I'm trying to do some confirmatory factor analysis on some data. My SEM
model solves in 22 iterations, but when I try to look at the modification
indices, using mod.indices, I get the following error message:
Error in solve.default(hessian) :
system is computationally singular: reciprocal condit
It's not an array, but a list:
lapply(2:n, function(x) matrix(1:x^2,x))
On Sat, Feb 20, 2010 at 3:17 PM, song song wrote:
> maybe its not an array
>
> like
>
> m[1]=matrix(1:4,2,2)
> m[2]=matrix(1:9,3,3)
> .
> .
> .
> m[k]=matrix(1:k^2,k,k)
>
> likes
>
> for (k in 1:n){
> s=matrix(1:k^2,k,k)
> }
On Fri, 19 Feb 2010, Jinfeng Liu wrote:
OS gmail.com> writes:
Hi,
I have some code I run interactively through the R interpreter and it works
fine. I then run it as a script with Rscript and I get an error. The error
is coming when Rscript builds a model matrix.
Output from Rscript with the e
maybe its not an array
like
m[1]=matrix(1:4,2,2)
m[2]=matrix(1:9,3,3)
.
.
.
m[k]=matrix(1:k^2,k,k)
likes
for (k in 1:n){
s=matrix(1:k^2,k,k)
}
how to store s like s[1], s[2] ... so that I can use it in other place?
thanks!
[[alternative HTML version deleted]]
_
Hello All,
I am learning mixed effects logistic regression (lmer in lme4), and I am
having problems deciphering the goodness of fit function
plot.logistic.fit.fnc (languageR package). The only information I can find
is that this function plots observed proportions against mean predicted
probabiliti
On Fri, Feb 19, 2010 at 4:44 AM, Paul Hiemstra wrote:
> Jyotirmoy Bhattacharya wrote:
>>
>> I could not find any documentation of how dot-dot-dot works when used
>> as an argument in a function call (rather than as a formal argument in
>> a definition). I would appreciate some references to the ru
Try
f <- function(nbr){
y<-rnorm(nbr)
y1 <- mean(y)
plot(y)
invisible( y1)
}
That will return y1 invisibly, so
f(100)
plots but returns nothing visible but
w<-f(100)
plots and places the return value in w
>>> Dennis Murphy 02/19/10 9:33 PM >>>
Hi:
Perhaps you want this:
f <-
Dear All:
For a few days now I have struggled with recreating a stable LaTeX/R/Sweave
environment on a new Windows XP SP3 machine. Unfortunately it seems that I
ran out of good ideas and I would like your expert opinion on how I should
approach this task and be successful. In essence, I install
I have tried to install the "genetics" package and am getting and error.
The log is below. For information, I am using R 2.10.1 on the Windows
XP operating system. The mirror site I'm using is Michigan Technological
University.
Your help in this matter will be greatly appreciated.
Murray M Coope
Hi,
I draw a surface corresponding to bivariate density of independent
variables (rho=0) using persp(). Then I add a contour line (i.e.,
circle in my case) at a particular density. Below is a minimal example
of what I have so far.
# Bivariate density
dnorm2d <- function(x, y, rho = 0) {
xoy = (
Hello,
I am attempting to write a script that adds error bars to a barchart. I
basing my attempt heavily on the following thread:
http://tolstoy.newcastle.edu.au/R/e2/help/06/10/2791.html
I can't seem to get around the problem that was discussed in the thread. The
following example should illust
Glad to know it isn't just me! I couldn't use Phil's data.frame method since
my real problem went from a POSIXct object to a large matrix where I used rbind
and then back to POSIXct. Jim's function worked great on converting the final
product back to the proper date.
Thanks!
