Hi:
It would really help to have a reproducible example to see exactly what
problems you're having, but here's a simple manufactured example to
illustrate how to produce a basic legend. The plot below is one with three
different 'y' variables against the same x. The x and y limits are made wide
On 2010-07-21 09:37, Tonja Krueger wrote:
Dear List
I’m using the ”lmomRFA” package to fit different distributions to my data
sample. To calculate the error bounds I used:
regsimq(…)
and
sitequantbounds(…)
So my questions are:
Are error bounds and confidence intervals the same
Hello,
HOw do I find limit of a function in R?
thanks,
Samit
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Is there a non animation version of newton.method
(http://bm2.genes.nig.ac.jp/RGM2/R_current/library/animation/man/newton.method.html)
for finding roots of a function? It should find the roots without showing it
on the GUI.
thanks,
Sam
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Hi,
Are there any other forums on R where I can get quick responses to my
questions? I've lot of questions as I am learning the basics and exploring
advanced computational functions and it is tough to wait for moderator
approval then a reply to each of those questions.
thanks!
Samit
--
View
I have heterogeneous variance in a two-way mixed effects ANOVA, with more than
2 groups in each factor. Is there something like oneway.test or a
Brown-Forsythe test that will let me test for differences in means?
Thanks!
-Keith
__
Lorenzo Isella lorenzo.isella at gmail.com writes:
Dear All,
I am a novice when it comes to time-series analysis and at the moment I
am actually interested in calculating the Hurst exponent of a time
series.
Some time ago I tested some of the classical chaotic time series
(such as the
Hi
r-help-boun...@r-project.org napsal dne 22.07.2010 05:58:54:
Hi,
Are there any other forums on R where I can get quick responses to my
questions? I've lot of questions as I am learning the basics and
exploring
advanced computational functions and it is tough to wait for moderator
Hi
maybe less number of lines?
regards
Petr
r-help-boun...@r-project.org napsal dne 22.07.2010 06:06:53:
Hi all,
I am have some difficulty with the legend function.
I need to add a legend to describe the different line types in a plot.
The
legend box is small.
It did not include
Hi
r-help-boun...@r-project.org napsal dne 21.07.2010 17:46:02:
Hi R-community,
I have the code as follows,i Fitted model as follows
lbeer-log(beer_monthly)
t-seq(1956,1995.2,length=length(beer_monthly)) #beer_monthly contains
400+
entries
not needed
t2=t^2
hi,
I get an Error when I try to run function help in R,
it says
starting httpd help server ...Error: cannot listen to TCP port 15259
I would be so glad if you help me
[[alternative HTML version deleted]]
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On Wed, 2010-07-21 at 20:58 -0700, saminny wrote:
Hi,
Are there any other forums on R where I can get quick responses to my
questions? I've lot of questions as I am learning the basics and exploring
advanced computational functions and it is tough to wait for moderator
approval then a reply
On 2010-07-21 22:06, li li wrote:
Hi all,
I am have some difficulty with the legend function.
I need to add a legend to describe the different line types in a plot. The
legend box is small.
It did not include sufficient length of each line type to help distinguish
the differnt line types.
The short-term work-around is probably to use the help_type= argument to
help, e.g.
help(help, help_type=text)
Hope this helps a little.
Allan
On 22/07/10 07:30, Na.Ebrahimi wrote:
hi,
I get an Error when I try to run function help in R,
it says
starting httpd help server ...Error: cannot
Hi,
I am making a mock user interface in tcltk and I would like to add a 'check
menu button' such as shown here:
http://zetcode.com/tutorials/pygtktutorial/images/checkmenuitem.png
Does anybody know how to do this? I am quite new to R.
Cheers!!
Albert-Jan
Dear R-user,
a few weeks ago I consulted the list-serve with a similar question.
However, my task changed a little but sufficiently to get lost again. So
I would appreciate any help on the following issue.
