Michael Lachmann wrote:
> Hi,
>
> R seems to have a feature that isn't used much, which I don't really
> now how to call. But, the dimnames function, can in addition to giving
> names to rows/columns/dim 3 rows/dim 4 rows... can also give labels to
> the dimensions themselves.
Actually, thi
David Winsemius wrote:
> On Jul 26, 2010, at 2:36 PM, xin wei wrote:
>
>> hi, this is more a statistical question than a R question. but I do
>> want to
>> know how to implement this in R.
>> I have 10,000 data points. Is there any way to generate a empirical
>> probablity distribution from it (
Hi Sherri,
There are examples of topographic maps which you have been pointed to, however,
I suspect that you want to know where you can obtain topographic data from
rather than a canned example.
There are quite a few intricacies to the process so I will go through them for
you.
(1)
Topograph
This is an example that I have found to be very useful example, and one that
I have adapted myself:
http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=1
On Sun, Jul 25, 2010 at 6:27 PM, David Winsemius wrote:
>
> On Jul 25, 2010, at 6:31 PM, sh...@ucar.edu wrote:
>
> Hi All-
>>
>> I w
Please type RSiteSearch("vignette") in the console and you will see what
vignette is. I am also new to R and only learnt this week that you could
always find out about these complicated jargon by typing
RSiteSearch("...") where ... would be whatever you want to search,
thanks to the discussions on
Hello,
My impression is that the OOP package is not used that often.
For alternatives (R.oo, proto) see the posts of Gabor Grothendieck:
http://www.mail-archive.com/r-help@r-project.org/msg73565.html
http://www.mail-archive.com/r-help@r-project.org/msg75308.html
Friedrich Schuster
Dompfaf
On Tue, Jul 27, 2010 at 12:41 AM, Deepayan Sarkar
wrote:
> On Mon, Jul 26, 2010 at 9:11 PM, Prasenjit Kapat wrote:
>> Hi,
>>
>> (please Cc me)
>>
>> In xyplot (), type = "l" (or one that includes "l", *el*) is
>> (generally) meaningful only when the 'x' variable is sorted. In
>> practice, one eit
On Mon, Jul 26, 2010 at 9:11 PM, Prasenjit Kapat wrote:
> Hi,
>
> (please Cc me)
>
> In xyplot (), type = "l" (or one that includes "l", *el*) is
> (generally) meaningful only when the 'x' variable is sorted. In
> practice, one either sorts the data frame before hand or writes a tiny
> panel funct
Isn't there a danger that what you teach will then be driven by what templates
you receive?
I would have thought this was an easy thing to do yourself, once you have
decided what you want to teach your students. Just write a scropt (e.g. using
the inbuilt script edit in R for Windows if you ca
try type = "a"
-Felix
On 27 July 2010 14:11, Prasenjit Kapat wrote:
> Hi,
>
> (please Cc me)
>
> In xyplot (), type = "l" (or one that includes "l", *el*) is
> (generally) meaningful only when the 'x' variable is sorted. In
> practice, one either sorts the data frame before hand or writes a tin
On Mon, Jul 26, 2010 at 6:54 PM, Paul Smith wrote:
> On Mon, Jul 26, 2010 at 11:42 PM, Gabor Grothendieck
> wrote:
>>> The following code should return 1, but it returns 0:
>>>
>>> source("http://r-bc.googlecode.com/svn/trunk/R/bc.R";)
>>> bc("9 % 2")
>>>
>>
>> See FAQ 2 on the r-bc package home
Hi,
(please Cc me)
In xyplot (), type = "l" (or one that includes "l", *el*) is
(generally) meaningful only when the 'x' variable is sorted. In
practice, one either sorts the data frame before hand or writes a tiny
panel function which sorts the supplied x and then calls the default
panel.xyplot(
I am introducing the scan() function to my class. Consider the following
file (Scanexamp.Rnw )
\documentclass[12pt]{article}
\begin{document}
<<>>=
height = scan()
64 62 66 65 62
69 72 72 70
part = scan(what = character(0))
"Soprano" "Soprano" "Soprano"
"Alto""Alto""Tenor"
"Tenor"
One method:
dd <- do.call(make.groups, mydata[,-1])
dd$X <- mydata$X
xyplot(data ~ X | which, dd)
Another method:
form <- paste(paste(colnames(mydata)[-1], collapse = " + "), "~ X")
xyplot(as.formula(form), mydata)
Yet another method:
library(latticeExtra)
xyplot.list(mydata[,-1], FUN = functi
plyr is a set of tools for a common set of problems: you need to break
down a big data structure into manageable pieces, operate on each
piece and then put all the pieces back together. For example, you
might want to:
* fit the same model to subsets of a data frame
* quickly calculate summary
Does anyone know of any tools to do multichannel(multivariate) singular
spectrum analysis (M-SSA) in R? Seems like this would be a useful thing to
have.
-Hilbert
[[alternative HTML version deleted]]
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https://
Please look at RExcel, available through the RExcelInstaller package from
CRAN or
included with a full R download from rcom.univie.ac.at (go to the download
page and take
the most recent RAndFriends).
You can use the supplementary book R through Excel by Erich Neuwirth
begin_of_the_skype_highlighti
On Mon, Jul 26, 2010 at 11:42 PM, Gabor Grothendieck
wrote:
>> The following code should return 1, but it returns 0:
>>
>> source("http://r-bc.googlecode.com/svn/trunk/R/bc.R";)
>> bc("9 % 2")
>>
>
> See FAQ 2 on the r-bc package home page:
> http://r-bc.googlecode.com
Thanks to all respondents f
Hi,
I am trying to create a spatial weight matrix based on the neighboring
relationship among Europan countries. Can I use the map function in R to
generate a European map and then use this map to build the w weight matrix? So
far, I cannot create a map for Finland using the map function
Yes, thanks. Just found out the solution. Thanks for the help.
