Dieter Menne wrote:
srinivasa raghavan gmail.com> writes:
I am interested to generate dashboard and reports based on data from MS
Access. These reports need to be posted on a weekly basis to the web. The
reporting interface should provide facilities for "what if" scenarios.
Is it possible
Hi, Gurus
Thanks to your good helps, I have managed starting the use of a text
mining package so called "tm" in R under the OS of Win XP.
However, during running the tm package, I got another mine like memory problem.
What is a the best way to solve this memory problem among increasing a
physica
There is a mechanism for testing code in R packages (R CMD check), see
the Writing R extensions manual. If you need more flexibility for your
tests, you could look at RUnit on CRAN, or svUnit on R-Forge
(http://r-forge.r-project.org, on CRAN soon). For the later one, you
install it by:
instal
Hi
r-help-boun...@r-project.org napsal dne 15.01.2009 04:47:39:
>
> Dear R-Users
> if I have a data frame (or zoo data) as follows:
>
> Run Bike Walk Drive
> 1 2 7 5
> 5 2 4 2
> 8 3 2 1
>
> How can I re-order it so that it reads:
>
> Drive Run
I think the OP was asking about test suites that test the software.
The R package structure includes a test/ directory which you can use
to put tests.
For example, in the onion package I check that I have got my
signs and multiplication table correctly implemented:
stopifnot(Hi*Hj == Hk)
sto
Use the is.na function to test for NA values. And do read about
vectorizing your code. You don't need loops for this problem. Without
loops your code will run more than 400 times faster!
> n <- 1000
> x <- matrix(data=rep(c(1,2,3,NA), n), ncol=n, nrow=n)
>
> system.time({
+ y <- matrix(data=0
For the record, I thought I'd share two findings:
First, the web of science website does seem to have some sort of API,
as discussed here:
http://scientific.thomson.com/support/faq/webservices/
It does not seem like a trivial thing to set up though.
Second, because I could not pass the searc
Hi Nathan,
in addition to what others have already mentioned there is some documentation
in the Writing R Extensions Manual:
- on the supported structure of packages, indicating where test code might be
added
http://stat.ethz.ch/R-manual/R-patched/doc/manual/R-exts.html#Package-subdirectories
-
Hi,
I'm trying to make a chromosomal map in R by using the locus. I have a list
of genes and their locus, and I want to visualise that so you can see if
there are multiple genes on a specific place on a chromosome. A example of
what I more or less want is below:
http://www.nabble.com/file/p214742
thank you,
it worked properly.
On Thu, Jan 15, 2009 at 4:53 AM, Marc Schwartz wrote:
> on 01/14/2009 03:32 PM John Lande wrote:
> > dear all,
> >
> > I want to plot a kaplan Meier plot with the following functions, but I
> fail
> > to produce the plot I want:
> >
> > library(survival)
> > tim <-
Whoops, it seems I could use some help with regular expressions...
Consider the following two functions, creating a search string, and
retrieving the content from the url,
makeURLsearch <- function(key, dates=c(NULL, NULL)){
base.search <- "http://scholar.google.co.uk/scholar
Ladies and gentlemen,
this email is directed to all German-speaking members of the r-help list.
We from the chair of statistics and quantitative methods on the Catholic
University of Eichstätt-Ingolstadt, Germany, are searching for a new
colleague. You will find the corresponding job offer
How about:
remove.constant.values<-function(x,MARGIN,value2remove) {
is.constant.line<-function(x,value2remove) { return(any(x!=value2remove)) }
return(unlist(apply(x,MARGIN,is.constant.line,value2remove)))
}
x[,remove.constant.values(x,2,0)]
Jim
__
Dear all,
I am trying to estimate a system of equations with a Seemingly Unrelated
Regression. However because of the characteristics of the data I'd like to
do it with a negative binomial estimation. Would anybody know how to
implement a Seemingly Unrelated Negative Binomial (SUNB) estimation in R
Sorry for my late reply.
Thank you so much Jim. This script of yours
is very2 useful. I have used it.
