
Hello everyone,
I have two problems which I am unable to solve :
1.I am trying to add the row labels (g1-g2000) to the very left of a data
table. The data
table is 2000 rows by 62 columns.I have used the following code.
read.table(file="C:\\Documents and Settings\\Owner\\My Documents\\colon
can
Daniel Folkinshteyn wrote:
works for me:
> sub('1.00', '1', '1.00E-20')
[1] "1E-20"
when i input what you wrote, i get the same result. but that doesn't
change the value for TreeTag at row 1501, it's just floating around in
space. if i try it for yr1bp$TreeTag[1501], which is 1.00E-20 i get thi
On Fri, Jun 6, 2008 at 6:23 PM, Achim Zeileis
<[EMAIL PROTECTED]> wrote:
> On Fri, 6 Jun 2008, Michael Friendly wrote:
>
>> In an R graphic, I'm using
>>
>> cond.col <- c("green", "yellow", "red")
>> to represent a quantitative variable, where green means 'OK', yellow
>> represents 'warning'
>> and
>> > install.packages("profr")
>> Warning message:
>> package 'profr' is not available
>
> I selected a different mirror in place of the Iowa one and it
> worked. Odd, I just assumed all the same packages are available
> on all mirrors.
The Iowa mirror is rather out of date as the guy who was loo
On Fri, 6 Jun 2008, Daniel Folkinshteyn wrote:
install.packages("profr")
library(profr)
p <- profr(fcn_create_nonissuing_match_by_quarterssinceissue(...))
plot(p)
That should at least help you see where the slow bits are.
Hadley
so profiling reveals that '[.data.frame' and '[[.data.fram
[EMAIL PROTECTED] MAKE=gmake
[EMAIL PROTECTED] R
.
> install.packages("Matrix")
--- Please select a CRAN mirror for use in this session ---
Loading Tcl/Tk interface ... done
trying URL 'http://bibs.snu.ac.kr/R/src/contrib/Matrix_0.999375-9.tar.gz'
Content type 'application/x-gzip' length 148367
Esmail Bonakdarian wrote:
hadley wickham wrote:
Hi,
I tried this suggestion as I am curious about bottlenecks in my own
R code ...
Why not try profiling? The profr package provides an alternative
display that I find more helpful than the default tools:
install.packages("profr")
> inst
hadley wickham wrote:
Hi,
I tried this suggestion as I am curious about bottlenecks in my own
R code ...
Why not try profiling? The profr package provides an alternative
display that I find more helpful than the default tools:
install.packages("profr")
> install.packages("profr")
Warnin
Daniel, allow me to step off the party line here for a moment, in a problem
like this it's better to code your function in C and then call it from R. You
get vast amount of performance improvement instantly. (From what I see the
process of recoding in C should be quite straight forward.)
H.
--
You may also want to look at the "show.colors" function in the "DAAG" package
to get candidate colors.
From: [EMAIL PROTECTED] [EMAIL PROTECTED] On Behalf Of Michael Friendly [EMAIL
PROTECTED]
Sent: Friday, June 06, 2008 2:37 PM
To: R-Help
Subject: [R] co
install.packages("profr")
library(profr)
p <- profr(fcn_create_nonissuing_match_by_quarterssinceissue(...))
plot(p)
That should at least help you see where the slow bits are.
Hadley
so profiling reveals that '[.data.frame' and '[[.data.frame' and '[' are
the biggest timesuckers...
i suppose
Try this (you need tcl 8.5 and the TeachingDemos package):
library(teachingDemos)
tmpplot <- function(col1='red', col2='yellow', col3='green'){
plot(1:10,1:10, type='n')
rect(1,1,4,4, col=col1)
rect(1,4,4,7, col=col2)
rect(1,7,4,10, col=col3)
rect(6,1,9,4, col=col2grey(col1))
rect(6,4,9,7, c
On Fri, 6 Jun 2008, Michael Friendly wrote:
In an R graphic, I'm using
cond.col <- c("green", "yellow", "red")
to represent a quantitative variable, where green means 'OK', yellow
represents 'warning'
and red represents 'danger'. Using these particular color names, in B/W, red
is darkest
and
on 06/06/2008 06:55 PM hadley wickham said the following:
Why not try profiling? The profr package provides an alternative
display that I find more helpful than the default tools:
install.packages("profr")
library(profr)
p <- profr(fcn_create_nonissuing_match_by_quarterssinceissue(...))
