On 10/06/2010 02:41 AM, james hirschorn wrote:
> Yes, your solution of setting quote="" would read the multi-word strings
> incorrectly. A more complicated version of your solution should work: First
> check which columns are identified as strings, and then apply your solution
> to
> the remain
PS ... actually, now I think about it...
> as.polylist(apply(m, 1, polynom))
List of polynomials:
[[1]]
1 + 5*x + 9*x^2 + 13*x^3
[[2]]
2 + 6*x + 10*x^2 + 14*x^3
[[3]]
3 + 7*x + 11*x^2 + 15*x^3
[[4]]
4 + 8*x + 12*x^2 + 16*x^3
This works with the PolynomF package because in that case the ob
I was unaware of this discussion till now, so I'm not up with what's been said
already.
Here is how I would go about the problem I think you are trying to solve.
> m <- matrix(1:16, 4, 4)
> require(PolynomF)
> (lop <- as.polylist(lapply(1:nrow(m), function(i) polynom(m[i,]
List of polynomia
Paula,
Just for fun, another way:
## Create 10 by 10 example matrix
set.seed(100)
a <- matrix(sample(c(0,1,2),100, replace=T), nrow=10, ncol=10)
a[1,] <- sample(c(0.25,0.5,0.75),ncol(a), replace=T)
## Use apply to substitiute matrix values by row
b <- apply(a, 1, function(x)
{
row <- x
row[
Dear Lemarian,
I was going to reply to say this is how I would do (almost) the same
thing with the prob package:
library(prob)
S <- rolldie(2, makespace = TRUE)
tmp <- sim(S, ntrials = 1)
with(tmp, mean(X1 == X2))
but then it occurred to me that Greg's "dice" was much faster than
"sim" (plus
Hi Alex,
I remember that you had the discussion recently about identifying grid
cells that were crossed by a line...
http://tolstoy.newcastle.edu.au/R/e11/help/10/09/8431.html
So, following on from those posts, for any cell crossed by a line you
will know the coordinates of the cell corners and
The key is in the help page for apply. It says (in part):
In all cases the result is coerced by Œas.vector¹ to one of the
basic vector types before the dimensions are set, so that (for
example) factor results will be coerced to a character array.
So although as.polynomial() returns
On Mon, Oct 4, 2010 at 12:39 PM, james hirschorn wrote:
> Suppose I have a data file (possibly with a huge number of columns), where the
> columns with factors are coded as "1", "2", "3", etc ... The default behavior
> of
> read.table is to convert these columns to integer vectors.
>
> Is there a
Thank you, Henrik! That makes more sense now.
You mentioned that every double value needs 8 bytes. So, in R, how many
decimal point, or any number smaller than, say 10^4 are considered as double
value ? (Sorry I don't have any C or Java language background, and couldn't
find it for R. )
I apprecia
Dear r-users!
I have another question regarding the dlm package and I would be very
happy if someone could give me a hint!
I am using the dlm package to get estimates for an endogenous rate of
capacity utilization over time. The general form of a state space model
is
(1) b_t = G * b_t-1 + w_t
On Oct 5, 2010, at 8:41 PM, james hirschorn wrote:
Yes, your solution of setting quote="" would read the multi-word
strings
incorrectly. A more complicated version of your solution should
work: First
check which columns are identified as strings, and then apply your
solution to
the remain
Hi,
I can see that there is a function for mahalanobis distance in stat
package. But I'm wondering if there is any implementation that can
take in a mix of ordinal data and numerical values together? If no,
would you mind pointing me to some reference for a methodology to do
this? Thank you in adv
Yes, your solution of setting quote="" would read the multi-word strings
incorrectly. A more complicated version of your solution should work: First
check which columns are identified as strings, and then apply your solution to
the remaining columns.
I'm a newbie at R, but it seems to me that t
On Tue, Oct 5, 2010 at 5:21 PM, Carrie Li wrote:
> I am sorry that it has been couple days.
> I've read the website you provided below, but still don't quite know if this
> is doable.
> The maximum vector length is 2^31-1, so here is what I tired, and it
> returned errors as below.
>
>
> P=2
>
I am sorry that it has been couple days.
I've read the website you provided below, but still don't quite know if this
is doable.