- Gundala Viswanath
Jakarta - Indonesia
On Wed, Jan 14, 2009 at 12:17 AM, jim holtman wrote:
> Here is a function I use to see how big the objects in my workspace are:
>
>> my.ls <-
> + funct
Hi,
here's a possibility!
Your problem can be restated as "given 2 observers giving 2 measures what
is their level of agreement" - the classical measure here is the Kappa (see
sec 10.5 of Categorical Data Analysis by Alan Agresti (in Ed 1 - Ed 2
should also have it!) and you can also model the si
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Sorry list, I guess I sent an html reply.
Hi Sake:
If you do not find an answer within the list, MapChart will probably do
what you want (without the fancy chromosomal bands, though). You can
find it at
http://www.biometris.wur.nl/uk/Software/MapChart/
Hope this helps
Sake escribió:
Hi,
I
Hello,
I create this array:
x <- cbind(c(1:4, rep(0,10)), c(rep(0,4), 1:2, rep(3,6), 4,5))
[,1] [,2]
[1,]10
[2,]20
[3,]30
[4,]40
[5,]01
[6,]02
[7,]03
[8,]03
[9,]03
[10,]03
[11,]03
[12,]0
Try this:
x[,2][x[,1][x[,1] > 0]] <- table(x[,2])[as.character(x[,1][x[,1] > 0])]
On Thu, Jan 15, 2009 at 10:36 AM, Guillaume Chapron <
carnivorescie...@gmail.com> wrote:
> Hello,
>
> I create this array:
>
> x <- cbind(c(1:4, rep(0,10)), c(rep(0,4), 1:2, rep(3,6), 4,5))
>
> [,1] [,2]
> [1
Thank you! This is exactly what I wanted. Could you please explain the
logic behind your code?
x[,2][x[,1][x[,1] > 0]] <- table(x[,2])[as.character(x[,1][x[,1] >
0])]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-he
hello.
I have a question on the interpretation of a logistic model.
is it helpful to exponentiate the coefficients (estimates)? I think I
once read something about that, but I cannot remember where.
if so, how would be the interpretation of the exp(estimate) ?
would there be a change of the i
Hi Sean,
as.layer() does not do what you think it does: it does not attempt to
plot things on a common scale, it simply draws the panels of two
lattice plots in the same space. Actually, it is not very useful on
its own.
What you want is doubleYScale() in the development version (0.5-5) of
lattic
Guillaume,
# First, you need "extract" the values of x[,1] greater than 0:
x[,1][x[,1] > 0]
# After, you want the number of times that appears in the second column:
table(x2)
# but only to values above:
table(x2)[as.character(x[,1][x[,1] > 0])]
# Now, just put the values in column 2:
x[,2][x[,1
Dear List, Dear Jim,
is it possible to draw multiple polygons with different line types?
lty=c or line.lty=c do not work with radial.plot (in the matrix case) as
well as add=TRUE.
Stefan
Jim Lemon schrieb, Am 14.11.2008 10:38:
Jeremy Claisse wrote:
Is it possible to plot multiple polygons
table(x2)[as.character(x[,1][x[,1] > 0])]
Why do I need as.character() here? I checked it does not work without,
but I don't see why. The help says "as.character attempts to coerce
its argument to character type".
Thanks very much!
Guillaume
__
Because we want the values in table(x[,2]) where the **names** are
as.character(x[,1][x[,1] > 0])], not the positions x[,1][x[,1] > 0])].
Thanks!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the p
Because we want the values in table(x[,2]) where the **names** are
as.character(x[,1][x[,1] > 0])], not the positions x[,1][x[,1] > 0])].
Best
On Thu, Jan 15, 2009 at 10:59 AM, Guillaume Chapron <
carnivorescie...@gmail.com> wrote:
> table(x2)[as.character(x[,1][x[,1] > 0])]
>>
>
> Why do I need
Hi
Is any multiple regression-like test with correction for autocorrelation ?