plot(p)
On Fri, Jun 6, 2008 at 3:37 PM, Michael Friendly <[EMAIL PROTECTED]> wrote:
> In an R graphic, I'm using
>
> cond.col <- c("green", "yellow", "red")
> to represent a quantitative variable, where green means 'OK', yellow
> represents 'warning'
> and red represents 'danger'. Using these particular co
thanks for the suggestions! I'll play with this over the weekend and see
what comes out. :)
on 06/06/2008 06:48 PM Don MacQueen said the following:
In a case like this, if you can possibly work with matrices instead of
data frames, you might get significant speedup.
(More accurately, I have had
On Fri, Jun 6, 2008 at 5:10 PM, Daniel Folkinshteyn <[EMAIL PROTECTED]> wrote:
> Hmm... ok... so i ran the code twice - once with a preallocated result,
> assigning rows to it, and once with a nrow=0 result, rbinding rows to it,
> for the first 20 quarters. There was no speedup. In fact, running wi
In a case like this, if you can possibly work with matrices instead
of data frames, you might get significant speedup.
(More accurately, I have had situations where I obtained speed up by
working with matrices instead of dataframes.)
Even if you have to code character columns as numeric, it can
I'll add my $0.02 as I've just gone thru a (painful) transition to Linux. In my
case Ubuntu didn't quite work for reason I'm still not sure (must be hardware +
driver issue). I eventually put on opensuse 10.3 and installed R in an rpm
pkgage on the command line. Getting R in was not simple. I go
FWIW, those who are curious about Linux but are not willing
or ready to abandon the Windows platform can now very easily
try out Ubuntu without having to repartition their hard drive.
Wubi is a project that installs Ubuntu under Windows so that it
can be uninstalled easily and requires no messing
Hmm... ok... so i ran the code twice - once with a preallocated result,
assigning rows to it, and once with a nrow=0 result, rbinding rows to
it, for the first 20 quarters. There was no speedup. In fact, running
with a preallocated result matrix was slower than rbinding to the matrix:
for prea
R works just fine on Fedora 9.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible cod
another vote for ubuntu here - works for me, and quite trouble-free. add
the r-project repositories, and you're sure to always have the latest,
too. (if you don't care for the latest R, you can of course also just
get R from the distro's repos as well)
on 06/06/2008 05:22 PM Abhijit Dasgupta s
works for me:
> sub('1.00', '1', '1.00E-20')
[1] "1E-20"
remember, according to ?sub, it's sub(pattern, repl, string)
try doing it step by step. first, see what yr1bp$TreeTag[1501] is.
then, if it's the right data item, see what the output of sub("1.00",
"1", yr1bp$TreeTag[1501]) is.
that'll l
I've had R on an Ubuntu system for about 18 months now, and getting R
up and running was a breeze. (I didn't realize it earlier, but Dirk
certainly gets my vote of thanks for his efforts in making this process
as easy as it is). Specially in terms of dependencies and the like, the
Ubuntu packa
steven wilson wrote:
I'm planning to install Linux on my computer to run R (I'm bored of
W..XP). However, I haven't used Linux before and I would appreciate,
if possible, suggestions/comments about what could be the best option
install,
Hi,
I have used Linux since the early 1990s starting wi
This is not sound advice. For 1GB yes, perhaps 2GB. Beyond that the
extra freedom in the address space of a 64-bit system pays off.
The user address space of a 32-bit Linux system is (in the examples I have
seen) 3 to 3.5Gb. See ?"Memory-limits" for why that is restrictive.
There are some
On 6 June 2008 at 16:18, Kevin E. Thorpe wrote:
| Any of the three distros mentioned are sure to be fine.
| Personally, I find the sysadmin tool in opensuse to be
| fantastic for a novice.
|
| It comes down to preference. Try some live versions of the distros to
| see what you like best.