Tim
Tim Clark
I have used the following function to convert to POSIXct from a numeric
without any problems:
> unix2POSIXct <- function (time) structure(time, class = c("POSIXt",
"POSIXct"))
>
> unix2POSIXct(946764000)
[1] "2000-01-01 17:00:00 EST"
>
On Fri, Feb 19, 2010 at 4:07 PM, Tim Clark wrote:
> Dear
OS gmail.com> writes:
> Hi,
> I have some code I run interactively through the R interpreter and it works
> fine. I then run it as a script with Rscript and I get an error. The error
> is coming when Rscript builds a model matrix.
>
> Output from Rscript with the error:
> running
> '/usr/lib/R/
Hello, I'm trying to add lme results in a table with lm coef results, but as
I know, estout or xtabel cannot support lme objects.
I'm a new in R and I'll appreciate some helpful comments.
--
Daniel Marcelino
Phone: (647) 8910939
[[alternative HTML version deleted]]
ggplot2
ggplot2 is a plotting system for R, based on the grammar of graphics,
which tries to take the good parts of base and lattice graphics and
avoid bad parts. It takes care of many of the fiddly details
that make plotting a hassle (l
See ?Syntax noting from the list of operators there that : is of lower
precedence than [.
On Fri, Feb 19, 2010 at 4:33 PM, Stuart Luppescu wrote:
> Hello, this is just a point of curiosity with me. I want a sequence of
> numbers from 64 to 70, omitting the 2nd and 4th numbers. I can do it
> these
Look at the circular and CircStats packages. Look at the kernel density plots.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project
You can probably adapt stars() to give something similar to what you want.
Otherwise, you need to give us examples of what you've tried and how it
diverges from what you need.
Sarah
On Fri, Feb 19, 2010 at 4:34 PM, T.D. Rudolph
wrote:
>
> In conducting studies of animal orientation and displace
Order of operations. Since you didn't specify the order you wanted by
using parentheses, R chose, and it wasn't what you wanted.
> 64:70[-c(2, 4)] # what you said
[1] 64 65 66 67 68 69 70
> (64:70)[-c(2, 4)] # what you thought you said
[1] 64 66 68 69 70
> 64:(70[-c(2, 4)]) # what R thought you
Hi:
Perhaps you want this:
f <- function(nbr){
y<-rnorm(nbr)
y1 <- mean(y)
plot(y)
list(y1 = y1)
}
f(100) prints out the mean and executes the plot
w <- f(100) executes the plot
> w$y1
[1] 0.06965205
returns the mean as a component of the object w.
HTH,
Dennis
On Fri, Feb 19,
In conducting studies of animal orientation and displacement, I need to
produce circular histograms of angles (bearings in radians 0-2pi) where the
centre of the circle indicates very few observations for a given bin of
angles and outwardly concentric circles indicate greater frequencies of
observ
Hello, this is just a point of curiosity with me. I want a sequence of
numbers from 64 to 70, omitting the 2nd and 4th numbers. I can do it
these ways:
> seq(64, 70)[-c(2, 4)]
[1] 64 66 68 69 70
> foo <- 64:70
> foo[-c(2, 4)]
[1] 64 66 68 69 70
But how come this doesn't work?
> 64:70[-c(2, 4)]
[
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of blue sky
> Sent: Friday, February 19, 2010 12:11 PM
> To: Peter Dalgaard
> Cc: r-h...@stat.math.ethz.ch
> Subject: Re: [R] What is the difference between expression
> and quote w
hi,
surprisingly, I haven't found an implementation of GEE2 in R, except in the
orth package but only for binary outcome. There was a post in 2004 that asked
the same question: is there any package in R can do the GEE2?
thanks in advance,
delphine
__
Dear List,
I am having a problem with dates and I would like to understand what is going
on. Below is an example. I can produce a date/time using as.POSIXct, but I am
trying to combine two as.POSIXct objects and keep getting strange results. I
thought I was using the wrong origin, but accord
Have you considered the implications of that solution?