Wojciech
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http:
I'm not getting that difference:
> .Machine$double.eps
[1] 2.220446e-16
# so I don't think that explains the different behavior.
> WB<-as.matrix(read.table(file.choose()))
> tcp1 <- tcrossprod(WB)
> tcp2 <- crossprod(t(WB))
> tcp1
[,1] [,2] [,3]
[1,] 1916061939 2281366606
Greg and Ben,
Thanks for the suggestions. I'll give it a try, and I'll also poke around to
find the r-sig-mixed-models list. What a wonderful world we live in that
such a thing exists!
--Lee
--
View this message in context:
http://www.nabble.com/power-analyses-for-mixed-effects-lmer-models-tp2
Dear Qing,
look at the file "svminternals.txt" in the /doc subdirectory of the
installed e1071 package (or inst/doc in the source package), that should
help.
David
>
> In the svm() function in the package e1071, is there anyway to get the
> signed distance of a sample point to the separating
Hello,
I will use the lca method in the e1071 package. But I get the following error:
Error in pas[j, ] <- drop(exp(rep(1, nvar) %*% log(mp))) :
number of items to replace is not a multiple of replacement length
Does anybody know this error and knows what this means?
Kind regards,
Tryntsje
_
Hi everyone:
I want to fit the truncated polynomial smoothing to the quantiles
instead of means, does someone know how to do it? I am thinking that
maybe I can use SemiPar package, but can not find how.
Thanks so many,
Suyan
__
R-help@r-project.
Hi,
I have a binary matrix and I would like from it make a cluster by
agglomerative method (like agnes), and then generate a ascii art dendrogram.
See the illustration:
binary matrixascii art dendrogram
ABCDEFGH
P1 0101011101
Elisa Lanzi unive.it> writes:
>
> Dear all,
> I am trying to estimate a system of equations with a Seemingly Unrelated
> Regression. However because of the characteristics of the data I'd like to
> do it with a negative binomial estimation. Would anybody know how to
> implement a Seemingly Unrel
If I want to make a numerical series, I can do so easily with:
series.numbers <- 1:10
But, I don't seem to be able to do the same with time. I want to create
a vector with 480 points that corresponds to the 480 minutes in a 8 hour
work day. Thus I want series.time to look something like t
On 13/01/2009, David Winsemius wrote:
> It's fairly clear from the documentation that approxfun() will not
> extrapolate.
>
> help.search("extrapolate")
> library(Hmisc)
> ?approxExtrap
>
> Some sort of minimization approach:
>
> > approxExtrap(x=c(0,5,10,15,20), y=c(16,45,77,101,125),xout=c(-4,0
Dear R-helpers,
I have a problem with the tcl "after" instruction. When I send:
> library(tcltk)
Loading Tcl/Tk interface ... done
> tcl("after",1000,cat("try tcl after\n"))
try tcl after
the tcl command works fine. Similarly, the tcl command:
> tcl("after",1000,plot(rnorm(100)))
works fine.
On Jan 15, 2009, at 1:10 AM, John Kerpel wrote:
Hi folks!
I run the following code to get a CI for a Poisson with lambda=12.73
library(MASS)
set.seed(125)
x <- rpois(100,12.73)
lambda_hat<-fitdistr(x, dpois, list(lambda=12))$estimate
#Confidence Intervals - Normal Approx.
alpha<-c(.05,.02
Try this:
library(chron)
times("9:00:00") + 0:479/(24*60)
Also see R News 4/1 and read the help pages in chron.
If you eventually want to use these times with corresponding series
data look at the zoo package.
On Thu, Jan 15, 2009 at 10:01 AM, Andrew Choens wrote:
> If I want to make a numeric
On 1/15/2009 10:18 AM, davide.massi...@unipd.it wrote:
Dear R-helpers,
I have a problem with the tcl "after" instruction. When I send:
library(tcltk)
Loading Tcl/Tk interface ... done
tcl("after",1000,cat("try tcl after\n"))
try tcl after
the tcl command works fine. Similarly, the tcl com
See also ll() in the R.oo package, e.g.
# To list all objects in .GlobalEnv:
ll()
member data.class dimension objectSize
1*tmp* Person 1 428
2 as.character.Person function NULL1208
3 country character 1
> I will use the lca method in the e1071 package. But I get the following
error:
> Error in pas[j, ] <- drop(exp(rep(1, nvar) %*% log(mp))) :
> number of items to replace is not a multiple of replacement length
>
> Does anybody know this error and knows what this means?
