Hello R mailing list
Is there a better way than this to see if an element exists *within* a
list object :
#generate "file.txt" using current routine
cat("var1=33\nvar2=TRUE",file="file.txt")
#load file to a list called "ipf"
f <- function(.file){source(.file,local=TRUE);as.list(environment())}
Try
abs(outer(xk,x,"-"))
(see ?outer)
--- On Wed, 30/7/08, dxc13 <[EMAIL PROTECTED]> wrote:
> From: dxc13 <[EMAIL PROTECTED]>
> Subject: [R] finding a faster way to do an iterative computation
> To: r-help@r-project.org
> Received: Wednesday, 30 July, 2008, 4:12 AM
> useR's,
>
> I am trying
Prof Brian Ripley wrote:
>
> A patch to do this was posted on 2007-09-29 by Glenn Davis. Some people
> not addicted to Matlab find the behaviour very inconvenient and prefer the
> getline/readline behaviour (triggered by ^R/^S) of Rterm and R on Unixen.
>
> On Tue, 29 Jul 2008, losemind wro
Hi all,
How do I operate R from within Excel?
I mean, highlight a bunch of cells, and send to R, and do some statistics in
R, and return back the numbers and do some plots in R.
For example, I have some parameters for Gaussian Random variable, and I
would find the most convenient way to send th
> legend("topleft",c("Your text here"),pch=1)
Thanks!
But is it possible to get rid of the legend-box and the box-shadow?
I don't find an argument in the manual for that.
Am 30.07.2008 um 00:08 schrieb Jorge Ivan Velez:
>
> Dear Jörg,
>
> Perhaps,
>
> plot(1:10)
> legend("topleft",c("Your t
Hi All:
I know this has been discussed at length already, but 1) I get R-Help
in digest and didn't see this until 3am Pacific time this morning, and
2) We, REvolution, have been discussing this as of late. I thought I
would pass on some of the knowledge we have recently obtained vis a
vi
In general there is no relation. The output of density gives you
something you can plot, essentially. So the x-values are simply a
series of points covering the range of the density and the y values are
the ordinates at those abscissae.
try
plot(density(rnorm(1000)^2)))
for example.
Bill Ven
If v is your vector of sample variances (and assuming that their distribution
is chi-square) you can define
f(df) <- sum(dchisq(v,df,log=TRUE))
and now you need to maximize f, which can be done using any optimization
function (like optim).
--- On Sat, 26/7/08, Julio Rojas <[EMAIL PROTECTED]> wr
On 30/07/2008, at 10:47 AM, Jörg Groß wrote:
legend("topleft",c("Your text here"),pch=1)
Thanks!
But is it possible to get rid of the legend-box and the box-shadow?
I don't find an argument in the manual for that.
You explicitly said that you *wanted* a box!!!
But be tha
On Jul 29, 2008, at 6:47 PM, Jörg Groß wrote:
legend("topleft",c("Your text here"),pch=1)
Thanks!
But is it possible to get rid of the legend-box and the box-shadow?
I don't find an argument in the manual for that.
Hi,
The documentation for the argument is right there - but as always i
Hi everyone,
I'm having trouble applying the Cor() function to two matrices, both of
which contain NAs. I am doing the following:
a<-cor(m1, m2, use="complete.obs")
... and I get the following error message:
Error in cor(m1, m2, use = "complete.obs") :
no complete element pairs
Does anyone
Hi Gabor,
Thanks for your reply. Assuming I have a time series that is ready made
(i.e. not constructed it using a zoo function) will the procedure below
still retain the dates in the matrix?
Thanks,
rcoder
quote author="Gabor Grothendieck">
rollapply along an index:
library(zoo)
z <- zoo(mat
Hi everyone,
I am trying to apply linear regression to adjacent columns in a matrix (i.e.
col1~col2; col3~col4; etc.). The columns in my matrix come with identifiers
at the top of each column, but when I try to use these identifiers to
reference the columns in the regression function using rollap
If your legal department is still concerned, have them look at the
license in Java. It most likely is GPL (unless you have a special
commercial version of Java installed) and thus makes any Java program
you create subject to the GPL in Java due to the JIT compiler used in
Java. If your legal
On 27/07/2008 3:10 PM, Stephen Tucker wrote:
Hi list,
I was using Sweave and was wondering if anyone has had any luck changing the
font colors of the code chunks. For instance, in my .Rnw preample I tried
including:
===
\usepackage[usenames]{colors}
\definecolor{darkred}{rgb}{0.545,0,0}
\defi
Gad Abraham wrote:
If your legal department is still concerned, have them look at the
license in Java. It most likely is GPL (unless you have a special
commercial version of Java installed) and thus makes any Java program
you create subject to the GPL in Java due to the JIT compiler used in
J
Hello Gad:
On Jul 29, 2008, at 4:41 PM, Gad Abraham wrote:
If your legal department is still concerned, have them look at the
license in Java. It most likely is GPL (unless you have a special
commercial version of Java installed) and thus makes any Java
program you create subject to the G
On 29-Jul-08 16:28:26, Edna Bell wrote:
> Hi again!