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Kum-Ho
The best validation of assumptions is a good knowledge of the origin of your
data. And with 18 bullet points below, if you do all of these every time you
are going to end up with a lot of false positives when all your assumptions are
met. Understanding your data so that you know which assumpti
Hi R enthusiasts,
I am using JRI which is great to run R commands from Java. (Please let
me know if this is the wrong forum for those questions). Its
straightforward to run single R commands via JRI, but what I really
would like to do is running existing R scripts from that interface.
Does any of
On Fri, Feb 19, 2010 at 12:39 PM, Peter Dalgaard
wrote:
> blue sky wrote:
>>
>> I made the following example to see what are the difference between
>> expression and quote. But I don't see any difference when they are
>> used with eval()? Could somebody let me know what the difference is
>> betwee
Hi all:
I've done this before with factors but can't figure how to do it with
a continuous variable. I am trying to reorder the sequence of my weeks
along the X axis. I want to start with week 27 to 52 and then 1 to 26.
I guess I could use levels along with seq() but doesn't seem to work for me.
Th
I still can't tell exactly what you want, but maybe something like this?
> legend(30,0.08, c("Normal density","Kernel
> density"),col=c("red","blue"),lty=1:1,lwd=1:1,bty="n")
> legend(30,0.06, c(paste("Mean =",round(mean(CDR3),1)),paste("Median
> =",median(CDR3))),col=c("red","black"),pch=19:19,
Hi,
Sorry to forget providing the data before.
I want to get a histogram with the legend for my data. I
drew a normal density curve and kernel density curve in the histogram,
and I also label mean and median in the X axis. From the code, I got
two legend: One shows "Normal Density" and "Kernel De
I think of quote(something) as being equivalent
to expression(something)[[1]].
eval() goes through the outermost expression layer
> eval(expression(expression(noSuchFunction(noSuchValue
expression(noSuchFunction(noSuchValue))
> eval(eval(expression(expression(noSuchFunction(noSuchValue))
> as.list(expression( 2*3))
[[1]]
2 * 3
> as.list(quote( 2*3))
[[1]]
`*`
[[2]]
[1] 2
[[3]]
[1] 3
> identical(as.list(expression( 2*3))[[1]],quote(2*3))
[1] TRUE
expression() wraps the call into an expression object (as pointed out to me
by Gabor Grothendieck).
Bert Gunter
Genentech Nonclinica
Hi,
I want to get a histogram with the legend for my data. I drew a normal density
curve and kernel density curve in the histogram, and I also label mean and
median in the X axis. From the code, I got two legend: One shows "Normal
Density" and "Kernel Density" and their corresponding lines, the
The replicate function will run code multiple times and return the output as a
list (or possibly something simplified). You can then use lapply, sapply, or
other tools to summarize those results.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.o
On Linux You can use Rkward http://rkward.sourceforge.net/ - very nice and
good graphical user interface for R.
Dne Pá 19. února 2010 19:39:53 xin...@stat.psu.edu napsal(a):
> Hi, I am using R in Linux environment. How can i make plot in Linux just
> like in windows?
>
> thanks
>
>
Thanks to Patrick Burns, Bert Gunter, Steve Lianoglou, and Professor Brian
Ripley.
2010/2/17 Prof Brian Ripley
> On Wed, 17 Feb 2010, Randall Wrong wrote:
>
> Dear R users,
>>
>> I have multi-argument returns in a function and I am warned by the program
>> they are deprecated.
>>
>
> Defunct as
On 02/19/10 xin...@stat.psu.edu wrote:
> Hi, I am using R in Linux environment. How can i make plot in Linux just
> like in windows?
There shouldn't be much difference, but we're going to need a lot more
information. What version of linux, what version of R, what exactly is the
problem you are h
Hi, I am using R in Linux environment. How can i make plot in Linux just
like in windows?
thanks
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.ht
blue sky wrote:
I made the following example to see what are the difference between
expression and quote. But I don't see any difference when they are
used with eval()? Could somebody let me know what the difference is
between expression and quote?
Expressions are vectors of unevaluated express
Marc,
A thoughtful, well reasoned discussion. I welcome this kind of analysis. (I was
not criticizing Bert, I was using his post as an example of an unreasonable
statement made to, but not by him, that can serve as an object lesson for all
of us.)