The error means that yo
hello friends
i have a created a database table in MYSQL consisting of 11 columns.
throught RMYSQL i managed to read the entire table in R. but i have few
qureries which i need solutions...here they are:
1. Using R how to read a one column alone from a database table from MYSQL.
2. Using R how t
(I'm quite afraid this is a dumb question, so sorry in advance)
I am trying to install the package rimage to R version 2.8.1 with Ubuntu
Intrepid. When I type "sudo R CMD INSTALL" I get the following error
message (pasted below). I know the same problem appears here
(http://www.r-project.org/nosv
Dear Sir/Madam,
Thanks for the software.
May I know that R can run on Vista system or not? Cause i donwload but cant
install.. =(
Thanks s much!!
--
Best Regards,
Ivy
[[alternative HTML version deleted]]
__
R-help@r-project.org m
On Jan 15, 2009, at 10:04 AM, e-letter wrote:
On 13/01/2009, David Winsemius wrote:
It's fairly clear from the documentation that approxfun() will not
extrapolate.
help.search("extrapolate")
library(Hmisc)
?approxExtrap
Some sort of minimization approach:
approxExtrap(x=c(0,5,10,15,20),
Duncan Murdoch wrote:
> On 1/15/2009 10:18 AM, davide.massi...@unipd.it wrote:
>> Dear R-helpers,
>> I have a problem with the tcl "after" instruction. When I send:
>>
>>> library(tcltk)
>> Loading Tcl/Tk interface ... done
>>
>>> tcl("after",1000,cat("try tcl after\n"))
>> try tcl after
>>
>>
>>
> Perhaps a coding error on my part (or on your part). Perhaps different
> methods (none of which you describe)?
>
> I suspect that my method only used the first two points (I just
> checked by plotting and -2.7 is closer to the paper and pen result I
> get than is -3.28. Perhaps you made an extra
On 1/15/2009 6:53 AM, Ziqi wrote:
Dear Sir/Madam,
Thanks for the software.
May I know that R can run on Vista system or not? Cause i donwload but cant
install.. =(
Thanks s much!!
R can run on Vista. See the FAQ (online at
http://cran.r-project.org/bin/windows/base/rw-FAQ.html, yo
Hi list,
I´m working on a predictive modeling task using the caret package.
I found the best model parameters using the train() and trainControl() command.
Now I want to evaluate my model and make predictions on a test dataset. I tried
to follow the instructions in the manual and the vignettes b
Prof Brian Ripley a écrit :
On Tue, 13 Jan 2009, Charles C. Berry wrote:
On Tue, 13 Jan 2009, Matthieu Stigler wrote:
Hello
I'm trying to run a fortran code which use LAPACK subroutines. I
think I should use some points shown in the manual 5.5 Creating
shared objects but it is too technica
On Jan 15, 2009, at 11:31 AM, e-letter wrote:
Perhaps a coding error on my part (or on your part). Perhaps
different
methods (none of which you describe)?
I suspect that my method only used the first two points (I just
checked by plotting and -2.7 is closer to the paper and pen result I
g
sankar82 tkk.fi> writes:
> i have a created a database table in MYSQL consisting of 11 columns.
> throught RMYSQL i managed to read the entire table in R. but i have few
> qureries which i need solutions...here they are:
>
> 1. Using R how to read a one column alone from a database table from MY
On 15 January 2009 at 03:06, abbethesieyes wrote:
|
| (I'm quite afraid this is a dumb question, so sorry in advance)
| I am trying to install the package rimage to R version 2.8.1 with Ubuntu
| Intrepid. When I type "sudo R CMD INSTALL" I get the following error
| message (pasted below). I know
It appears the answer to your goal after a discursive exploration of
"interpolation", which was really extrapolation, is that you need to
look at the predict methods for linear (and other sorts as well) models.