I use the plm package and work with firm-level data in a panel. I would
like to eliminate
Dear David and the list,
Thanks for your comment.
Unfortunately, I don't have an instructor on R.
And I usually don't deal with contingency table analysis.
I have consulted more than two books on R already, but I found no good answer.
Could you point me to some references about log linear
Hi all,
My question is really more statistical theory than about R, however I would
greatly appreciate if someone can provide guidance.
I am trying to measure distance between points given j number of variables,
a subset of which is latitude and longitude. To make the distance
scale-invariant, I
Hello,
### I created two classes A and B. A is the superclass of B.
setClass(A, representation(s1=numeric),prototype=prototype(s1=8))
setClass(B,contains=A,representation(s2=character),prototype=list(s2=hi))
myA=new(A)
myB=new(B)
I created functions for A and B
f1=function(x,...)
hi,
When I try to use the function coordProj {mclust}
coordProj(diabetes[,-1],dimens=c(2,3),what=uncertainy,uncertainty=diabetesModel$uncertainty,parameters=diabetesModel$parameters)
to identify uncertainty, it errors and send this warning message:
Warning message:
In
Sorry, I'm getting error messages on your email?
Anyhows... Spellcheck. You're using uncertainy and uncertainTy (the
latter being correct but the former being what you've programmed).
Good luck!
Siri.
Siterer Na.Ebrahimi na.ebrah...@mail.sbu.ac.ir:
hi,
When I try to use the function
Please help me with this i need to submit my thesis .
Thanks In advance
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Dear R Users,
If we issue simple plot command in R we don't get legend of the plot
automatically.
For example, following lines plots two curves, but to write a legend of
these two curves there is no simple command. I checked with ?legend but
it seems bit complicated for me. Does anyone know how
Am 22.07.2010 11:18, schrieb Christian Schoder:
I use the plm package and work with firm-level data in a panel. I would
like to eliminate all firms that do not fulfill the requirement of
having an observation in every variable used for at least x consecutive
years.
There are probably more
One option:
/## Your data:/
data - structure(list(id = structure(c(1L, 2L, 2L, 3L, 3L, 3L, 3L, 4L,
4L, 4L, 4L, 4L), .Label = c(a, b, c, d), class = factor),
year = c(2000L, 2000L, 2001L, 1999L, 2000L, 2001L, 2002L,
1998L, 1999L, 2000L, 2001L, 2002L), y = c(1L, NA, 3L, 1L,
2L, 4L,
Hi,
There is no way to obtain an automatical legend with plot. The choice
you have is to draw the legend by yourself or to use another graphical
package but it is not necessary easier...
For example, the code for the package ggplot2:
donnees - data.fram(x=x,y=y,y1=y1)
melt(donnees,id=x) -
Also:
1. Have you used help.start()? This opens a display in a
browser with links to the available manuals.
2. There are multiple mailing lists and multiple search engines
listed in www.r-project.org.
3. Have you subscribed? I had the impression that moderator
Have you tried something like the following:
library(sos)
H - ???Hurst
summary(H)
H
This identified 50 links in 15 packages, and displayed the
results in a table in a web browser with links to the best match in the
best package first. The installPackages and writeFindFn2xls
On Thu, Jul 22, 2010 at 4:07 AM, Gavin Simpson gavin.simp...@ucl.ac.uk wrote:
On Wed, 2010-07-21 at 20:58 -0700, saminny wrote:
Hi,
Are there any other forums on R where I can get quick responses to my
questions? I've lot of questions as I am learning the basics and exploring
advanced
try this:
Dat - read.table(textConnection(
id year y z
1 a 2000 1 1
2 b 2000 NA 2
3 b 2001 3 3
4 c 1999 1 1
5 c 2000 2 2
6 c 2001 4 NA
7 c 2002 5 4
8 d 1998 6 5
9 d 1999 5 NA
10 d 2000 6 6
11 d 2001 7 7
12 d 2002 3 6
), header = TRUE)
On 07/22/2010 02:06 PM, li li wrote:
Hi all,
I am have some difficulty with the legend function.