I feel unhappy about posts which attack SAS, R o
Try this:
nbr <- 30
lapply(body(x), eval)[[grep("y1", body(x))]]
On Fri, Feb 19, 2010 at 3:39 PM, threshold wrote:
>
> Hi, say I got the function:
>> x=function(nbr){y<-rnorm(nbr);y1 <- mean(y);plot(y)}
>
> how can I retrieve value of y1, when I need it.
>
> I don't want:
>> x=function(nbr){y<-r
Hi Colin,
You'll get a better answer if you provide an actual example, including what
you tried that didn't work. It looks to me like you want to subset a
data frame based
on column names, though I'm not sure what A(i, J) is supposed to be.
if so, then this is one approach:
> A <- data.frame(A =
Thank you for response. The problem is that using return(y1) in my function
formula always returns y1, but what I want is to return it only when I wish,
like p.value in
t.test(rnorm(100),rnorm(100))$p.value
robert
--
View this message in context:
http://n4.nabble.com/retrieve-from-function-tp15
Gentlemen:
Thankyou for your comments. I know I promised not to respond, but I just
wanted to clarify something that Dieter mentioned below, not defend my
remarks.
I understand "validate code" per 21 CFR part 11 complance to mean that for
**ANY** code that is written, be it in SAS, R, VB, or what
hello all,
thank you for taking the time
I have a matrix A that have column names (let say n columns), I want to
reduce the matrix to have just a few of those column (p colums, this is
trivial), but for the lines I want only the lines such that A(i,J) is part
of a list (J is fixed and known)
I a
On Feb 19, 2010, at 6:56 AM, John Sorkin wrote:
> Bert,
> There is a lesson here. Just as intolerance of any statistical analysis
> program (or system) other than SAS should lead to our being drive crazy, so
> to should intolerance of
> any statistical analysis program (or system) other than R.
Hi Robert,
You need to modify your function to return a value. Something like
x <- function(nbr){
y<-rnorm(nbr)
y1 <- mean(y)
plot(y)
return(y1)
}
-Ista
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https://stat.ethz.ch/mailman/listinfo/r-help
P
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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.
Hi, see the code below. Next time, please provide self-contained code to
generate the matrix (read the posting guide) so that we can just copy-paste
it into R to get the matrix.
x=rep(c(0,1),4)
y=rep(c(0,0,1,1),2)
z=rep(c(0,1),each=4)
m=cbind(z,y,x)
w=rowSums(m)
m=cbind(m,w)
m
m[order(w,x,y,z
I made the following example to see what are the difference between
expression and quote. But I don't see any difference when they are
used with eval()? Could somebody let me know what the difference is
between expression and quote?
expr=expression(2*3)
quo=quote(2*3)
eval(expr)
str(expr)
class(
Hi, say I got the function:
> x=function(nbr){y<-rnorm(nbr);y1 <- mean(y);plot(y)}
how can I retrieve value of y1, when I need it.
I don't want:
> x=function(nbr){y<-rnorm(nbr);y1 <<- mean(y);plot(y)}
> y1
I want someting like:
"x$y1" and then I get the value
Many thanks, robert
--
View
Dear users,
I have a matrix b as
[,1] [,2] [,3] [,4]
[1,]0000
[2,]0011
[3,]0101
[4,]0112
[5,]1001
[6,]1012
[7,]1102
[8,]1113
with last column the rowSums(b)
I w
On 17.02.2010 22:13, Juan Santiago Ramseyer wrote:
==> CRASH INSTALLING PACKAGES CHRON IN:
Intel(R) Core(TM) i7 CPU 860 @ 2800GHz
==> IT'S COMPILED DIRECTIVE WRONG? REGARD YOUR ATENTION
==> I RUN:
SO:Linux 2.6.31.12-174.2.19.fc12.x86_64 x86_64 Fedora release 12
(Constantine)
==> AND RE
On 19.02.2010 17:44, Andrew Yee wrote:
I was wondering if someone could help with the antialias option in tiff().
