See ?fivenum in the stats package. If you just type
stats::fivenum
you will get the code. The crucial calculations are in the last few
lines.
Simon.
On Mon, 2008-08-04 at 16:19 +0930, Fernando Marmolejo Ramos wrote:
> Dear people
>
> I've learnt that by using the "boxplot.stats" command in th
Hello,
Is ?segments what you are looking for ?
François
-Message d'origine-
De : [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] De la part de elyakhlifi
mustapha
Envoyé : vendredi 1 août 2008 17:15
À : r-help@r-project.org
Objet : [R] graph
Hello.
I don't know how to do to ouput segment
Dear colleagues,
I know SPSS can not compute linear mixed models. I used 'R' before for
computing multivariate analyses. But, I never encountered such a major
difference in outcome between SPSS and 'R':
In SPSS the Pearson correlation between variable 1 and variable 2 is 31%
p<0.001.
In
Dear Sorn,
It's hard to guess what your problem is, as you don't provide any sample
code. My guess is that the graphics are empty. Did you use
print(qplot(...)) or just qplot(). The latter won't work. You need
print(qplot(...))
HTH,
Thierry
-
Andrew Robinson wrote:
Dear R colleagues,
a friend and I are trying to develop a modest workflow for the problem
of decomposing tests of higher-order terms into interpretable sets of
tests of lower order terms with conditioning.
For example, if the interaction between A (3 levels) and C (2 leve
Draga, R. wrote:
.
Would somebody know how it is possible that in SPSS we get p<0.001 and
in R we get p=0.24?
And, in 'R' the 95% CI of the Odds Ratio is 6.2-14.1. Why is the
p-value=0.24?
You asked this before and you are still not providing sufficient
information.
Harold alre
On Mon, 2008-08-04 at 17:00 +1000, Simon Blomberg wrote:
> See ?fivenum in the stats package. If you just type
>
> stats::fivenum
>
> you will get the code. The crucial calculations are in the last few
> lines.
That will only give the code to calculate the five number summary, but
Fernando want
Hi Everyone,
I'd omitted the non-optional 'parameters' argument to selfStart. Making this
change to SSbatch gives the same (successful) result from the two calls to
nls.
SSbatch<-selfStart(
model=function(Batch, Coeffs)
{
Coeffs[Batch]
}
,initial=function(mCall, data, LHS)
{
# Estimate coeffic
I think I have found the answer myself.
If you have something better, please write it.
Pierre8r
My R code :
---
library(quantmod)
library(xts)
Lines <-
"2008.07.01,02:00,1.5761,1.5766,1.5760,1.5763,65
2008.07.01,02:15,1.5762,1.5765,1.5757,1.5761,95
2008.07.01,02:30,1.5762,1.5765,1.575
> "SM" == Steven McKinney <[EMAIL PROTECTED]>
> on Fri, 1 Aug 2008 18:38:54 -0700 writes:
SM> Thanks to Duncan Murdoch and Marc Schwartz for their
SM> excellent help.
SM> As we don't yet have the apache http: interface to svn
SM> running yet, 'svn export' is the access
Checkout this one:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/82452.html
On Mon, Aug 4, 2008 at 6:24 AM, Henning Wildhagen <[EMAIL PROTECTED]> wrote:
> Dear list,
>
> for a data structure like in df:
>
> set.seed(100)
> Treatment<-rep(c("Nitrogen","Carbon", "Sulfur"),each=9)
> week<-rep(c(1,
I wrote a "program" truc.r with emacs
In emacs, I start the buffer R, then I eval the buffer truc.r (C-c C-b)
All is correct and I have my results.
But, when I return to Console R, and make:
> source ("truc.r"), I obtain an error:
Erreur dans source("truc.r") :
invalid multibyte character in p
Dear list,
for a data structure like in df:
set.seed(100)
Treatment<-rep(c("Nitrogen","Carbon", "Sulfur"),each=9)
week<-rep(c(1,5,9),3,each=3)
genes<-rep(c("18s", "EF1b", "NR"),9)
copies<-rnorm(27, 100,40)
df<-data.frame(Treatment,week,genes,copies)
i wrote this code for a xyplot:
libra
pir2.jv wrote:
I wrote a "program" truc.r with emacs
In emacs, I start the buffer R, then I eval the buffer truc.r (C-c C-b)
All is correct and I have my results.