?predict
?predict.lm
> y <- c(16,45,77,101,125)
> x <- c(0,5,10,15,20)
>
> lmmo
On Sun, Jan 11, 2009 at 11:05 AM, Kingsford Jones
wrote:
> I hope that Marc doesn't mind, but I felt that part of his recent post
> was important enough to deserve it's own subject line rather then
> being lost in a 60-msg-long thread...
I also wanted to thank Marc for this wealth of information
On Thu, 15 Jan 2009, dos Reis, Marlon wrote:
I'm using:
"solve(a, b, tol, LINPACK = FALSE, ...)"
Therefore,tol ==.Machine$double.eps
.Machine$double.eps
[1] 2.220446e-16
It explains why 'solve' works for 'tcp1' but not for 'tcp2':
eigen(tcp1)$values
[1] 5.208856e+09 2.585816e+08 -3.660671e
Here's another help file question.
Some context: University setting wherein R is installed for
availability to students and course instructors across campus in
various PC labs. Windows Vista environment.
Goal: To maximize flexibility and functionality of installing add-on
packages and as
Hi Stephen,
> #I am putting a test together for an introductory biology class and I
> would like to put different cross hatching inside of each bar for the
> bar plot below
ggplot2 uses the grid package to do all the drawing, and currently
grid doesn't support cross-hatching, so unfortunately the
This will be fixed in the next version, but until then you can do
title = "Aquarium\n"
Hadley
On Wed, Jan 14, 2009 at 9:24 PM, stephen sefick wrote:
> Also notice that the q in Aquarium is hidden. Is there a way to make
> this not happen?
> thanks
>
> Stephen Sefick
>
> On Wed, Jan 14, 2009 at
I have a question on whether a warning message is valid or if I just don't
understand the process. Let me illustrate via some R code:
x <- 1:20
i <- x %% 2 > 0
y <- rep(1,20)
x[i] <- y
Warning message:
In x[i] <- y :
number of items to replace is not a multiple of replacement length
But it st
The lengths are different, particularly the length of subsetted x[i]
> x <- 1:20
> i <- x %% 2 > 0
> y <- rep(1,20)
> length(x)
[1] 20
> length(i)
[1] 20
> length(x[i])
[1] 10
> length(y)
[1] 20
You happened to be lucky and got what you wanted, but a more reliable
approach is:
> x[i] <- y[i]
S
This was just an illustration. It is the warning message that I don't
understand. The warning says "number of items to replace is not a multiple of
replacement length". The way I look at it 10 is a multiple of 20.
Kevin
Sarah Goslee wrote:
> The lengths are different, particularly the le
rkevinbur...@charter.net wrote:
> This was just an illustration. It is the warning message that I don't
understand. The warning says "number of items to replace is not a
multiple of replacement length". The way I look at it 10 is a multiple
of 20.
Um, with a multiplier of 0.5 ?
You're trying to p
on 01/15/2009 11:42 AM David M Smith wrote:
> On Sun, Jan 11, 2009 at 11:05 AM, Kingsford Jones
> wrote:
>> I hope that Marc doesn't mind, but I felt that part of his recent post
>> was important enough to deserve it's own subject line rather then
>> being lost in a 60-msg-long thread...