While
Dear all,
a related follow up -- with the hope for some feedback from the specialists.
Is the following general advice justified:
=
If one has not more than 4GB RAM and one wants to run primarily R on
one's Linux machine, it is a good idea
hi there:
please refer to:
http://www.math.usu.edu/~adele/forests/cc_home.htm
and
http://www.math.usu.edu/~minnotte/S5600S07/R17.txt
thanks
BertrandM wrote:
>
> Hello
>
> Is there exists a package for multivariate random forest, namely for
> multivariate response data ? It seems to be imposs
In an R graphic, I'm using
cond.col <- c("green", "yellow", "red")
to represent a quantitative variable, where green means 'OK', yellow
represents 'warning'
and red represents 'danger'. Using these particular color names, in B/W,
red is darkest
and yellow is lightest. I'd like to find color de
I have both Debian, Ubuntu, RedHat and CentOS systems, and primary run R
on the Debian and RedHat machines. I have encountered few problems
running R on RedHat/CentOS, but I do think the Debian/Ubuntu package
management system, combined with the kind provision of packages, makes
life a lot simpler.
Any of the three distros mentioned are sure to be fine.
Personally, I find the sysadmin tool in opensuse to be
fantastic for a novice.
It comes down to preference. Try some live versions of the distros to
see what you like best.
Douglas Bates wrote:
> On Fri, Jun 6, 2008 at 1:13 PM, steven wilso
John Nolan wrote:
I am trying to call a precompiled C function that uses a struct as one of
it's arguments.
I could write a wrapper function in C, but I was hoping there is some way
to
pack fields into an array of type raw that could be passed directly to the
function.
Here is some more detail.
Many thanks to everyone who replied,
Tolga
William Pepe <[EMAIL PROTECTED]>
06/06/2008 20:52
To
<[EMAIL PROTECTED]>,
cc
Subject
RE: [R] col.names ?
As a very simple example:
TolgaData<-data.frame(A=c(1,2),B=c(3,4))
names(TolgaData )<- c( "newA", "newB" )
> TolgaData
Column names s
As a very simple example:
TolgaData<-data.frame(A=c(1,2),B=c(3,4))names(TolgaData )<- c( "newA", "newB"
)> TolgaData Column names should now be newA and newB Best, Bill> To:
r-help@r-project.org> From: [EMAIL PROTECTED]> Date: Fri, 6 Jun 2008 20:06:41
+0100> Subject: [R] col.names ?> > Dear R
Dear Paul,
Try also:
ddTable <-
data.frame(Id=c(1,1,2,2),name=c("Paul","Joe","Bob","Larry"))
ddTable[unique(ddTable$Id),]
HTH,
Jorge
On Fri, Jun 6, 2008 at 9:35 AM, Emslie, Paul [Ctr] <[EMAIL PROTECTED]> wrote:
> I want to take the first row of each unique ID value from a data frame.
> For
On 6/6/2008 3:06 PM, [EMAIL PROTECTED] wrote:
Dear R Users,
A bit of an elementary question, but somehow, I haven't been able to
figure it out. I'd like to changes the column names of a data frame, so I
am looking for something like col.names (as in row.names). Could someone
please show me h
See ?names
On 6/6/08, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
>
> Dear R Users,
>
> A bit of an elementary question, but somehow, I haven't been able to
> figure it out. I'd like to changes the column names of a data frame, so I
> am looking for something like col.names (as in row.names). C
Try this:
matplot(t(m), type='l', lty = 'solid', col='black')
On 6/6/08, Alberto Monteiro <[EMAIL PROTECTED]> wrote:
>
> Suppose that I have a matrix like:
>
> m <- rbind(c(1,2,3,4), c(2,3,2,1))
>
> Is there any way to efficiently plot the _lines_ as if
> I was doing:
>
> plot(m[1,], type="l")
On 6/6/2008 3:01 PM, Alberto Monteiro wrote:
Suppose that I have a matrix like:
m <- rbind(c(1,2,3,4), c(2,3,2,1))
Is there any way to efficiently plot the _lines_ as if
I was doing:
plot(m[1,], type="l")
points(m[2,], type="l", col="red")
(of course, in the "real world" there much more than
On Fri, Jun 6, 2008 at 1:13 PM, steven wilson <[EMAIL PROTECTED]> wrote:
> Dear all;
> I'm planning to install Linux on my computer to run R (I'm bored of
> W..XP). However, I haven't used Linux before and I would appreciate,
> if possible, suggestions/comments about what could be the best option
Dear R Users,
A bit of an elementary question, but somehow, I haven't been able to
figure it out. I'd like to changes the column names of a data frame, so I
am looking for something like col.names (as in row.names). Could someone
please show me how to change the column names of a data frame ?