But, when I return to Console R, and make:
> source ("truc.r"), I obtain an error:
Erreur dans source("truc.r") :
invalid multibyt
Hi John
I don't get an error message but a warning
> write.table(myclara$clustering,"cluster.dat",append=TRUE)
Warning message:
In write.table(myclara$clustering, "cluster.dat", append = TRUE) :
appending column names to file
>
Here it is the output of str(myclara), it looks "strange" to me. I
Dear HelpeRs,
I am trying to write axis labels with some letters in cursive for later
inclusion in a LaTeX document. The following code does what I want with
latin letters (c cursive and the rest not cursive):
plot(1:10, 1:10, ylab = expression(italic(c)*(MB)))
or
plot(1:10, 1:10, ylab = expr
Many thanks
I was sure it was simple. That was exactly what I wanted.
I should have clarified that I was looking for a box-plot.
Thanks to all who responed.
Sandy
S Ellison wrote:
Sandy,
You can re-order a factor with
df$Eyeball<-factor(df$Eyeball, levels=c("Normal", "Mild", "
You could piece it together using the Hershey fonts for the italic rho:
op <- par(xpd = NA)
plot(1, type = "n", ylab = "")
u <- par("usr")
text(u[1] - .1 * diff(u[1:2]), 1, "\\*r", vfont = c("serif", "italic"))
par(op)
Now add the rest.
?Hershey
?par
?strwidth
On Mon, Aug 4, 2008 at 7:12 AM, Lui
On Mon, Aug 04, 2008 at 10:17:38AM +0200, Peter Dalgaard wrote:
> Andrew Robinson wrote:
> >Dear R colleagues,
> >
> >a friend and I are trying to develop a modest workflow for the problem
> >of decomposing tests of higher-order terms into interpretable sets of
> >tests of lower order terms with co
> Hi,
>
> yesterday i had the surprise not to be able to load the package "ca" on R
> 2.7.0 saying that cannot find required package rgl although it was there. So
> today i've upgraded to 7.2.1. patched and i got the following error:
>
>> local({pkg <- select.list(sort(.packages(all.available =
Hi,
Sorry for that . Today R did load both ca and rgl packages with no problems
... the single thing i've done since Friday was to put "defrag" on drive C, i
even didn't re-start the computer ...actually it was re-started several time
with same weird problem ... so i suppose something was
Andrew Robinson wrote:
On Mon, Aug 04, 2008 at 10:17:38AM +0200, Peter Dalgaard wrote:
Andrew Robinson wrote:
Dear R colleagues,
a friend and I are trying to develop a modest workflow for the problem
of decomposing tests of higher-order terms into interpretable sets of
tests of lower o
On Mon, Aug 04, 2008 at 02:51:48PM +0200, Peter Dalgaard wrote:
> Andrew Robinson wrote:
> >On Mon, Aug 04, 2008 at 10:17:38AM +0200, Peter Dalgaard wrote:
> >
> >>Andrew Robinson wrote:
> >>
> >
> >That is a neat idea, thanks, Peter, but it doesn't quite fit the bill.
> >The summary provides
Gavin Simpson wrote:
On Mon, 2008-08-04 at 17:00 +1000, Simon Blomberg wrote:
See ?fivenum in the stats package. If you just type
stats::fivenum
you will get the code. The crucial calculations are in the last few
lines.
That will only give the code to calculate the five number summary, but
Greetings,
I recently discovered R and how to tackle my own functions.
I'm blocked on a simple barrier I guess. I call a function that recognizes a
string and return the adequate formula in a main function. This is done
through a script.
When I type the formula in the main function, it works fin
Hi all,
I have this figure:
http://docs.google.com/Doc?id=df5zfsj4_103rjt2v4d5
created with the following steps:
> x
[1] 90.4 57.8 77.0 103.7 55.4 217.5 68.1 85.3 152.0 113.0 97.1 89.9
[13] 68.1 83.7 77.4 34.5 104.9 170.3 88.6 88.1 108.8 77.4 85.6 82.7
[25] 81.3 108.0 49
bbouling wrote:
Greetings,
I recently discovered R and how to tackle my own functions.
I'm blocked on a simple barrier I guess. I call a function that recognizes a
string and return the adequate formula in a main function. This is done
through a script.