I need to add a legend to describe the different line types in a plot. The
legend box is small.
It did not include sufficient length of each line type to help distinguish
the differnt line
On Thu, Jul 22, 2010 at 5:18 AM, Christian Schoder
schoc...@newschool.edu wrote:
Dear R-user,
a few weeks ago I consulted the list-serve with a similar question.
However, my task changed a little but sufficiently to get lost again. So
I would appreciate any help on the following issue.
I
On 07/22/2010 09:39 PM, Jim Lemon wrote:
On 07/22/2010 02:06 PM, li li wrote:
Hi all,
I am have some difficulty with the legend function.
I need to add a legend to describe the different line types in a plot.
The
legend box is small.
It did not include sufficient length of each line type to
There is no simpler way. It's not automatic but it is fairly simple at the most
basic. It can get a bit complicated later.
ltxt - c(cat,dog)
x - 1:5
y - 1:5
plot (x,y, type='n', ann=FALSE)
lines(x,y,col=1,lty=solid)
points(x,y,pch=16)
legend(1,4,ltxt, pch=16)
--- On Thu, 7/22/10, Yogesh
Dear R users,
I want to aggregate data in the following way:
###
X - data.frame(u = c(T1,T1,T1,T2), v=c(a,a,b,a))
X
library(sqldf)
sqlOut - sqldf(select count(distinct(v)) from X group by u)
sqlOut
###
Now I want to get the same result without using SQL. How can I achieve
that ?
Thanks for
HI list,
I want to know whether tsdiag uses k-(p+q) as the lag in ljung box
test. How is it possible to save those values
nuncio
--
Nuncio.M
Research Scientist
National Center for Antarctic and Ocean research
Head land Sada
Vasco da Gamma
Goa-403804
[[alternative HTML version
There are so many ways Here is one:
aggregate(v ~ u, data=X, function(...) length(unique(...)))
#u v
# 1 T1 2
# 2 T2 1
Hope this helps
Allan.
On 22/07/10 12:52, Gildas Mazo wrote:
Dear R users,
I want to aggregate data in the following way:
###
X- data.frame(u = c(T1,T1,T1,T2),
Got it already. I should have rtfm'd. Sorry to bother you.
Here's a modified example from a very neat webpage I found:
require(tcltk)
tt - tktoplevel()
topMenu - tkmenu(tt)
tkconfigure(tt,menu=topMenu)
fileMenu - tkmenu(topMenu,tearoff=FALSE)
openRecentMenu - tkmenu(topMenu,tearoff=FALSE)
Dear all,
I use Sweave to create my reports. Unfortunately my script crashes whenever I
my R code contains special characters like umlauts.
Is there a way to to escape special characters in Sweave... This is the line
that crashes Sweave:
gl_bybranch = ddply(new_wans,.(period,Branchen),
Among those users of Primer, stress values greater than 0.3 are
interpreted as questionable. Using both isoMDS and metaMDS (vegan
package), the stress values returned are much higher using my own data
and using examples provided in R Help. For example Rstress = 8.3, and
the stressplot r2 = 0.99
On 22/07/2010 8:19 AM, Bunny, lautloscrew.com wrote:
Dear all,
I use Sweave to create my reports. Unfortunately my script crashes whenever I my R code contains special characters like umlauts.