I am trying to call a precompiled C function that uses a struct as one of
it's arguments.
I could write a wrapper function in C, but I was hoping there is some way
to
pack fields into an array of type raw that could be passed directly to the
function.
Here is some more detail. The C struct is s
On 6/6/2008 11:43 AM, Marco Chiapello wrote:
Hi,
I have a simple question. If I have a table and I want to have the mean
for each row, how can I do?!
Es:
c1 c2 c3 mean
1 12 13 14 ??
2 15 24 10 ??
...
Thanks,
Marco
VADeaths
Suppose that I have a matrix like:
m <- rbind(c(1,2,3,4), c(2,3,2,1))
Is there any way to efficiently plot the _lines_ as if
I was doing:
plot(m[1,], type="l")
points(m[2,], type="l", col="red")
(of course, in the "real world" there much more than
just 2 lines and 4 columns...)
Alberto Monteir
dear R users,
the data frame (read in from a csv) looks like this:
TreeTag CensusStage DBH
1 CW-W740 2001 juvenile 5.8
2 CW-W739 2001 juvenile 4.3
3 CW-W738 2001 juvenile 4.7
4 CW-W737 2001 juvenile 5.4
5 CW-W736 2001 juvenile 7
I still like the number 4 option, so I think we need to come up with a formal
definition for a "junk" of data. I read somewhere that Tukey coined the word
"bit" as it applies to computers, we can share the credit/blame for "junks" of
data.
My proposal for a statistical/data definition of the w
See also
?rowMeans
On Fri, Jun 6, 2008 at 1:56 PM, <[EMAIL PROTECTED]> wrote:
>> TABLE<-matrix(data=c(12,13,14,15,24,10),byrow=T,nrow=2,ncol=3)
>> TABLE
>
> [,1] [,2] [,3]
> [1,] 12 13 14
> [2,] 15 24 10
>>
>> apply(TABLE,1,mean)
>
> [1] 13.0 16.3
>
> Chunhao
>
>
> Quoti
OK,
The original ggplot() construct (below) on the following two dataframes
(test1, test2) generate different outputs, which I have attached. The
output that I expect is that shown in test2.png. My expectations are
that I have set the plotting limits with 'scale_x_continuous(lim = c(0,
360)) + sca
TABLE<-matrix(data=c(12,13,14,15,24,10),byrow=T,nrow=2,ncol=3)
TABLE
[,1] [,2] [,3]
[1,] 12 13 14
[2,] 15 24 10
apply(TABLE,1,mean)
[1] 13.0 16.3
Chunhao
Quoting Marco Chiapello <[EMAIL PROTECTED]>:
Hi,
I have a simple question. If I have a table and I want to have
> -Original Message-
> From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
> Sent: Friday, June 06, 2008 12:33 PM
> To: Greg Snow
> Cc: Patrick Burns; Daniel Folkinshteyn; r-help@r-project.org
> Subject: Re: [R] Improving data processing efficiency
>
> On Fri, Jun 6, 2008 at 2:28 PM, Greg Sn
Dear all;
I'm planning to install Linux on my computer to run R (I'm bored of
W..XP). However, I haven't used Linux before and I would appreciate,
if possible, suggestions/comments about what could be the best option
install, say Fedora, Ubuntu or OpenSuse which to my impression are the
most popul
Hello
Is there exists a package for multivariate random forest, namely for
multivariate response data ? It seems to be impossible with the
"randomForest" function and I did not find any information about this
in the help pages ...
Thank you for your help
Bertrand
___
My guess is that number 2 is closest to the mark.