When I type the formula in the main funct
Gundala Viswanath wrote:
Hi all,
I have this figure:
http://docs.google.com/Doc?id=df5zfsj4_103rjt2v4d5
created with the following steps:
x
[1] 90.4 57.8 77.0 103.7 55.4 217.5 68.1 85.3 152.0 113.0 97.1 89.9
[13] 68.1 83.7 77.4 34.5 104.9 170.3 88.6 88.1 108.8 77.
Hi Gundala,
You have to reorder the points, like below:
#(your code ...)
g <- cbind(x,g.pdf)[order(x),]
points(x=[,1],y=g[,2],type='l',col="red")
Cheers,
gary
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Gundala Viswanath
Sent: Monday, August 04, 2008
Hi R people
Pulling my hair out here trying to get something very simple to work.
Have a data frame of 774 rows and want to plot first and second half
on same axes with different colours. A variable is present call 'row'
created like this and checked to be OK:
row <- seq( len=le
Hi R people [duplicate - sorry, just posted HTML by mistake]
Pulling my hair out here trying to get something very simple to work.
Have a data frame of 774 rows and want to plot first and second half
on same axes with different colours. A variable is present call 'row'
created like this
Hi Michael,
>> Pulling my hair out here trying to get something very simple to work. ...
I can't quite see what you are trying to do [and I am not sure that you
clearly state it], but you could make things easier and simpler by (1)
creating a factor to identify your groups of rows more cleanly a
With the R prompt still active, data may be cleanly viewed (not
edited) in a separate window by a simplification of function xless in
Hmisc by Frank Harrell:
look. <- function (x, ...,
title = substring(deparse(substitute(x)), 1, 20))
{
page(x, method = "print", title = title,opti
This is a useful tip, but there is some confusion in the thread that makes it
hard to follow. Just clarifying, I hope, so that folks can easily follow it
in the future.
The original posting had a set of encounter histories with descriptions that
did not match the histories. There were 2 males and
I am try to plot a data frame, and encounter some difficulties.
my x axis range is from -60 to 80 and unit length is 10
so x axis looks like
-60, -40, -20, .., 40,60,80
But I want chang unit length from 10 to 5
-60, -50, -40, .., 60,70,80
Does anyone know how to do it?
Hi, all,
I would like to know if there is any gui interface out there
(academic or commercial) that allows one to edit R-language generated
graphs (e.g positioning x axis labels.) It would be nice to have
something like the user interface of Igor or Origin. I have already
used JGR and
You can always create the axis yourself:
x <- seq(-60,80)
plot(x, seq_along(x), xaxt='n')
axis(1, at=pretty(x,10))
On Mon, Aug 4, 2008 at 12:47 PM, Qian R <[EMAIL PROTECTED]> wrote:
>
>
>
>
>
>
>
>
> I am try to plot a data frame, and encounter some difficulties.
>
> my x axis range is from -60
No. Can't be. Editable graphs require that the graph be produced via code
that produces changeable components. All R graphs are essentially static.
That said, caveats: graphs drawn via the grid package functionality -- for
example lattice graphs -- **are** produced via changeable code. If you read
My goal is to prepare a thematic map of the US, with states shaded
according to their values for a variable of interest. I would like to
include an inset for Alaska in the upper left and an inset for Hawaii in
the lower left. If possible, I'd like to use Albers conic projection, or
something s
On 04/08/2008 12:50 PM, Arthur Roberts wrote:
Hi, all,
I would like to know if there is any gui interface out there
(academic or commercial) that allows one to edit R-language generated
graphs (e.g positioning x axis labels.) It would be nice to have
something like the user interface of
A colleague of mine, quite by accident, discovered that Adobe Illustrator
can manipulate plots made by base graphics, and when you do, many pieces of
the plot are separate items that can be manipulated with Illustrator. He
cuts and pastes from a Quartz window on his Mac, into Illustrator.
Apparent
Bert Gunter wrote:
No. Can't be. Editable graphs require that the graph be produced via code
that produces changeable components. All R graphs are essentially static.
Well, there's the xfig() device whose output can be edited with xfig
This was originally written (by me, for S-PLUS, back
Microsoft Word's graphics editor can edit R graphics saved in
metafile format, wmf. That includes x axis labels, etc.