The maximum vector length is 2^31-1, so here is what I tired, and it
returned errors as below.
P=2
D=matrix(rep(0, P*P), nrow=P)
Error: cannot allocate vector of s
Yes Hmisc was loaded. Thanks for the insight. I usually pay
attention to which objects get "masked" ... but reorder was not in
that list. Anyway thanks.
Darin
On Tue, Oct 05, 2010 at 03:50:35PM -0700, Phil Spector wrote:
> Is it possible that the original poster had the Hmisc package loaded?
>
>>
Is it possible that the original poster had the
Hmisc package loaded?
Hmisc::reorder.factor
function (x, v, FUN = mean, ...)
ordered(x, levels(x)[order(tapply(v, x, FUN, ...))])
Without that package, reorder.default gets called:
getAnywhere('reorder.default')
A single object matching ‘reo
> sessionInfo()
R version 2.11.1 (2010-05-31)
i486-pc-linux-gnu
> x <- factor(1:5)
> x.ro <- reorder(x, rnorm(5))
> is.ordered(x.ro)
[1] FALSE
> x.ro <- reorder(x, rnorm(5), ordered=FALSE)
> is.ordered(x.ro)
[1] FALSE
On Tue, Oct 5, 2010 at 3:14 PM, Darin A. England wrote:
>
> Or at least is se
On Oct 5, 2010, at 6:14 PM, Darin A. England wrote:
Or at least is seems that way to me. It's not a big problem, but the
behavior doesn't match the documentation. (I think r-help is the
place to report this. )
x <- factor(1:5)
is.ordered(x.ro) # should be FALSE according to ?reorder
[1] TRU
R-bugs is the appropriate place to report this.
It may be particular to your system. Here's what happens on mine (Windows XP,
R 2.11.1)
> f <- factor(1:5)
> g <- reorder(f, rnorm(5))
> is.ordered(g)
[1] FALSE
> g
[1] 1 2 3 4 5
attr(,"scores")
1 2 3 4
Is there a trick to posting links here or is wikipedia blocked?
I see in my sent folder that I included this wikipedia
link on threaded b-trees but it didn't show up in
the mail i got back from the list? Sorry,
I'm new to this list :)
http:// en.wikipedia.org /wiki/Threaded_binary_tree
---
?par
lood at the 'las' parameter.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of ANJAN PURKAYASTHA
Sent: Wednesday, 6 October 2010 8:13 AM
To: r-help@r-project.org
Subject: [R] Plotting x-axis labels perpendicular to the axis
Hi
On Oct 5, 2010, at 6:13 PM, ANJAN PURKAYASTHA wrote:
Hi,
The following command: axis(1, at= 1:50, labels= sampleNames(data))
produces
an X-axis with 50 ticks and labels drawn from the sampleNames
namespace.
Problem is the labels at each tick are printed parallel to the x-
axis and
thus run
Or at least is seems that way to me. It's not a big problem, but the
behavior doesn't match the documentation. (I think r-help is the
place to report this. )
> x <- factor(1:5)
> x.ro <- reorder(x, rnorm(5))
> is.ordered(x.ro) # should be FALSE according to ?reorder
[1] TRUE
>
> x.ro <- reorder(
Hi,
The following command: axis(1, at= 1:50, labels= sampleNames(data)) produces
an X-axis with 50 ticks and labels drawn from the sampleNames namespace.
Problem is the labels at each tick are printed parallel to the x-axis and
thus run into each other. How do I specify for the labels to be printed
> Date: Tue, 5 Oct 2010 14:57:40 -0700
> From: matl...@cs.ucdavis.edu
> To: r-help@r-project.org
> Subject: Re: [R] binary tree construction in R
>
> MK wrote:
>
> > Hi all,
> >
> > I'm very new to R and I'm trying to construct a threaded binary tree
MK wrote:
> Hi all,
>
> I'm very new to R and I'm trying to construct a threaded binary tree using
> recursive functions.
>
> I'm very confused was wondering if anyone had any R sample code they would
> share. I've come across a lot of C++ code(nothing in R) and this is not
> helping.