> Suppose I have the following:
>
>> xy <- round(rexp(20),1)
>> xy
> [1] 0.1 3.4 1.6 0.4 1.0 1.4 0.2 0.3 1.6 0.2 0.0 0.1 0.1 1.0 2.0 0.9
> 2.5 0.1 1.5 0.4
>> table(xy)
> xy
> 0 0.1 0.2 0.3 0.4 0.9 1 1.4 1.5 1.6 2 2.5 3.4
> 1 4 2 1
Gad and David - would you be so kind as to include me in any off-line
discussion of this vital topic of R and GPL?
Many thanks.
Charles Annis, P.E.
[EMAIL PROTECTED]
phone: 561-352-9699
eFax: 614-455-3265
http://www.StatisticalEngineering.com
-Original Message-
From: [EMAIL PROTECTED
I am searching for a way to get nearest position of a number in a
vector from a search-value.
So perhaps if I have this vector:
1.0 1.2 1.4 1.6 1.8 2.0
I want to get the position of a special number.
with match(1.4 ,x), I get "3" for position three.
But now I want the nearest numbe
methods(rollapply)
shows which classes have rollapply methods.
On Tue, Jul 29, 2008 at 5:48 PM, rcoder <[EMAIL PROTECTED]> wrote:
>
> Hi Gabor,
>
> Thanks for your reply. Assuming I have a time series that is ready made
> (i.e. not constructed it using a zoo function) will the procedure below
> st
Read the last line of every message to r-help.
On Tue, Jul 29, 2008 at 6:15 PM, rcoder <[EMAIL PROTECTED]> wrote:
>
> Hi everyone,
>
> I am trying to apply linear regression to adjacent columns in a matrix (i.e.
> col1~col2; col3~col4; etc.). The columns in my matrix come with identifiers
> at the
Thank you to all who applied. These all seem to work the way I want them to.
The outer function seems really useful, I probably could use that for a lot
of my work.
Thanks!
Rolf Turner-3 wrote:
>
>
> On 30/07/2008, at 6:12 AM, dxc13 wrote:
>
>>
>> useR's,
>>
>> I am trying trying to find out
?findInterval
> x <- c(1.0,1.2,1.4,1.6,1.8,2.0)
> findInterval(1.4,x)
[1] 3
> findInterval(1.5,x)
[1] 3
> findInterval(10,x)
[1] 6
On Tue, Jul 29, 2008 at 8:32 PM, Jörg Groß <[EMAIL PROTECTED]> wrote:
> I am searching for a way to get nearest position of a number in a vector
> from a search-valu
Is this what you want:
> "var3" %in% names(ipf)
[1] FALSE
> "var1" %in% names(ipf)
[1] TRUE
>
On Tue, Jul 29, 2008 at 6:20 PM, <[EMAIL PROTECTED]> wrote:
>
> Hello R mailing list
>
> Is there a better way than this to see if an element exists *within* a
> list object :
>
> #generate "file.txt"
Regarding dubuggers, RSiteSearch("debug", "fun") produced 183
hits for me just now. The first two reference a "debug" package, while
the third describes the "debug" function in the "base" package.
The "debug{base}" function is great, but not as good as the Matlab
debugger. The "de
Hi R users,
I google the website and I found that there are three ways to perform
repeated measure ANOVA: aov, lme and lmer.
http://www.mail-archive.com/[EMAIL PROTECTED]/msg58502.html
But the questions are which one is good to use and how to do post-hoc test?
I use the example that is provide
Hi Chunhao,
>> I google the website and I found that there are three ways to perform
>> repeated measure ANOVA: aov, lme and lmer.
It's also a good idea to search through the archives.
>> I use the example that is provided in the above link and I try
>> > tt<-aov(p.pa~group*time+Error(subject
Hi Yunlei,
Is your problem constrained or not?
If it is unconstrained and your matrix is not positive definite, the minimum is
unbounded (unless you are extremely lucky and the matrix is positive
semi-definite and the vector which multiplies the unknowns is exactly
perpendicular to all the eige
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