On Mon, Aug 4, 2008 at 12:50 PM, Arthur Roberts <[EMAIL PROTECTED]> wrote:
> Hi, all,
>
>I would like to know if there is any gui interface out there
> (academic or commer
On 04/08/2008 1:14 PM, John P. Burkett wrote:
My goal is to prepare a thematic map of the US, with states shaded
according to their values for a variable of interest. I would like to
include an inset for Alaska in the upper left and an inset for Hawaii in
the lower left. If possible, I'd like
Peter Dalgaard wrote:
Bert Gunter wrote:
No. Can't be. Editable graphs require that the graph be produced via
code
that produces changeable components. All R graphs are essentially
static.
Well, there's the xfig() device whose output can be edited with
xfig This was originally written (
R 2.6
Windows XP
I have a 100x4 matirx
data<-matrix(nrow=100,ncol=4)
I would like to sort the entire matrix by column two, i.e. data[,2]
I looked at the help page for sort() but can not determine how I can use it to
sort a matrix on one of the matrix's columns.
Thanks,
John
John David Sorkin
On Mon, 4 Aug 2008, John Sorkin wrote:
R 2.6
Windows XP
I have a 100x4 matirx
data<-matrix(nrow=100,ncol=4)
I would like to sort the entire matrix by column two, i.e. data[,2]
I looked at the help page for sort() but can not determine how I can use it to
sort a matrix on one of the matrix's
probably you need order(), e.g.,
data[order(data[, 2]), ]
I hope it helps.
Best,
Dimitris
John Sorkin wrote:
R 2.6
Windows XP
I have a 100x4 matirx
data<-matrix(nrow=100,ncol=4)
I would like to sort the entire matrix by column two, i.e. data[,2]
I looked at the help page for sort() but c
Dear R Users,
Can anyone point me to a package for R vrsion 2.7.1 which implements some
Hurst exponent estimation methods ?
Thanks in advance,
Tolga
Generally, this communication is for informational purposes only
and it is not intended as an offer or solicitation for the purchase
or sale of an
Well, just goes to show you how much I know! Glad you were able to get some
help.
-- Bert
-Original Message-
From: Arthur Roberts [mailto:[EMAIL PROTECTED]
Sent: Monday, August 04, 2008 12:05 PM
To: Bert Gunter
Subject: Re: [R] Are there any guis out there,which will allow editing of
t
On Mon, Aug 4, 2008 at 3:36 AM, Gabor Grothendieck
<[EMAIL PROTECTED]> wrote:
> Checkout this one:
>
> http://finzi.psych.upenn.edu/R/Rhelp02a/archive/82452.html
And there's a wrapper for this in the latticeExtra package:
library(latticeExtra)
useOuterStrips(xyplot(data=df, copies~week|Treatment+
Hi, all,
All your comments have been very useful. I was wondering if there was
a package that can make pretty R tables to pdf. I guess I could use
xtable, but I would like something a little more elegant. Your input
is greatly appreciated.
Best wishes,
Art
On Mon, Aug 4, 2008 at 8:44 AM, Mark Difford <[EMAIL PROTECTED]> wrote:
>
> Hi Michael,
>
>>> Pulling my hair out here trying to get something very simple to work. ...
>
> I can't quite see what you are trying to do [and I am not sure that you
> clearly state it], but you could make things easier a
Dear list,
After searching many old posts, I can't find the solution to a simple problem.
can someone tell me how to create a character string with multiple backslashes,
as in:
file_dir <- c("C:\files\data\")
I need to create this string and then paste it to many files names for batch
Dear R Users,
I am using code from the following links to do Hurst exponent estimation.
link:
http://r-forge.r-project.org/plugins/scmsvn/viewcvs.php?rev=1&root=rmetrics&view=rev
file: LongRangeDependence.R
Take a look at the following run:
> x<-(cos((1:200)/10))
> rsFit(diff(x,15))@hurst$H
[1]
Try:
file_dir <- "C:\\files\\data\\"
On Mon, Aug 4, 2008 at 5:02 PM, zack holden <[EMAIL PROTECTED]> wrote:
>
> Dear list,
>
> After searching many old posts, I can't find the solution to a simple problem.
>
> can someone tell me how to create a character string with multiple
> backslashes, as i
Hi Arthur,
>> I was wondering if there was a package that can make pretty R tables to
>> pdf.