Typing too fast is unfortunately not one of my
habitual attributes.
Gabor Grothendieck wrote:
On Fri, Jun 6, 2008 at 2:28 PM, Greg Snow <[EMAIL PROTECTED]> wrote:
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On
Hello
Is there exists a package for multivariate random forest, namely for
multivariate response data ? It seems to be impossible with the
"randomForest" function and I did not find any information about this
in the help pages ...
Thank you for your help
Bertrand
___
I have installed R (D)COM on a (windows) machine that is part of a windows
domain. if I run the test file in a local (log into this machine)
administrative account it works fine. If I run the test file on a domain
account with administrative rights it will not connect to the server, even
is I chan
The path to R/bin is in the Windows PATH variable. Yet I get this
error.
On Jun 6, 10:37 am, "Dumblauskas, Jerry" <[EMAIL PROTECTED]
suisse.com> wrote:
> Try and make sure that R is in your windows Path variable
>
> I got your message when I first did this, but when I did the about it
> then work
Hi,
I have a simple question. If I have a table and I want to have the mean
for each row, how can I do?!
Es:
c1 c2 c3 mean
1 12 13 14 ??
2 15 24 10 ??
...
Thanks,
Marco
__
R-help@r-pro
Hello
Is there exists a package for multivariate random forest, namely for
multivariate response data ? It seems to be impossible with the
"randomForest" function and I did not find any information about this
in the help pages ...
Thank you for your help
Bertrand
___
On Fri, Jun 6, 2008 at 2:28 PM, Greg Snow <[EMAIL PROTECTED]> wrote:
>> -Original Message-
>> From: [EMAIL PROTECTED]
>> [mailto:[EMAIL PROTECTED] On Behalf Of Patrick Burns
>> Sent: Friday, June 06, 2008 12:04 PM
>> To: Daniel Folkinshteyn
>> Cc: r-help@r-project.org
>> Subject: Re: [R] Im
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Patrick Burns
> Sent: Friday, June 06, 2008 12:04 PM
> To: Daniel Folkinshteyn
> Cc: r-help@r-project.org
> Subject: Re: [R] Improving data processing efficiency
>
> That is going to be situation depende
Please read the help for par(mfrow)! AFAICS this is nothing to do with
boxplot().
In a layout with exactly two rows and columns the base value
of '"cex"' is reduced by a factor of 0.83: if there are three
or more of either rows or columns, the reduction factor is
Cool, I do have an upper bound, so I'll try it and how much of a
speedboost it gives me. Thanks for the suggestion!
on 06/06/2008 02:03 PM Patrick Burns said the following:
That is going to be situation dependent, but if you
have a reasonable upper bound, then that will be
much easier and not f
> Does the difference have something to do with ggplot() using ranges
> derived from the data?
> When I modify my original 'test' dataframe with two extra rows as
> defined below, I get expected results in both versions.
Order shouldn't matter - and if it's making a difference, that's a
bug. But
On Fri, 6 Jun 2008, Luca Mortarini wrote:
Hi,
I am new to R and i am looking for a way to extract a subset from a
vector.
I have a vector of number oscillating around zero (a decreasing
autocorrelation function) and i would like to extract only the first
positive part of the function (from z
That is going to be situation dependent, but if you
have a reasonable upper bound, then that will be
much easier and not far from optimal.
If you pick the possibly too small route, then increasing
the size in largish junks is much better than adding
a row at a time.
Pat
Daniel Folkinshteyn wrot
Thanx Thierry,
Suggestion #1 had no effect.
I have been playing with variants on #2 along the way.
DaveT.