>
> best,
On Tue, 2010-10-05 at 13:56 -0700, 1Rnwb wrote:
> Hello List,
> I have a longitudinal samples on multiple individuals, for initial analysis,
> i am doing some plots to see how the variable changes with the age. In some
> of the individuals one or two data points are missing,
>IL1Ra
This should work:
> test <-subset(wild,ID=="2830" & Date==as.POSIXlt("2010-08-17"))
If not, here's another solution:
> dates <- c("2010-05-28","2010-08-17")
> dates <- as.POSIXlt(dates)
> id <- c("2830","2830")
> data <- data.frame(id,dates)
> test <- data[data$id == "2830" & data$dates == as.POS
If you are refering to the file GRASS_0.3-9.tar.gz which
you may have downloaded from a CRAN archive, then you normally
do not need to decompress and untar it. Type
R CMD INSTALL GRASS_0.3-9.tar.gz
at the command line (with the working directory set to where
you've downloaded the file.)
Even b
Hello List,
I have a longitudinal samples on multiple individuals, for initial analysis,
i am doing some plots to see how the variable changes with the age. In some
of the individuals one or two data points are missing,
IL1Ra Age.at.Sample.Collection Subject.ID
6.575
Hi everybody,
I'm quite new to ubuntu and tar.gz packages, anyway following these
instructions https://help.ubuntu.com/community/CompilingEasyHowTo I tried to
install the R Grass package.
Everything worked nice until I did:
$ ./configure
checking for gcc... gcc -std=gnu99
checking for C comp
The regsubsets does work.
I also notice the bestglm, will see how it work too.
Thank you.
--
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Hello friends,
I would like to thank you all for your advices. Rcmdr is gread.
All the best,
Mehdi
* Mehdi Zarrei, PhDPostdoctoral
fellowUniversity of Toronto
Dept. of Ecology & Evolutionary Biology
Royal Ontario Museum
Department of Natural Histo
Thanks for the feedback, this is really helpful.
4) Your solution works like a charm. I opted for reference by column number
since I had so many.
ynFields = c(12:22,58:229)
ynLabel = c("No","Yes")
X[ynFields] <- lapply(X[ynFields], factor, levels=0:1, labels=ynLabel)
whenFields = c(24:56)
when
Jared -
The value "2010-08-17" is a character value, and your dates
are stored as POSIXlt values. So you'd need to use
Date == as.POSIXlt("2010-08-17")
in your subset statement.
- Phil Spector
Statistic
Hello All
I, for the life of me, can't figure out to subset my dataframe by my Date
field. The field has been formatted as POSIXlt.
str(wild)
..
$ Date: POSIXlt, format: "2010-05-28" "2010-05-28" "2010-05-28" ...
This doesn't work:
test <-subset(wild,ID=="2830" & Date=="2010-08-17"
Hi,
On Tue, Oct 5, 2010 at 3:52 AM, Trying To learn again
wrote:
>
> Hi all,
>
> Reading more I have find a partial solution on a part of the proble in some
> part of the code it should appea something like:
>
> # NC: All the potential combinations 3^15
>
> if
>
> NC[price(i,j)=="1" & price(i,j)=
On Oct 5, 2010, at 2:46 PM, QiJun Fung wrote:
Does any one know how to optimize the following function w.r.t to
beta? The difficulty here is the beta is a matrix not a vector. and f
is also a function of beta and an element of the objective function. I
just want to know for this complicated sit
Thank you for your help!
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__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/list
Does any one know how to optimize the following function w.r.t to
beta? The difficulty here is the beta is a matrix not a vector. and f
is also a function of beta and an element of the objective function. I
just want to know for this complicated situation, is there any
packages or function to estim
On Sat, 02-Oct-2010 at 11:07PM +0400, Aleksey Noskov wrote:
|> Hello, I am new in R
|>
|> I try to build R on CentOS 64bit. *./configure --prefix=$HOME/soft *executed
|> without any errors. But make command print that error:
I use R on CentOS without root permission. It's always worked if I
ru
On Tue, 5 Oct 2010, Richard R. Liu wrote:
Hi,
I have a function f <- function(..., func){ something }, where func is a
function of the form function(...). ??I would like to pass func all the
arguments
passed to f except the last. ??I know that I can manipulate the variable number
of arguments
On 05/10/2010 2:23 PM, Richard R. Liu wrote:
Hi,
I have a function f<- function(..., func){ something }, where func is a
function of the form function(...). I would like to pass func all the arguments
passed to f except the last. I know that I can manipulate the variable number
of arguments
Hi,
I have a function f <- function(..., func){ something }, where func is a
function of the form function(...). I would like to pass func all the arguments
passed to f except the last. I know that I can manipulate the variable number
of arguments passed to f by converting ... to a list, i.e.,
Someone said that I need to define "intrinsic dimensionality" in my question.