You got through TeX/LateX, but PDF could be your terminus. Package Hmisc:
? summary.formula
and its various arguments and options. You can't get much better.
http://cran.za.r-project.org/doc/contrib/
Thanks Gary.
That package is a bit weird. When one installs and loads it up, you don't
actually get any of those functions. One has to go to the following link:
http://r-forge.r-project.org/plugins/scmsvn/viewcvs.php?rev=1&root=rmetrics&view=rev
and then download and source the file LongRangeD
Hi Arthur,
Sorry, sent you down the wrong track: this will help you to get there:
http://biostat.mc.vanderbilt.edu/twiki/pub/Main/StatReport/summary.pdf
Regards, Mark.
Arthur Roberts wrote:
>
> Hi, all,
>
> All your comments have been very useful. I was wondering if there was
> a package
You have to escape every backslash by a second backslash - try this:
f <- "C:\\files\\data\\" # create character string
write(f, "") # write to file
system("open -t ") # see what happen to the character string
> file_dir <- c("C:\files\data\")
__
On 8/1/08 1:13 PM, "Richard M. Heiberger" <[EMAIL PROTECTED]> wrote:
> I meant 5a 5b 5c. Multiple-line commands are handled correctly.
> What is is doing is looking for "> " and " +" prompts. Anything else
> is removed.
When I said "5c) prune any lines that don't have assignment operators" I
Given four known and fixed vectors, x1,x2,x3,x4, I am trying to
generate a fifth vector,z, with specified known and fixed partial
correlations.
How can I do this?
In the past I have used the following (thanks to Greg Snow) to
generate a fifth vector based on zero order correlations---howe
Hi,
I guess my question is really more about the nested for loop construct and
whether it is doing what I intend it to do in my code in the previous post.
I would be grateful if anyone who has used nested loops could let me know if
I am doing something wrong.
Thanks,
rcoder
rcoder wrote:
>
Dear R Experts,
I am wondering if anyone has a solution to the following:
I am creating some cross-tabulation tables and want to input the results
from these tables into a new set of functions. Is there a way to turn
the cross-tab table into a new dataset? I haven't yet been able to
accomplis
I have another questions. How can I type specific names into strips of the
resulting plot?
For instance, in the resulting figure from the attached code, instead of
'umbrella(d)', I want have 'UMBRELLA' in the strip.
library(lattice)
flat <- function(d) 0 * d
linear <- function(d) -(1.65/8)
Given four known and fixed vectors, x1,x2,x3,x4, I am trying to
generate a fifth vector,z, with specified known and fixed partial
correlations.
How can I do this?
In the past I have used the following (thanks to Greg Snow) to
generate a fifth vector based on zero order correlations---howe
>
There is the 'fdim' package that computes the fractal dimension D.
Between D and the Hurst exponent H there should be a relation
D = 2 - H
I wonder if this is true when computing D and H with different
approaches
Regards,
Hans Werner Borchers
ABB Corporate Research
jpmorgan.com> write
Hi all,
I'd like to fit a multivariate regression with the variance of the error term
porportional to the predictors, like the WLS in the univariate case.
y_1~x_1+x_2
y_2~x_1+x_2
var(y_1)=x_1*sigma_1^2
var(y_2)=x_2*sigma_2^2
cov(y_1,y_2)=sqrt(x_1*x_2)*sigma_12^2
How can I specify thi
On 5/08/2008, at 1:31 AM, Frank E Harrell Jr wrote:
I wonder why we don't just use the exact nonparametric confidence
interval for the median, which is just as easy to compute. Also,
it will be asymmetric if the data are skewed, as it should be.
The boxplot.stats() fun
Thanks.
Sincerely,
Yanwei Zhang
Department of Actuarial Research and Modeling
Munich Re America
Tel: 609-275-2176
Email: [EMAIL PROTECTED]
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: Monday, August 04, 2008 5:15 PM
To: Zhang Yanwei - Princeton-MRAm
Subject
Hi all,
I'd like to fit a multivariate regression with the variance of the error term
porportional to the predictors, like the WLS in the univariate case.
y_1~x_1+x_2
y_2~x_1+x_2
var(y_1)=x_1*sigma_1^2
var(y_2)=x_2*sigma_2^2
cov(y_1,y_2)=sqrt(x_1*x_2)*sigma_12^2
How can I specify thi
Duncan Murdoch stats.uwo.ca> writes:
>
> On 04/08/2008 12:50 PM, Arthur Roberts wrote:
> > Hi, all,
> >
> > I would like to know if there is any gui interface out there
> > (academic or commercial) that allows one to edit R-language generated
> > graphs (e.g positioning x axis labels.)