>-Original Message-
>From: ONKELINX, Thierry [mailto:[EMAIL PROTECTED]
>Sent: June 6, 2008 04:02 AM
>To: Thompson, David (MNR); hadley wickham
>Cc: r-help@r-project.org
>Subject: RE:
The interesting thing about R is that there are several ways to "skin the
cat"; here is yet another solution:
> do.call(rbind, by(ddTable, ddTable$Id, function(z) z[1,,drop=FALSE]))
Id name
1 1 Paul
2 2 Bob
>
On Fri, Jun 6, 2008 at 9:35 AM, Emslie, Paul [Ctr] <[EMAIL PROTECTED]> wrote:
> I
Ok, sorry about the zip, then. :) Thanks for taking the trouble to clue
me in as to the best posting procedure!
well, here's a dput-ed version of the small data subset you can use for
testing. below that, an updated version of the function, with extra
explanatory comments, and producing an ext
On Fri, 6 Jun 2008, Nanye Long wrote:
Hi all,
Does anyone know where to download the "BRugs" package? I did not find
it on r-project website. Thanks.
It is Windows-only, and you download it from 'CRAN (extras)' which is part
of the default repository set on Windows versions of R. So
insta
One simple way is to do something like:
> fit <- lm(y ~ I(x1-x2) + x3, data=mydata)
The first coeficient (after the intercept) will be the slope for x1, the slope
for x2 will be the negative of that. This model is nested in the fuller model
with x1 and x2 fit seperately and you can therefore t
just in case, uploaded it to the server, you can get the zip file i
mentioned here:
http://astro.temple.edu/~dfolkins/helplistfiles.zip
on 06/06/2008 01:25 PM Daniel Folkinshteyn said the following:
i thought since the function code (which i provided in full) was pretty
short, it would be reaso
I think the posting guide may not be clear enough and have suggested that
it be clarified. Hopefully this better communicates what is required and why
in a shorter amount of space:
https://stat.ethz.ch/pipermail/r-devel/2008-June/049891.html
On Fri, Jun 6, 2008 at 1:25 PM, Daniel Folkinshteyn <
ci = rainbow(7)[c(4:7, 1:3)]
on 06/06/2008 01:02 PM avilella said the following:
Hi,
I want to reorder the colors given by rainbow(7) so that the last half
move to the first 4.
For example:
ci=rainbow(7)
ci
[1] "#FFFF" "#FFDB00FF" "#49FF00FF" "#00FF92FF" "#0092"
"#4900"
[7] "#FF
thanks for the tip! i'll try that and see how big of a difference that
makes... if i am not sure what exactly the size will be, am i better off
making it larger, and then later stripping off the blank rows, or making
it smaller, and appending the missing rows?
on 06/06/2008 11:44 AM Patrick Bu
Dear Dani,
I intend at some point to extend the effects package to linear and
generalized linear mixed-effects models, probably using lmer() rather
than lme(), but as you discovered, it doesn't handle these models now.
It wouldn't be hard, however, to do the computations yourself, using
the coeff
i thought since the function code (which i provided in full) was pretty
short, it would be reasonably easy to just read the code and see what
it's doing.
but ok, so... i am attaching a zip file, with a small sample of the data
set (tab delimited), and the function code, in a zip file (posting
On Fri, 6 Jun 2008, Neil Gupta wrote:
Hello R-users,
I have a very simple problem I wanted to solve. I have a large dataset as
such:
Lag X.Symbol Time TickType ReferenceNumber Price Size X.Symbol.1
Time.1 TickType.1 ReferenceNumber.1
1 ES 3:ESZ7.GB 08:30:00B74390987 15
Hi,
I am new to R and i am looking for a way to extract a subset from a
vector.
I have a vector of number oscillating around zero (a decreasing
autocorrelation function) and i would like to extract only the first
positive part of the function (from zero lag to the lag where the function
invert
The package has a plot() method for random-effects meta-analyses as well,
either those produced by meta.DSL or meta.summaries.
There are examples on the help page for meta.DSL.
-thomas
On Tue, 27 May 2008, Shi, Jiajun [BSD] - KNP wrote:
Dear all,
I could not draw a forest plot fo
Hi,
I want to reorder the colors given by rainbow(7) so that the last half
move to the first 4.
For example:
> ci=rainbow(7)
> ci
[1] "#FFFF" "#FFDB00FF" "#49FF00FF" "#00FF92FF" "#0092"
"#4900"
[7] "#FF00DBFF"
I would like "#FFFF" "#FFDB00FF" "#49FF00FF" to be at the end of
ci,
Perhaps this was too big a question, so I'll ask something shorter:
I have fit a linear model, and want to use its prediction intervals to
calculate the sum of many individual predictions.