Basically, this is the topological dimension of a set of points that are
represented in a space of possibly higher dimension. For example, your data
might describe a corkscrew line through a 30-dimensional space (you mea
Thank you, Ista!
you are right - in the paper I was refering to it's "measured" but in
?melt it's measure.vars!
Dimitri
On Tue, Oct 5, 2010 at 12:17 PM, Ista Zahn wrote:
> Hi Dimitri,
> The argument names may have changed. Notice that
>
> melt.data.frame(smiths, measured = c("age", "weight", "h
I meant to say "took me"
On Tue, Oct 5, 2010 at 1:01 PM, Dimitri Liakhovitski
wrote:
> BTW - I did not install it in any special Windows area - I installed
> it wherever the default too me.
>
> On Tue, Oct 5, 2010 at 11:42 AM, Prof Brian Ripley
> wrote:
>> See
>> http://cran.r-project.org/bin/wi
BTW - I did not install it in any special Windows area - I installed
it wherever the default too me.
On Tue, Oct 5, 2010 at 11:42 AM, Prof Brian Ripley
wrote:
> See
> http://cran.r-project.org/bin/windows/base/rw-FAQ.html#Does-R-run-under-Windows-Vista_003f
>
> which does cover this. Most likely
Correct - only to update packages, not for regular use.
Thanks a lot for the references!
Dimitri
On Tue, Oct 5, 2010 at 12:29 PM, Henrik Bengtsson
wrote:
> Just to clarify to the OP (and please correct me if I'm wrong),
>
> having to run as admin should only be needed when you have to *update*
>
On Tue, 2010-10-05 at 08:11 -0700, Filoche wrote:
> Hi everyone.
>
> I'm trying to do a loess with missing value on independant variable.
>
>
>
> doc = c(2.27904, 2.59536, 7.44696, NA, 6.24264, 4.58400, 5.79192, 5.39502,
> 7.41216, 4.09440, 4.22868, 4.24620, 5.43804, 1.95528);
> distance = c(26
Forgot ... The best thing with emacs , it that it has modes for close
to everything, so you don'nt need to
learn new editors for whatever strange projecy you start.
By the way, i installed eclipse to try it out. It is way of biggish...
Kjetil
On Tue, Oct 5, 2010 at 12:30 PM, Kjetil Halvorsen
wr
Paula-
Look at this example. R in this case is your data matrix.
-tgs
R<-matrix(sample(0:2,100,replace=T),nrow=10)
R[1,]<-round(runif(10),3)
f<-function(x){
show(x)
y<-x[-1]
y[y!=2]<--999
y[y==2]<-x[1]
c(x[1],y)
}
apply(R,2,f)
On Tue, Oct 5, 2010 at 11:54 AM, Paula Fergnani Salvi
I really like emacs+ess, so getting that to work again
is the preferred solution...
Kjetil
On Tue, Oct 5, 2010 at 11:50 AM, Erik Iverson wrote:
>
>> Another alternative is to use Geany [1]. It would save you the trouble
>> of learning Emacs,
>
> /s/save/deprive
> /s/trouble/thrill
>
> :)
>
> ___
Just to clarify to the OP (and please correct me if I'm wrong),
having to run as admin should only be needed when you have to *update*
(update.packages()) the so called recommend packages (comes with R)
that sits under 'Program Files'. If you don't run as admin and try
to *install* (install.pack
Does anyone know of any packages or have any loose code for estimating the
intrinsic dimension of a data set?
Thanks for any leads,
Jack
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On 05.10.2010 18:19, Neeti wrote:
It would be nice if you suggest me the steps to check whether svmlight is
present in my system or not?
If you have not installed it, it is probably not present yet. How to
check it: Well, that depends on your OS that is unstated so far.