Is there a way to to escape special characters in Sweave... This is the line that crashes Sweave:
A standalone example is always helpful. The following works for me, so
I am probably not understanding your problem:
---[BEGIN: umlaut.Rnw]---
\documentclass[a4paper]{article}
\usepackage{ucs}
\usepackage[utf8x]{inputenc}
\begin{document}
test,echo=TRUE=
x - data.frame(Geschäftslage=1:10)
If you want to *see* the contributions of the cells to the association
between lang and cons, try
library(vcd)
mosaic(~lang+cons, data=langcons.table, shade=TRUE)
Tsunhin John Wong wrote:
Dear R users,
I have a question of how to do some specific cell to cell comparisons
on a R x C
In http://tolstoy.newcastle.edu.au/R/e7/help/09/06/1900.html
Jim Holtman wrote:
Here is the way that I do it instead of creating a package of my
functions. I 'source' the file into an environment and then attach the
environment to keep the global from being clustered:
# read my functions
On Wed, Jul 21, 2010 at 11:48 PM, saminny jain.sa...@gmail.com wrote:
Hello,
HOw do I find limit of a function in R?
thanks,
There are examples in the sympy help file in the rSymPy package and in
the Ryacas vignette in the Ryacas package.
They have home pages at:
On 22/07/2010 8:58 AM, Michael Friendly wrote:
In http://tolstoy.newcastle.edu.au/R/e7/help/09/06/1900.html
Jim Holtman wrote:
Here is the way that I do it instead of creating a package of my
functions. I 'source' the file into an environment and then attach the
environment to keep the global
Dear List,
I have posted my questions to the r-devel list as well, but I think this
list is read by more users, and maybe someone from this list may point
me in the right direction. Here is the corresponding thread:
https://stat.ethz.ch/pipermail/r-devel/2010-July/058003.html
I am sorry for
Sorry all,
for not posting a minimal example. I am running R on Mac OS X snow leopard with
Komodo edit / Sciviews-R. The problem is that the code does not work any more
if there is an umlaut in my R code. The error message is some strange mixture
of german and english, so this what the error
On 22/07/2010 9:39 AM, Bunny, lautloscrew.com wrote:
Sorry all,
for not posting a minimal example. I am running R on Mac OS X snow leopard with
Komodo edit / Sciviews-R. The problem is that the code does not work any more
if there is an umlaut in my R code. The error message is some strange
On 22/07/2010 9:31 AM, Thomas Etheber wrote:
Dear List,
I have posted my questions to the r-devel list as well, but I think this
list is read by more users, and maybe someone from this list may point
me in the right direction. Here is the corresponding thread:
Hi guys,
I hope you can help me with this.
Here is the problem:
I have some data (myData) that looks similar to this:
[,1] [,2] [,3] [,4]
[1,] A A B B
[2,] B B B B
[3,] C C C C
When I build a contingency table for the first and second row using:
tb - table(myData[1, ],
Have you tried the following:
library(sos)
(lj - findFn('ljung box'))
This just identified for me 32 matches in 13 packages, at least
one of which should do what you want. In particular, I wrote the
AutocorTest function in FinTS to deal with issues that sound similar
to what
On Jul 22, 2010, at 9:39 AM, Bunny, lautloscrew.com wrote:
Sorry all,
for not posting a minimal example. I am running R on Mac OS X snow
leopard with Komodo edit / Sciviews-R. The problem is that the code
does not work any more if there is an umlaut in my R code. The error
message is
Am 22.07.2010 15:53, schrieb Duncan Murdoch:
On 22/07/2010 9:31 AM, Thomas Etheber wrote:
Dear List,
I have posted my questions to the r-devel list as well, but I think
this list is read by more users, and maybe someone from this list may
point me in the right direction. Here is the
* Please cc me if you reply as I am a digest subscriber *
Hi,
I am wondering how I can run a multilevel survival model in R? Below is
some of my data.
head(bi0.test)
childid famid lifedxm sex age delta
1 22.0222 CONTROL MALES 21.36893 0
2 13.0213 MAJOR MALES
This message is for those familiar with the survey package. I need to fit a
weighted Cox model to accommodate the sampling weights as I have a
case-control study with controls sampled at random from a database in a
ratio 2:1 to cases (whom were all sampled). I want to make sure I am using
the
Hello,
I am trying to create a plot often seen in hydrodynamic work than includes a
contour plot representing the water speed with arrows pointing in the
direction of flow. Does anyone have any idea how I might add arrows based on
wf$angle (in the example below) to the plot below?