I'm running R 2.9.2 on a Linux machine.
I'm working on creating a tiff file and have tried different antialias
parameters, e.g. default, none, gray, and subpixel, but don't seem
On 18.02.2010 11:24, Oduor Olande wrote:
Hello
I am using the following command but not able to text the values on the graph
can
someone please make suggestions for improvement
Perhaps dist.r does not have row.names(dist.r) nor colnames(dist.r)?
We cannot know since we do not have dist.r.
On 19.02.2010 01:29, Amy Hessen wrote:
Hi
Thank you very much for your reply.
I meant I got different RMSE with different runs to the same program without
any change for the dataset or the parameters of SVM.
This is my code:
library(e1071)
readingmydata<- as.matrix(read.delim("mydataset.tx
I was wondering if someone could help with the antialias option in tiff().
I'm running R 2.9.2 on a Linux machine.
I'm working on creating a tiff file and have tried different antialias
parameters, e.g. default, none, gray, and subpixel, but don't seem to be
seeing a difference in the output. Or
If you only need it while in R you can use the RENVIRON file or the
.Rprofile or Rprofile.site file. Use help in R to find more on this.
environment variables can also be set through the control panel.
Google to get instructions.
Another possibility is the free command line Windows utility seten
That is a big help Chris - thank you very much for your assistance. I
also ran your code (below) for the example in Caswell (2001, ~p300) and
got matching results.
I'd like to add my vote for including this method in the next version of
popbio - I think it would be a very useful addition.
Th
Thanks Ken, that was it.
I've updated to R version 2.11.0 Under development (unstable) (2010-02-18
r51149)
x86_64-apple-darwin9.8.0
and everything looks fine now. Thanks too to Simon for sorting this out!
Roger
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Thanks,
I ended up using that method!
On Fri, Feb 19, 2010 at 7:53 AM, Dieter Menne
wrote:
>
>
> Paul Rigor (ucla) wrote:
> >
> > I have two data frames that share a common column, data1.name and
> > data2.name.
> > How would I be able to order the other based on an integer column from
> > eithe
Hi,
I'm trying to generate a mixture from 2 univariate normal distributions and
would like to put a constrained on the means (Equal) of the two distributions.
Here is my shot at the code:
library(nor1mix)
X <- norMix(mu=c(50, 60 ), sig2=c(5,10), w=c(0.5, 0.5))
mixData <- rnorMix(1000, X)
p
Paul Rigor (ucla) wrote:
>
> I have two data frames that share a common column, data1.name and
> data2.name.
> How would I be able to order the other based on an integer column from
> either data frame?
>
> For example, how do I order data2 based on data1.age? Or data1 based on
> data2.salary
Hi,
I'm an R nooob! So please pardon this easy question!
I have two data frames that share a common column, data1.name and data2.name.
How would I be able to order the other based on an integer column from
either data frame?
For example, how do I order data2 based on data1.age? Or data1 based
Hello,
Thank you but I think not what I would like to get as an answer is the list
("x","v","n","m") + what you gave me could work for 2 fields but if I have
200...
What I want is a vectorize way to do
bindlists <- function(x){
output = c();
for (i in 1:length(x))
{
DarioAustralia wrote:
>
>
> I have a bunch of histogram bars that I'd like the first to be a certain
> colour, second to be another colour, third to be a third colour, and
> repeat for all my 39 bars.
>
>> levels(forPlot$type)
> [1] "Blah" "Blah 2" "Blah 3"
>
> trellis.par.set("plot.polygon$
Bravo baptiste it works
what does do.call do exactly ??
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htt
On Feb 19, 2010, at 9:49 AM, statquant wrote:
Hello,
Thank you but I think not what I would like to get as an answer is
the list
("x","v","n","m") + what you gave me could work for 2 fields but if
I have
200...