>
> I al
dear All,
i'm trying to calculate the number of Mondays, Tuesdays, etc that each
month within a date range has. I have time series data that spans 60
months and i want to calculate the number of Mondays, Tuesdays, Wed,
etc of each month. (I want to control for weekly seasonality but my
dat
?chron() in particular day.of.week
-Roy
On Jan 15, 2009, at 11:28 AM, Carlos Hernandez wrote:
dear All,
i'm trying to calculate the number of Mondays, Tuesdays, etc that
each month within a date range has. I have time series data that
spans 60 months and i want to calculate the number of M
On Jan 15, 2009, at 2:28 PM, Carlos Hernandez wrote:
dear All,
i'm trying to calculate the number of Mondays, Tuesdays, etc that
each month within a date range has. I have time series data that
spans 60 months and i want to calculate the number of Mondays,
Tuesdays, Wed, etc of each month
Try this:
> library(zoo) # as.yearmon
> dd <- seq(as.Date("2000-01-01"), as.Date("2004-12-31"), "day")
> dow <- as.numeric(format(dd, "%w"))
> ym <- as.yearmon(dd)
> tab <- do.call(rbind, tapply(dow, ym, table))
> rownames(tab) <- format(as.yearmon(as.numeric(rownames(tab
> head(tab)
Carlos Hernandez wrote:
> dear All,
> i'm trying to calculate the number of Mondays, Tuesdays, etc that each
> month within a date range has. I have time series data that spans 60
> months and i want to calculate the number of Mondays, Tuesdays, Wed, etc
> of each month. (I want to control for week
Thanks, I knew it was something simple :)
Petr Pikal wrote:
>
> Hi
>
> r-help-boun...@r-project.org napsal dne 15.01.2009 04:47:39:
>
>>
>> Dear R-Users
>> if I have a data frame (or zoo data) as follows:
>>
>> Run Bike Walk Drive
>> 1 2 7 5
>> 5 2 4 2
>>
This helps a lot. I have options(htmlhelp=TRUE) and
options(chmhelp=FALSE) (else problems). Now, ?, help(), and
help.search() seem to work well. But, help.start() appears restricted
to packages in the site library. I see, in the Help/FAQ on R for
Windows/ Sections 4.3 and 4.4, that I ca
On 1/15/2009 3:17 PM, Jarrett Barber wrote:
This helps a lot. I have options(htmlhelp=TRUE) and
options(chmhelp=FALSE) (else problems). Now, ?, help(), and
help.search() seem to work well. But, help.start() appears restricted
to packages in the site library. I see, in the Help/FAQ on R f
Hi there,
Thanks for the help.
I see now where my results are coming from.
Marlon.
-Original Message-
From: Charles C. Berry [mailto:cbe...@tajo.ucsd.edu]
Sent: Friday, 16 January 2009 6:26 a.m.
To: dos Reis, Marlon
Cc: David Winsemius; r-help@r-project.org
Subject: RE: [R] Precision in
Whoops! Yes, my last "help.search()" should have been
"help.start()". Thanks again. -- Jarrett
On Jan 15, 2009, at 1:27 PM, Duncan Murdoch wrote:
On 1/15/2009 3:17 PM, Jarrett Barber wrote:
This helps a lot. I have options(htmlhelp=TRUE) and
options(chmhelp=FALSE) (else problems). Now
Or for those not allergic to reading help, see ?weekdays .
Just how hard do you have to work to miss that? E.g. ??day works.
On Thu, 15 Jan 2009, Peter Dalgaard wrote:
Carlos Hernandez wrote:
dear All,
i'm trying to calculate the number of Mondays, Tuesdays, etc that each
month within a date
On 1/15/2009 1:32 PM, David Winsemius wrote:
On Jan 15, 2009, at 12:25 PM, Charles C. Berry wrote:
This is what I get on windows XP:
tcp1-tcp2
[,1] [,2] [,3]
[1,] -2.861023e-06 -4.768372e-07 -4.768372e-07
[2,] -4.768372e-07 -3.814697e-06 2.622604e-06
[3,] -4.
Indeed, i overlooked weekdays.
Thank you all for your replies!
On Jan 15, 2009, at 21:23 , Prof Brian Ripley wrote:
Or for those not allergic to reading help, see ?weekdays .
Just how hard do you have to work to miss that? E.g. ??day works.
On Thu, 15 Jan 2009, Peter Dalgaard wrote:
Carl
On Jan 15, 2009, at 12:25 PM, Charles C. Berry wrote:
This is what I get on windows XP:
tcp1-tcp2
[,1] [,2] [,3]
[1,] -2.861023e-06 -4.768372e-07 -4.768372e-07
[2,] -4.768372e-07 -3.814697e-06 2.622604e-06
[3,] -4.768372e-07 2.622604e-06 -5.960464e-08
bu
On 16/01/2009, at 1:50 AM, gregor rolshausen wrote:
hello.
I have a question on the interpretation of a logistic model.
is it helpful to exponentiate the coefficients (estimates)? I think I
once read something about that, but I cannot remember where.
if so, how would be the interpretation of
It's the other way around.