Thanks in
On Jul 22, 2010, at 4:00 PM, Liat wrote:
Hi guys,
I hope you can help me with this.
Here is the problem:
I have some data (myData) that looks similar to this:
[,1] [,2] [,3] [,4]
[1,] A A B B
[2,] B B B B
[3,] C C C C
When I build a contingency table for the first
Hi,
I am trying to write some code that tracks changes that may have been made
after a dataframe has been edited using fix(). But if I edit only the first
cell (row 1, col 1) of the dataset below, it is like many records were edited.
What is the explanation for this error?
Hi, try ?levels
myData - matrix(sample(c(LETTERS[1:10],NA),100,replace=T),nrow=25)
table(factor(as.vector(myData),levels=LETTERS[1:26]),useNA=ifany)
-
A R learner.
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Thank you!!!
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R-help@r-project.org mailing list
Thanks Peter!
that worked, and was so easy. LOL.
I tried playing with factor and levels before I wrote here, but obviously I
didn't do it right.
Thanks again,
Liat.
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Dear list,
I'm using the Sweave function in order to get some report.
Here one chunk:
echo=F,results=tex=
report-lapply(repor, function(x) {
(print(xtable(data.frame(x[1:2,]), align=|l|rrr|),floating=FALSE,tabular.
environment=longtable,include.colnames=FALSE,size=\\small))
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Liat
Sent: Thursday, July 22, 2010 8:41 AM
To: r-help@r-project.org
Subject: Re: [R] how to force a table to be square?
Thanks Peter!
that worked, and was so easy. LOL.
I
On 7/22/2010 5:01 AM, Allan Engelhardt wrote:
There are so many ways Here is one:
aggregate(v ~ u, data=X, function(...) length(unique(...)))
# u v
# 1 T1 2
# 2 T2 1
Hope this helps
Here is one other way, using the plyr package (which is very good for
taking a data structure
I'd like to train a decision tree on a set of weighted data points. I looked
into the rpart package, which builds trees but doesn't seem to offer the
capability of weighting inputs. (There is a weights parameter, but it seems to
correspond to output classes rather than to input points).
I'm
Hi William,
I'm curious about that you used d[] - lapply(d, factor...
Could you please tell me if there are any differences between d[] and d?
Thank you.
-
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-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Wu Gong
Sent: Thursday, July 22, 2010 9:44 AM
To: r-help@r-project.org
Subject: Re: [R] how to force a table to be square?
Hi William,
I'm curious about that you used d[]
On 2010-07-22 10:43, Wu Gong wrote:
Hi William,
I'm curious about that you used d[]- lapply(d, factor...
Could you please tell me if there are any differences between d[] and d?
Thank you.
Why not try it both ways and inspect the result.
d[] -
d -
Obvious?
-Peter Ehlers
I have data as below.Please let me know how the ACF and Pacf used to
determine the order od arima model.
Is there any rules need to be followed to determine order.Please advise
turkey.price.ts
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
2001 1.58 1.75 1.63 1.45 1.56 2.07
[...]
compo[[ which.max(sapply(compo, vcount)) ]]
That's better command for obtaining Largest Component
Sometimes, I get this error while reading pajek file. Does anybody know what
this error message means ?
g - read.graph (F://data/pfu_raw.net, pajek)
Error in read.graph.pajek(file, ...) :
I have been trying to get the lme4 package installed on Mac OS X... with no
success. The Mac OS binary is not available on any CRAN, and I also can¹t
install the package from old source. Has anyone found a solution to this
problem? I am happy to use nlme for now, but I tend to prefer to do my
Hi All,
I'm using R 2.11.1 on 64 bit windows XP. The little function I wrote below
I use often to import the first 1001 lines in an excel sheet to R. This
works fine on the 32 bit version of R but fails on the 64 bit [both on the
same machine, using the same function, importing the same .xls
On Jul 22, 2010, at 11:34 AM, Nicholas Griffin wrote:
I have been trying to get the lme4 package installed on Mac OS X...
with no
success. The Mac OS binary is not available on any CRAN, and I also
can’t
install the package from old source. Has anyone found a solution to
this
problem?