1) Some of the lower prediction intervals are negative, which is
non-sensical. Should I just set all negati
Dear NL.
BRugs is available from the CRAN extras repository hosted by Brian Ripley.
install.packages("BRugs")
should install it as before (for R-2.7.x), if you have not changed the
list of default repositories.
Best wishes,
Uwe Ligges
Nanye Long wrote:
Hi all,
Does anyone know where t
Hello,
I am trying to perform a fit.contrast() on a lme object with this code:
attach(error_DB)
model_temperature <- lme(Error ~ Temperature, data = error_DB,random=~1|ID)
summary(model_temperature)
fit.contrast(model_temperature, "Temperature", c(-1,1), conf.int=0.95 )
detach(error_DB)
but I go
You can write your own function, something about like this:
read.table2 <- function(file, ...)
{
x <- read.table(file, ...)
attributes(x)[["file_name"]] <- file
return(x)
}
mydata <- read.table2("Run224_v2_060308.txt", sep = "\t", header = TRUE)
myfile <- attr(x, "file_name")
On Fri, Jun 6, 2008
well, where are you getting the filename in the first place? are you
looping over a list of filenames that comes from somewhere?
generally, for concatenating strings, look at function 'paste':
write.table(myoutput, paste(myfilename,"_out.txt", sep=''),sep="\t")
on 06/06/2008 11:51 AM DAVID ARTE
That is the last line of every message to r-help.
On Fri, Jun 6, 2008 at 12:05 PM, Gabor Grothendieck
<[EMAIL PROTECTED]> wrote:
> Its summarized in the last line to r-help. Note reproducible and
> minimal.
>
> On Fri, Jun 6, 2008 at 12:03 PM, Daniel Folkinshteyn <[EMAIL PROTECTED]>
> wrote:
>>
Its summarized in the last line to r-help. Note reproducible and
minimal.
On Fri, Jun 6, 2008 at 12:03 PM, Daniel Folkinshteyn <[EMAIL PROTECTED]> wrote:
> i did! what did i miss?
>
> on 06/06/2008 11:45 AM Gabor Grothendieck said the following:
>>
>> Try reading the posting guide before posting.
Is there a way to set up a regression in R that forces two coefficients
to be equal but opposite in sign?
I'm trying to setup a model where a subject appears in a pair of
environments where a measurement X is made. There are a total of 5
environments, one of which is a baseline. But each o
i did! what did i miss?
on 06/06/2008 11:45 AM Gabor Grothendieck said the following:
Try reading the posting guide before posting.
On Fri, Jun 6, 2008 at 11:12 AM, Daniel Folkinshteyn <[EMAIL PROTECTED]> wrote:
Anybody have any thoughts on this? Please? :)
on 06/05/2008 02:09 PM Daniel Folki
Hi all,
Does anyone know where to download the "BRugs" package? I did not find
it on r-project website. Thanks.
NL
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PLEASE do read the posting guide http://www.R-project.o
Hi list,
Is it possible to save the name of a filename automatically when
reading it using read.table() or some other function?
My aim is to create then an output table with the name of the original
table with a suffix like _out
example:
mydata = read.table("Run224_v2_060308.txt", sep = "\
Try reading the posting guide before posting.
On Fri, Jun 6, 2008 at 11:12 AM, Daniel Folkinshteyn <[EMAIL PROTECTED]> wrote:
> Anybody have any thoughts on this? Please? :)
>
> on 06/05/2008 02:09 PM Daniel Folkinshteyn said the following:
>>
>> Hi everyone!
>>
>> I have a question about data pro
One thing that is likely to speed the code significantly
is if you create 'result' to be its final size and then
subscript into it. Something like:
result[i, ] <- bestpeer
(though I'm not sure if 'i' is the proper index).
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-s
> It's too obvious, so I am positive that there is a good reason for not doing
> this, but still:
> why is it not possible, to have an "outlier" output in stat_boxplot that can
> be used at geom_text()?
>
> Something like this, with "upper":
> > dat=data.frame(num=rep(1,20), val=c(runif(18),3,3.5
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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