Asking a search en
Dear Gabor and Jim
You both gave me amazing solutions.
I will use.
Thanks!
Nilza
On Tue, Oct 5, 2010 at 2:11 AM, Gabor Grothendieck
wrote:
> On Sat, Oct 2, 2010 at 11:31 PM, Nilza BARROS
> wrote:
> > Dear R-users,
> >
> > I would like to know how could I read a file with different lines
> length
It would be nice if you suggest me the steps to check whether svmlight is
present in my system or not?
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___
We really cannot help to solve your ssh problems.
Uwe Ligges
On 04.10.2010 18:06, wesley mathew wrote:
Dear all
I am implement one grid computing system using GridR package. I am
working on windows platform. I did all the configuration according to
the GridR tutorial.
But it has some errors
Hi Dimitri,
The argument names may have changed. Notice that
melt.data.frame(smiths, measured = c("age", "weight", "height"))
gives a message saying "Using subject as id variables". This is
because measured variables need to be specified as measure.vars (or an
abbreviation of that: even m will wo
try this:
> matr
[,1] [,2]
[1,] 0.1 0.5
[2,] 1.0 1.0
[3,] 0.0 2.0
[4,] 2.0 2.0
> result <- apply(matr[-1,], 1, function(.row){
+ ifelse(.row == 2, matr[1,], -999)
+ })
> # data back together
> rbind(matr[1,], t(result))
[,1] [,2]
[1,]0.10.5
[2,] -999.0 -999.0
[3,]
Probably you do not have svmlight installed on your machine.
Uwe Ligges
On 05.10.2010 12:42, Neeti wrote:
Hi !
Right now I am learning to use svm functions available in R and trying to
use these function with given example. I was stuck with svmlight function
which is available in klaR pac
Note that we now have a constrained optimization problem in (n + 3) variables:
y, A, B, and C; but we also have `n' constraints among these n+3 variables.
Ravi.
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric M
I would do it like this:
library(TeachingDemos)
tmp <- dice(1, 2)
with(tmp, c(sum(Red==Green),mean(Red==Green)) )
plot(head(tmp,28))
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-h
On 05.10.2010 17:51, Erik Iverson wrote:
1) Please use an informative subject line
2) The answer may depend on the package, and can be easy
to difficult. Which package are you trying to install?
In any case, the answer is in the manual "R Installation and
Administration".
Uwe Ligges
w
Hello, I’m new at programming and I will greatly appreciate if you can help me
with this.
I have a very large matrix (hundreds of rows and columns), with the first raw
filled with different numbers (between 0 and 1). The rest of the matrix is
filled with values 0, 1, 2. What I need is to repl
Here is a way of putting "Order" on your data:
> x
V1 V2 V3 V4 V5
1 1 12345678 Soc101 34 02-04-2003
2 2 12345678 Soc101 62 31-11-2004
3 3 12345678 Psy104 63 03-05-2003
4 4 23456789 Soc101 73 02-04-2003
5 5 23456789 Psy104 76 25-02-2004
> x$order <- ave(x$V1, x$V2, x$V3, FU
1) Please use an informative subject line
2) The answer may depend on the package, and can be easy
to difficult. Which package are you trying to install?
wenyue sun wrote:
*Dear All, *
**
*I want to install tar.gz in R on a Windows operating system. I know that
R in Win accepts packages in
Another alternative is to use Geany [1]. It would save you the trouble
of learning Emacs,
/s/save/deprive
/s/trouble/thrill
:)
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http:
Thierry,
your ggplot solution really looks great and very parsimonious!
May I ask a couple of questions?
1. Is there a way to make the grid appear on top of the colored areas
instead of under them - like I did in the plot solution?
2. In your code line Molten2$variable <- factor(Molten2$variable,
l
*Dear All, *
**
*I want to install a tar.gz package in R on a windows operating system.
Since R in Windows accepts only packages in zip file for installation,
I tried to decompress the tar.gz file and create a zip file using 7 zip for
a subsequent installation, yet this didn't work. Can anyone hel
*Dear All, *
**
*I want to install tar.gz in R on a Windows operating system. I know that
R in Win accepts packages in zip file for installation. Therefore,
I**tried to decompress the tar.gz file and create a zip file using
7zip but
the installation of the resulting zip files didn't work. Can an
On Tue, Oct 5, 2010 at 4:52 PM, Mehdi Zarrei wrote:
> I am looking for an editor to be able to execute commands into R in Linux
> ubuntu. Is there any suggestion?