What I want is a vectorize way to do
bindlists <- function(x){
outpu
Hi,
Try this,
unlist(test)
or
do.call(c, test)
HTH,
baptiste
On 19 February 2010 15:19, statquant wrote:
>
> Hello all
> I am new in R and so easy stuff are difficult...
> let say that I have a list
> test <- list(a=c("x","v"),b=c("n","m"))
> how can I without a loop get test$a bind with te
Hello,
Thank you but I think not what I would like to get as an answer is the list
("x","v","n","m") + what you gave me could work for 2 fields but if I have
200...
What I want is a vectorize way to do
bindlists <- function(x){
output = c();
for (i in 1:length(x))
{
On Feb 19, 2010, at 2:20 AM, Jeremie Smaga wrote:
Good morning,
I am currently displaying a time series with the time on the Y axis,
and the
values on the X axis.
The problem is that I wanted to reverse the Y axis (that is, to have
the
latest date the closest to the X axis).
So I looked
Roger Koenker uiuc.edu> writes:
> I'm trying to redo an old plot with:
> > sessionInfo()
> R version 2.11.0 Under development (unstable) (2010-02-09 r51113)
> x86_64-apple-darwin9.8.0
> When I do:
>
> pdf("lty.pdf",height = 6, width = 8)
> u <- 1:100/100
> y <- matrix(rep(1:10,each = 100),100)
Just to say that I found the place to set the environment variables
(Computer/Properties/Advanced/Environment variables). The warning is still
there but the connection seems ok.
Concerning the warning, do you think I have to install an older version of
mysql ?
By the way, I also copied RMySQL.dl
JustADude wrote:
>
>
> ...
>
> By any chance is there more documentation out there on lists and this
> behavior, as I would like to try to better understand what is really going
> on and why one approach works and another doesn't.
>
> ...
> Example reproduced below
>
You forgot an assig
is this what you're looking for?
test <- data.frame(a=c("x","v"),b=c("n","m"))
test
statquant wrote:
Hello all
I am new in R and so easy stuff are difficult...
let say that I have a list
test <- list(a=c("x","v"),b=c("n","m"))
how can I without a loop get test$a bind with test$b (obviously in
I'm trying to redo an old plot with:
> sessionInfo()
R version 2.11.0 Under development (unstable) (2010-02-09 r51113)
x86_64-apple-darwin9.8.0
When I do:
pdf("lty.pdf",height = 6, width = 8)
u <- 1:100/100
y <- matrix(rep(1:10,each = 100),100)
matplot(u,y,lwd = 2,type ="l")
dev.off()
the l
Hello all
I am new in R and so easy stuff are difficult...
let say that I have a list
test <- list(a=c("x","v"),b=c("n","m"))
how can I without a loop get test$a bind with test$b (obviously in real life
their would be many fields)
Cheers and thanks
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http://n4.na
Hello again,
After 'googling', I set the environment variable MYSQL_HOME value to the
path to MYSQL with the command below.
I get a warning (is it a problem or not ???) but I managed to connect to my
databases.
But when I close and open again R, the connection is lost. I have to run the
command
Göran, thanks for the update, I'm just about to install it!
Just wanted to drop you a short line about performance (as you once
requested):
aftreg takes ages on my windows machine to calculate a small set of
7 observations which are not even grouped together by "id". To be a
bit more precise, it
Schmitt, H. (Heike) wrote:
>
> ..
> Trying to associate Rgui.exe with the Rdata files via control panel /
> default programs does not work - Rgui is not added to the list of
> recommended programs.
> ..
>
When your right-click/Open With on an rdata-file in Explorer, there is a
"Search" (Duch
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I used both BMDP and SAS in my earlier years, side by side. At that
time the BMDP statistical methods were much more mature and
comprehensive: we treated them as the standard when the two packages
disagreed. (It was a BMDP manual that clearly explained to me what the
hypothesis of "Yate's weigh
Jim Lemon wrote:
Hi Peter,
The two par arguments "fin" and "pin" allow a solution. What you want is
for the second values in "pin" (Plot dimensions in INches) to be the
same for all your plots. You can get an approximation by using the
layout function instead of mfrow and setting the height v
> I could not find any documentation of how dot-dot-dot works when used
> as an argument in a function call (rather than as a formal argument in
> a definition).