Dear list,
I'd like to do several t-test within a for loop. Example follows:
data1 - rnorm(1:25)
data2 - rnorm(1:25)
vars - c(data1, data2)
for (i in vars) {
t.test(i)
}
Unfortunately, it does not work because of the quotes in the vars-vector
(running t.test(data1) by hand
Dear all,
Is there anyway I can fix slope in some value and only estimate
intercept for a linear regression? I understand lm(y~1, data) will NOT
have a slope at all. This is not what I want. Thanks.
Jun
__
R-help@r-project.org mailing list
Hello, and sorry for not providing an example.
I run a regular linear regression (using lm) and use weights with it
(weights = ...).
I use QuantPsych package, its function lm.beta to extract
standardized regression weights from my lm regression object.
When I don't use weights, everything is
On Jul 22, 2010, at 1:28 PM, Arne Schulz wrote:
Dear list,
I'd like to do several t-test within a for loop. Example follows:
data1 - rnorm(1:25)
data2 - rnorm(1:25)
vars - c(data1, data2)
for (i in vars) {
t.test(i)
}
Unfortunately, it does not work because of the quotes in
Try this:
apply(Vectorize(get)(ls(patt = 'data')), 2, t.test)
On Thu, Jul 22, 2010 at 2:28 PM, Arne Schulz
arne.sch...@student.uni-kassel.de wrote:
Dear list,
I'd like to do several t-test within a for loop. Example follows:
data1 - rnorm(1:25)
data2 - rnorm(1:25)
vars - c(data1, data2)
saminny wrote:
Is there a non animation version of newton.method
(http://bm2.genes.nig.ac.jp/RGM2/R_current/library/animation/man/newton.method.html)
for finding roots of a function? It should find the roots without showing
it on the GUI.
Have a look at package nleqslv
/Berend
--
View
On 2010-07-22 11:28, Arne Schulz wrote:
Dear list,
I'd like to do several t-test within a for loop. Example follows:
data1- rnorm(1:25)
data2- rnorm(1:25)
vars- c(data1, data2)
for (i in vars) {
t.test(i)
}
Unfortunately, it does not work because of the quotes in the
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Jun Shen
Sent: Thursday, July 22, 2010 10:30 AM
To: R-help
Subject: [R] How to fix slope and estimate intercept
Dear all,
Is there anyway I can fix slope in some value and
Dear list,
I have a matrix, spc, that should row-wise be interpolated:
E.g.
spc - matrix (1:1e6, ncol = 250, nrow = 4000, byrow = TRUE)
spc [1:10, 1:10]
shifts - seq_len (nrow (spc))
wl - seq_len (ncol (spc))
interpolate - function (spc.row, shift, wl)
spline (wl + shift, spc.row, xout =
Hi,
I have a global data-frame in my R script.
At some point in my script, I want to update certain columns of this
data-frame by calling in an update function.
The function looks like this:
# get events data. This populates a global event data frame in the R-script
events - getEvents(con,
I'm parallelizing some computation on hierarchical data, and would
find it natural to do something like this (where a call to parLapply
is embedded in outer call to parLapply):
cl - makeCluster(rep.int('localhost', 5),
type='SOCK')
clusterExport(cl, 'cl')
parLapply(cl,
R Community,
I have a stupid little barplot I am trying to construct (Windows XP, R.11.1,
32-bit). Whenever I run it my bars run below the horizontal axis. Can anyone
(1) reproduce this and (2) offer a solution? Rather simplistic code follows
(I am new to the community) Thanks for your help!