Hi, all,
I have been trying to use adimpro. I have installed ImageMagick via
fink and also through MacPorts. I have tried them installed
separately and together, but R can't seem to recognize that they are
there. Perhaps, a Mac user out there that also uses R would be able
to help.
Mu
Hi All,
I have a X,Y and Z text file. I wish modify the X value in a logarithmic
value and to plot all log(Z) with Histogram. I try different code but I
have a problem to find the solution.
Thank you very much
Ale
[[alternative HTML version deleted]]
___
On Mon, 4 Aug 2008, Alessandro wrote:
Hi All,
I have a X,Y and Z text file. I wish modify the X value in a logarithmic
value and to plot all log(Z) with Histogram. I try different code but I
have a problem to find the solution.
My package HyperbolicDist has a log histogram function.
Davi
In the context of calling R from another program (namely gretl,
http://gretl.sourceforge.net ) I'm trying to understand the
interactions of the R init file (corresponding to the environment
variable RPROFILE) and the source() function.
I'll illustrate my problem with the following simplified co
Hello,
I am trying to run a constrained optimization in R. "constrOptim" is really
useful and has helped me a lot, but unfortunately, it doesn't provide the
hessian. Is there a solution to this problem?
I've tried "optim" with penalty-functions and "L-BFGS-B", but it doesn't help.
Alan.
--
Hi david,
I tried HyperbolicDist but I have this problem (I wish transform Z value)
Library (HyperbolicDist)
Data (testground) # X,Y and Z value in txt file
changetestground <- testground[-length(testground)]/testground[-1]
hist(change)
Error in hist.default(change) : 'x' deve essere numeric
logH
On Mon, Aug 4, 2008 at 1:39 PM, John Smith <[EMAIL PROTECTED]> wrote:
> I have another questions. How can I type specific names into strips of the
> resulting plot?
>
> For instance, in the resulting figure from the attached code, instead of
> 'umbrella(d)', I want have 'UMBRELLA' in the strip.
>
>
Exactly what problem are you having? There is nothing wrong with
nested for loops, so what is leading you to believe you have a
problem? I ran your program and it seems to terminate. Most of the
time seems to have been spent in the following statement:
if(!all(is.na(sel_col)))
Hi all,
Given a string list,
> paste("A",1:5,sep="")
[1] "A1" "A2" "A3" "A4" "A5"
I would like to create an empty data frame using that list as the
header, so I can access my data frame column using,
> df [ list [ i ] ]
Anyone ?
Thanks,
Etienne
[[alternative HTML version
Dear Thierry,
Your guess was correct. I was not aware that I needed print( ). It was not
needed with plot( ). I have yet to try xyplot( ) but I suspect that the
print( ) might be needed here as well. Thank you very much for your help.
Cheers,
Sorn
"ONKELINX, Thierry" <[EMAIL PROTECTED]>
0
Hi Etienne
It is not so elegant, bu I think that work.
colname.list<-paste("A",1:5,sep="")
df<-data.frame(matrix(matrix(rep(1,length(colname.list)),1),1))
df
colnames(df)<-colname.list
df
df<-df[-1,]
df
Cheers,
Miltinho Astronauta
Brazil
On 8/4/08, Etienne Bellemare Racine <[EMAIL PROTECTE
Actually now that I read it closer, I see what your problem is. what
did you think the statement:
Preg[,k]<-coef(lm(tt~sel_col))
was going to do? Preg is a 200x100 matrix and you are only storing
two values (the coefficients) so they will be repeated 100 times in
the column. So there is nothin
on 08/04/2008 07:40 PM Etienne Bellemare Racine wrote:
Hi all,
Given a string list,
> paste("A",1:5,sep="")
[1] "A1" "A2" "A3" "A4" "A5"
I would like to create an empty data frame using that list as the
header, so I can access my data frame column using,
> df [ list [ i ] ]
Anyon
Even if it's not so elegant, I will bind it in a function, so I don't have to
bother with that anymore. I think there should be a simple function in R to
initialize an empty data frame. From what I've read, it is a recurrent
question on that list.