You might also be interested in other thread regarding this problem:
http://groups.google.com/group/r-help-archive/msg/5c6ea5eb593337b4
Dear moderators,
when I installed the latest version of R (R2.10.1) under Microsoft Vista
(service pack 2), I ran into problems associating Rgui.exe with .RData files.
On doubleckicking such files in explorer, they open in a Rterm window. Trying
to associate Rgui.exe with the Rdata files via co
Bert,
There is a lesson here. Just as intolerance of any statistical analysis program
(or system) other than SAS should lead to our being drive crazy, so to should
intolerance of
any statistical analysis program (or system) other than R.
John
>>> Dieter Menne 2/19/2010 3:46 AM >>>
Bert,
I
chipmaney wrote:
Thanks, as a follow-up, how do i extract the list element name (ie, 4-2 or 44-1)
Look at names(your_list)
cheers,
Paul
thanks,
chipper
Date: Thu, 18 Feb 2010 11:56:45 -0800
From: ml-node+1560750-540257540-69...@n4.nabble.com
To: chipma...@hotmail.com
Subject: Re: Extracti
Through help from the list and a little trial and error (mainly error) I think
I figured out a couple of ways to append to a list. Now I am trying to access
the data that I appended to the list. The example below shows where I'm
trying to access that information via two different methods. I
Jyotirmoy Bhattacharya wrote:
I could not find any documentation of how dot-dot-dot works when used
as an argument in a function call (rather than as a formal argument in
a definition). I would appreciate some references to the rules
governing situations like:
f1<-function(x,y,...){
print(x)
}
It likely has to do with the fact that barchart uses unevaluated
arguments. Look at its source by entering its name without arguments:
barchart
This does work:
library(lattice)
f.barchart2 <- function(...) eval.parent(substitute(barchart(...)))
x <- data.frame(a = c(1,1,2,2), b = c(1,2,3,4), d
Thanks to you both,
Both methods work, I was trying to do what you have suggested earlier Paul,
but got in a mess, thanks for the code I can see why it was all going wrong
now!
Kenny
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I could not find any documentation of how dot-dot-dot works when used
as an argument in a function call (rather than as a formal argument in
a definition). I would appreciate some references to the rules
governing situations like:
f1<-function(x,y,...){
print(x)
}
f2<-function(...){
f1(...)
}
KennyL wrote:
Hi All,
I have a slight issue getting R to plot a series of tables automatically.
Essentially I have a series of tables that I wish to plot. They are named
on_2, on_3 etc. based on the file name when they were read in. I have
filelist <- list.files() to give me list of the table na
(Note that you did not follow the posting guide - please always start a new
thread/question by creating a new message - do not simply reply to an
existing posting erasing the subject; such a message stays in the old thread
and is hard to find in the archives)
See package RgoogleMaps on your neare
On 02/19/2010 10:32 PM, KennyL wrote:
Hi All,
I have a slight issue getting R to plot a series of tables automatically.
Essentially I have a series of tables that I wish to plot. They are named
on_2, on_3 etc. based on the file name when they were read in. I have
filelist<- list.files() to give
Hi All,
I have a slight issue getting R to plot a series of tables automatically.
Essentially I have a series of tables that I wish to plot. They are named
on_2, on_3 etc. based on the file name when they were read in. I have
filelist <- list.files() to give me list of the table names. I wish to
The problem is related to the processing of the 'groups' argument,
which is subject to non-standard evaluation. The groups argument is
substitute()d and evaluated in a special context ('data' etc). But
this gets messed up if you are not passing the symbol directly. The
work-around is to rewrite the
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