You are trying to replace 10 elements (x[i]) with 20 elements (y). R
makes a "best
guess" as to how you want to do that. 10 is not a multiple of 20.
If you were trying to replace 20 elements with 10, then R would recycle them
because 20 _is_ a multiple of 10.
The safes
Hello,
in a desperate desire of using partial function application in R I fried out
the following piece of code:
bind <- function( f, ... ) {
args <- list(...)
function(...) f( ..., unlist(args) )
}
Its purpose, if not clear, is to return a function with part of its
arguments bound to speci
On 1/15/2009 12:50 PM, Jarrett Barber wrote:
Here's another help file question.
Some context: University setting wherein R is installed for
availability to students and course instructors across campus in
various PC labs. Windows Vista environment.
Goal: To maximize flexibility and functi
How is function() not the correct approach?
> plot_lines <- function(x, ...) plot(x, type="l", ...)
>
> plot_lines(1:10, xlim = c(1,5))
> plot_lines(1:10, 11:20, xlim = c(1,5))
Still seems to get the unnamed optional y argument to the plotting
machinery.
--
David Winsemius
On Jan 15, 2009
Have a look at the setDefaults package. It will set the default
arguments of a function to whatever you specify so that
if you omit them then those are the values you get for them.
On Thu, Jan 15, 2009 at 4:25 PM, nosek wrote:
>
> Hello,
>
> in a desperate desire of using partial function applic
Well,
it looks like it's a perfectly correct approach to bind functions writing
their wrappers by hand.
But I don't want to write them by hand every time I need them.
Being lambda expression, function() is most general, but there must be some
kind of shorter way for such a common task as partial
Hi,
Today I came across the R application and I will admit I am not a
Statistician. However, I think this application will be useful for me
at work. I am a Network/System Security Engineer trying to make sense
of the huge security data I collect. I am trying to visualize the
traffic on our netw
One other idea. The proto package also does currying. If f a method
(i.e. an R function
that takes an object as arg1 then p$f, i.e. the $.proto function,
returns function(...) f(p, ...).
Looking at the code for setDefaults as in my prior response and/or
proto should give
you some ideas.
On Thu,
Hello, I am getting requests to place our PDF slides (output from
beamer) into Microsoft Powerpoint formats (.ppt). What's the best
practice or any recommended software packages (any success with open or
commercial) that we can use to convert PDF slides into an EDITABLE
powerpoint deck?
Tha
Dear R-Users
I have 2 questions to do with XYplot.
1)
I am trying to use the XYplot function to generate multiple line graphs with
the legend outside the plot.
I am using the following loop for each graph:
library(lattice)
for (i in x.sp){
xyplot(Catch~Year, df, groups = Stock, type="a",aut
Subba Rao wrote:
> Hi,
>
> Today I came across the R application and I will admit I am not a
> Statistician. However, I think this application will be useful for me
> at work. I am a Network/System Security Engineer trying to make sense
> of the huge security data I collect. I am trying to visu
Hello, I'm EDWIN, I create (make) GUI, with call many function
but I don't know why when I call function I can't. if without function, Yes
I can..
can you help me ? can you make this, become true with full code?
Can you help me to
create data.entry with interface
LM - BETA1.HAT - BETA2.HAT SD.BET
On Jan 15, 2009, at 9:27 PM, jimdare wrote:
Dear R-Users
I have 2 questions to do with XYplot.
1)
I am trying to use the XYplot function to generate multiple line
graphs with
the legend outside the plot.
I am using the following loop for each graph:
library(lattice)
for (i in x.sp){
xy
Short course: Statistical Learning and Data Mining III:
Ten Hot Ideas for Learning from Data
Trevor Hastie and Robert Tibshirani, Stanford University
Sheraton Hotel
Palo Alto, CA
March 16-17, 2009
This two-day course gives a detailed overview of statistical models
for
data mining, in
czesc,
looks like you want some sort of currying, or maybe partial currying,
right? anyway, here's a quick guess at how you can modify your bind,
and it seems to work, as far as i get your intentions, with the plot
example you gave:
bind = function(f, ...) {
args = list(...)
function(...)
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