>
There is a wiki page on the subject [1].
Everyone: Please contribute your favorite editor to that list.
Regards
Liviu
[1] http://
You can solve this as a (nonlinearly) constrained optimization problem:
Given: x, y.obs; both in R^n
Minimize the objective function: sum( (y - y.obs)^2 )
where `y' is such that it satisfies the constraints:
(A+B+C)*log((B+C+y)/C)-A*log((B-y)/B)=(B+C)*D*x
Do you have other constraints on A,
See
http://cran.r-project.org/bin/windows/base/rw-FAQ.html#Does-R-run-under-Windows-Vista_003f
which does cover this. Most likely you need to 'Run as
Administrator'.
There's lot of other useful information in that FAQ that you would do
to familiarize yourself with.
On Tue, 5 Oct 2010, Dim
On Tue, Oct 5, 2010 at 5:43 PM, Christopher W Ryan wrote:
> As an alternative to emacs-ess, you could try gedit with the R plugin.
>
Another alternative is to use Geany [1]. It would save you the trouble
of learning Emacs, and I find it better designed for code editing than
Gedit. If you're a begi
Hi all,
I've found a lot of helpful info regarding identifying and deleting duplicates
but I'd like to do something a little different - I'd like to identify the
duplicate values but instead of deletion, label them with a value.
I am working with historical data regarding school courses:
Hi there.
One more thing you could try is to avoid evaluation of length(CRX) every
loop.
Assign the length to a variable before the loop.
Regards,
Phil
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Hi everyone.
I'm trying to do a loess with missing value on independant variable.
doc = c(2.27904, 2.59536, 7.44696, NA, 6.24264, 4.58400, 5.79192, 5.39502,
7.41216, 4.09440, 4.22868, 4.24620, 5.43804, 1.95528);
distance = c(26.5,56.5, 90.3, 123.0, 147.5, 176.0, 215.7, 229.3, 252.0,
325.3, 362
> Date: Tue, 5 Oct 2010 08:18:10 -0700
> From: djmu...@gmail.com
> To: tott...@yahoo.com
> CC: r-help@r-project.org
> Subject: Re: [R] R-help
>
> Hi:
>
> This problem is a useful lesson in the power of vectorizing calculations in
> R.
>
> A remanufactu
Hellow!
I am replicating the example givien in "Reshaping Data with the
reshape Package" (http://www.jstatsoft.org/v21/i12 - see "download"
link on the right), p. 2-3.
library(reshape)
data(smiths)
str(smiths)
The text says: "If you specify only one of measured and identifier
variables, melt ass
Write a function that does your analysis and returns the values of interest in
a vector. Then run the function multiple times using replicate or sapply and
the results will be put into a matrix for you.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@
Dear Dimitri,
Ggplot2 solves your problem with the gridlines and requires much less
code. You only need to reshape your data somewhat.
library(ggplot2)
#changing the dataset
my.data2 <- my.data
my.data2$x <- my.data$x + my.data$a
my.data2$z <- my.data$y + my.data$z
Molten2 <- melt(my.data2, id.va
Hi:
This problem is a useful lesson in the power of vectorizing calculations in
R.
A remanufacturing of your function:
dubloop <- function(nsim) {
count <- 0
for(j in 1:nsim){ #begin loop
die1 <- sample(1:6,1)
die2 <- sample(1:6,1)
if(die1 ==
Barry - thanks a lot, it's great. Now, the gridlines "stick" to the tick marks!
Just wondering why the product of "grid" does not.
D.
On Tue, Oct 5, 2010 at 11:07 AM, Barry Rowlingson
wrote:
> On Tue, Oct 5, 2010 at 3:37 PM, Dimitri Liakhovitski
> wrote:
>> my.data$date
>
> I wouldn't use grid -
THanks a lot, Thomas - I am 100% clear on the legend symbol now, very helpful.
With legend colors - I just tried to have more colors in all.colors
than there are variables to plot on the graph. Did it on purpose - so
that there are always more colors to pick from.