#Create an empty data frame from a header list
em
See dataFrame() in R.utils. It was design for the purpose of
allocating empty data frames in an efficient way.
Example:
> library(R.utils);
> df <- dataFrame(colClasses=c(a="integer", b="double"), nrow=10);
> str(df)
'data.frame': 10 obs. of 2 variables:
$ a: int 0 0 0 0 0 0 0 0 0 0
$ b: n
Dear All,
I have a dataset (X,Y and Z) with no-normalize distribution (I saw with
qqnorm) but with Gaussian shape. I am a not good statistic and I wish to
know a method to normalize Z values.
Thanks for help. It's very useful for my PhD
ALe
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Dear R users,
I´m trying to optimize simultaneously two binomials inequalities (used in
acceptance sampling) which are nonlinear solution, so there is no simple
direct solution. Please, let me explain shortly the the problem and the
question as following.
The objective is to obtain the smallest v
Hi, all,
Does anyone know of a way to get a title on an array of plots? I have
tried to use title, mtext, and text to get the title on the array, but
to no avail. I would like something like the following.
title("Big Array")
par(mfrow=c(2,4))
plot(x1,y1)
plot(x2,y2)
..
..
..
Much apprec
This relates to the order of things being done. See ?Startup.
Note that the default packages are not loaded until after the Rprofile
file has been run, and postscript() is in package grDevices, not in
graphics. Using grDevices::postscript will always work.
It looks to me that you are trying
On Mon, 4 Aug 2008, Etienne B. Racine wrote:
Even if it's not so elegant, I will bind it in a function, so I don't have to
bother with that anymore. I think there should be a simple function in R to
initialize an empty data frame. From what I've read, it is a recurrent
question on that list.
par(mfrow=c(2,4), oma=c(0,0,1,0))
plots ...
title("Big Array", outer=TRUE)
You need to set up an outer margin. See 'An Introduction to R' for that
topic.
You probably want to reset the margins (mar=) for such a large array of
plots.
On Mon, 4 Aug 2008, Arthur Roberts wrote:
Hi, all,
Doe
Hi there,
I'm plotting some glass RI values just by plotting
plot(x)
then I put on my lowess smoother
lines(lowess(x))
now I want to put on some 95% Confidence Interval bands of the lowess
smoother, but don't know how??
Thanks
--
Gareth Campbell
PhD Candidate
The University of Auckland
P
gmx.net> writes:
>
> Hello,
>
> I am trying to run a constrained optimization in R. "constrOptim" is really
> useful and has helped me a lot, but unfortunately, it doesn't provide the
> hessian. Is there a solution to this problem?
You didn't provide an example to understand why 'optim' do
On Tue, 5 Aug 2008, Gareth Campbell wrote:
Hi there,
I'm plotting some glass RI values just by plotting
plot(x)
then I put on my lowess smoother
lines(lowess(x))
now I want to put on some 95% Confidence Interval bands of the lowess
smoother, but don't know how??
You can have pointwise con
Does anybody know if there is a way to include exogenous X-variables in an
ARFIMA model?
It appears the ARFIMA function in fracdiff does not support this.
Supposing there is nothing already written, and I wanted to modify the existing
arfima function, how would I go about this?
Would I need to
Gabor Grothendieck gmail.com> writes:
>
> Microsoft Word's graphics editor can edit R graphics saved in
> metafile format, wmf. That includes x axis labels, etc.
>
In theory, yes, in practice: not really. Vertical texts are horizontal after
editing, and clipped points may magically re-appear.
Hello there. I uses the following codes for the purpose of variable
selection.
> lmModel <- lm(y~.,data.frame(y=y, x=x))
> step <- stepAIC(lmModel, direction="both")
> step$anova
Stepwise Model Path
Analysis of Deviance Table
Initial Model:
y ~ x.Market.Price + x.Qua
Hi there,
I would like to design a graphical user interface to provide the functionality
of my R-scripts to other users (not knowing anything about R...)
Can anybody give me some hints which possibilities I have (advantages /
disadvantages)?
What do you use to build applications?
Ciao,
Antje
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