Here is the function that makes the data.frames in the list:
funweek - function(df)
if (length(df$elapsed_time) 5) {
res = fitdist(df$elapsed_time,exp)
year = df$sale_year[1]
sample = df$sale_week[1]
mid = df$m_id[1]
estimate = res$estimate
sd = res$sd
samplesize =
Looks like your event id is unique. If that is so, why not just do
##Not checked
events - events[sort(events$event),]
dataF - dataF[sort(data$event),]
if(doUpdate == 1){
if(!is.null(dataF) nrow(dataF) 0){
events[events$eventid %in% dataF$event, c(timestamp,
isSynchronized,timediff)]
Here is a function I use to return the non-NULL elements of a list:
delete.NULLs - function (x.list)
{
x.list[unlist(lapply(x.list, length) != 0)]
}
On Thu, Jul 22, 2010 at 2:22 PM, Ted Byers r.ted.by...@gmail.com wrote:
Here is the function that makes the data.frames in the list:
Thanks, that seems to have cleared it up.
(I did cut and paste from the console - I didn't think most of what I tried
was very going to be very helpful and didn't want to drown people in
spurious info. Sorry for the mistakes)
--
View this message in context:
Hi everyone, I am plotting a boxplot with main title as main =
bquote(paste(.(ts.ind[s]), : , bar(zeta), Boxplot from 2001 to 2009, sep =
)) but it doesn't work. The program said they cannot find the function
bar. Does anyone know how to do it correctly? Thanks.
tin
Hi, Bill,
Thanks. This is what I am looking for.
Jun
On Thu, Jul 22, 2010 at 1:54 PM, William Dunlap wdun...@tibco.com wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Jun Shen
Sent: Thursday, July 22, 2010 10:30 AM
To:
On Jul 22, 2010, at 2:13 PM, Gregory Gilbert wrote:
R Community,
I have a stupid little barplot I am trying to construct (Windows XP,
R.11.1,
32-bit). Whenever I run it my bars run below the horizontal axis.
Can anyone
(1) reproduce this and (2) offer a solution?
I'm not sure I should
Dear All,
I met problems when doing contrast and now really need some help in the
model below:
Fit=gam(y~treat+SEQUENCE+PERIOD+SEX+s(x),data=dat,
random=list(SUBJID=~1),correlation=corAR1(form=~1|SUBJID))
And error message keeps coming out when I want to compare the differences
between
Spencer Graves spencer.graves at structuremonitoring.com writes:
Have you tried something like the following:
library(sos)
H - ???Hurst
summary(H)
H
This identified 50 links in 15 packages, and displayed the
results in a table in a web browser with links to the best match in
Dear all,
i think I am a step further with the issue. I am pretty sure now, that it is
not a Sweave problem. It's indeed an encoding problem with my Mac. If i source
the file without any sweave
source(myRcode.R) I already get an error message, despite the fact that the
code is correct. If I
On 2010-07-22 11:44, Marcus Liu wrote:
Hi everyone, I am plotting a boxplot with main title as main =
bquote(paste(.(ts.ind[s]), : , bar(zeta), Boxplot from 2001 to 2009, sep = )) but
it doesn't work. The program said they cannot find the function bar. Does anyone know how to do it
Hi everyone.
I'm trying to display the r^2 of a linear regression on a plot using
text(...). I first build the string to display. However, I can't find why
this is not working (it display literally r^2 instead of r superscript 2).
r2string = expression(paste(r^2), = , r2);
Any help would be
Hi All,
I'm trying to use the sample function within a loop where the
vector being sampled from (the first argument in the function) will
vary in length and composition. When the vector is down in size to
containing only one element, I run into the undesired behaviour
acknowledged in the
Try this:
x - 10
sample(x, 1, prob = c(rep(0, x - 1), 1))
On Thu, Jul 22, 2010 at 5:31 PM, Jon BR jonsle...@gmail.com wrote:
Hi All,
I'm trying to use the sample function within a loop where the
vector being sampled from (the first argument in the function) will
vary in length and
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