But if I do names(my.data) - then
Hello,
I want to perform a nonlinear curve fit in order to obtain parameter
estimates from experimentally determined data (y in dependence of x),
but with an implicit function, thus, a function of which I cannot
isolate y on the left-hand side of the equation. As far as I understand,
the function
On Tue, Oct 5, 2010 at 3:37 PM, Dimitri Liakhovitski
wrote:
> my.data$date
I wouldn't use grid - it suggests you try abline, so something like:
> abline(v=my.data$date,lty=2)
- to put vertical lines at your data points. For the horizontal grid
lines, we can just use the axis tick marks:
> a
Dimitri-
Regarding question 1, it is the pch argument that controls the legend
symbol. In my configuration, pch=15 gives a colored, filled-in box. To see
the possible pch symbols, run the following code:
plot(0,0,xlim=c(0,9), ylim=c(0,9),type="n", axes=F,xlab="",ylab="")
for(i in 0:9){
for(j in
On Oct 5, 2010, at 16:29 , Kjetil Halvorsen wrote:
> see below.
>
> On Tue, Oct 5, 2010 at 10:25 AM, Laurent Gatto
> wrote:
>> Dear Kjetil,
>>
>> On 5 October 2010 15:12, Kjetil Halvorsen
>> wrote:
>>> emacs23+ess is great BUT there is a very large problem which is
>>> driving me away from i
On 5 October 2010 15:29, Kjetil Halvorsen
wrote:
> see below.
>
> On Tue, Oct 5, 2010 at 10:25 AM, Laurent Gatto
> wrote:
>> Dear Kjetil,
>>
>> On 5 October 2010 15:12, Kjetil Halvorsen
>> wrote:
>>> emacs23+ess is great BUT there is a very large problem which is
>>> driving me away from it:
>>
As an alternative to emacs-ess, you could try gedit with the R plugin.
--Chris Ryan
I am looking for an editor to be able to execute commands into R in
Linux ubuntu. Is there any suggestion?
[[alternative HTML version deleted]]
__
R-help
Hello!
If you run the whole code below, it'll produce a stacked diagram. And
it looks good - because the tick-marks are aligned with the grid.
However, if I stretch the graph window, grid becomes misaligned with
the tickmarks. Or, rather, it seems aligned for the first and the last
tick mark, but
Thanks a lot, Thomas.
May I ask 2 questions:
1. Is there a way to "fill" the legend rectangles with color
(currently, they just contain a colored cross)?
2. Currently, my graph grabs the colors automatically (as I loop
through var, it grabs col=all.colors[which(names(my.data) %in% var)].
I am wond
Hello
I would like to calculate a weighted line integral.
The integral is calculated by the cells that this lines trasverses (the small
cells belong to matrix (m*n) that represent the value that a specific area has.
I need to calculate the weights by finding out how much the line touches or
i
see below.
On Tue, Oct 5, 2010 at 10:25 AM, Laurent Gatto wrote:
> Dear Kjetil,
>
> On 5 October 2010 15:12, Kjetil Halvorsen
> wrote:
>> emacs23+ess is great BUT there is a very large problem which is
>> driving me away from it:
>> (on ubuntu) since version 23, emacs has a very problematic bug,
Thnaks a lot, Duncan!
Running R as administrator solved the problem!
Dimitri
On Tue, Oct 5, 2010 at 9:07 AM, Duncan Murdoch wrote:
> On 05/10/2010 8:58 AM, Dimitri Liakhovitski wrote:
>>
>> Hello!
>>
>> I've just installed R 2.11.1 on my new home PC. Just the base R. My PC
>> has Windows 7.
>> R
Dear Kjetil,
On 5 October 2010 15:12, Kjetil Halvorsen
wrote:
> emacs23+ess is great BUT there is a very large problem which is
> driving me away from it:
> (on ubuntu) since version 23, emacs has a very problematic bug, which
> I have reported as:
>
> bug#6193: emacs 23 in ubuntu 10.04 gnu linux
I think that the problem is probably that y and lag(y,k=1) just get coerced
back to numeric (i.e. their time-series attributes ar lost). So your model
formula is equivalent to y~ + y, which is probably not what you wanted.
If y is in time order, I would do something like
bellver$y.1 <- c
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