Just started R. Not familiar with its environment.
G
On Mon, Jul 26, 2010 at 5:08 PM, jim holtman wrote:
> Is this what you want:
>
> > equated<-c(111.0,112.06, 112.9, 113.8, 115.0, 116.2, 117.0, 118.0, 120.5,
> + 120.5, 120.5)
> >
"when the first level denotes failure and all others success"
Yes, I saw this sentence in the glm help file, but I hadn't understood it
this way... Anyway I checked this with a few examples and this is exactly
what it does.
Thanks a lot for your help !!!
I can go back now to the polr function a
Thanks, it works. I found out the solution a moment ago. The 2nd one works.
But, the weird thing is that if I use 'x', it works. If I use 'equated', it
didn't work. Not sure why.
Thanks again.
G
On Mon, Jul 26, 2010 at 5:04 PM, Joshua Wiley wrote:
> What about these two options?
>
> #One way
>
Hi
I am looking for R templates to introduce the R to my students at
Seton hall university. The templates are predefined scripts in R that
will retain its primary intent when individually customized with their
own variable data or text. In this case, my students at Seton Hall
University. For exampl
On Mon, Jul 26, 2010 at 10:00 PM, Ben Bolker wrote:
> Gabor Grothendieck gmail.com> writes:
>
>> See FAQ 2 on the r-bc package home page:
>> http://r-bc.googlecode.com
>
> This is getting a little bit self-referential -- that FAQ refers to
> the previous message in this thread ...
>
The above w
Hi, I have a data.frame with columns named X, D1, D2, D3
I know I can get a single plot with 3 curves by doing
xyplot(D1 + D2 + D3 ~ X, data)
but in some cases I might have columns D1 ... D10.
Is there a way to plot all 10 columns without having to specify each
individual term?
(By analogy wi
Gabor Grothendieck gmail.com> writes:
> See FAQ 2 on the r-bc package home page:
> http://r-bc.googlecode.com
This is getting a little bit self-referential -- that FAQ refers to
the previous message in this thread ...
__
R-help@r-project.org mailin
Hi, folks,
x=1:10
y=rep(2:6,2)
lin=lm(y~x)
x=3:12
new=predict(lin,se.fit=T)
#se.fit: the standard error of the predicted means, namely, the square root
of Var( E[y|x] | x)
# How can I generate the variances of the new observations? Namely the
square root of var(y|x),
## Which I think should be mu
On Jul 26, 2010, at 3:22 PM, vacas wrote:
Hi I am new to R.
I am having this problem
t1 <- read.csv("myfile.csv")
t2 <- data.frame(t1)
which have 10 row and 10 columns
t2[1,1] does not give the first element but it gives the levels, how
can I
fix it.
It gives you both. Factors are general
On Jul 26, 2010, at 3:22 PM, vacas wrote:
Hi I am new to R.
I am having this problem
t1 <- read.csv("myfile.csv")
t2 <- data.frame(t1)
which have 10 row and 10 columns
t2[1,1] does not give the first element but it gives the levels, how
can I
fix it.
It gives you both. Factors are general
Dear R-list members,
Once a gam (package gam) model has been fitted with family=poisson,
is there some R function that could extract the diagonal elements
of the smoother matrix S, to be used in a cross-validation for the
selection of the best smoothing parameter, following equation 3.19
on page
Hi Alexandre,
You can either initialize the table (using, maybe, all NA entries)
and then put a for loop around your analysis code and write to
different entries or (and this is probably the better way to do it,
but might be more time consuming the first time you try) put the
matrices into an arr
Dear Friends,
Using package Vegan, I need to calculate Shannons Diversity index
and Pielou's Evenness for a set of 20 study areas. Each area is
represented by a matrix of 25 sample plots x tree species. The code is
as following, where data stands for the data matrix of any of the 20
areas:
S
On Jul 26, 2010, at 2:36 PM, xin wei wrote:
hi, this is more a statistical question than a R question. but I do
want to
know how to implement this in R.
I have 10,000 data points. Is there any way to generate a empirical
probablity distribution from it (the problem is that I do not know
w
On Jul 26, 2010, at 10:56 AM, Steffen Uhlig wrote:
Dear David, Petr, and Alain,
thank you very much for your fast responses. It's a typical
"handbook-not-read-error" at my side. I will dig deeper into the
plot-functions and the assignment of data. I was not aware of that
the vector "a" i
On Mon, Jul 26, 2010 at 1:44 PM, Paul Smith wrote:
> Dear All,
>
> The following code should return 1, but it returns 0:
>
> source("http://r-bc.googlecode.com/svn/trunk/R/bc.R";)
> bc("9 % 2")
>
See FAQ 2 on the r-bc package home page:
http://r-bc.googlecode.com
Paul Smith gmail.com> writes:
>
> Dear All,
>
> The following code should return 1, but it returns 0:
>
> source("http://r-bc.googlecode.com/svn/trunk/R/bc.R";)
> bc("9 % 2")
>
> Do you confirm this bug?
>
It's not a bug in r-bc, it's a misfeature (?) in bc.
It has to do with the 'scale'
Hi all,
I googled this but no luck.
I am using R.app 2.10.1 on Mac OSX 10.4.
Here's the problem:
When I type at the R.app command line and hit the carriage
return "Enter" (right pinky, "Return" on some keyboards), it
just adds a blank line.
To actually get the command to execute, I have to
Hi,
Armadillo (http://arma.sourceforge.net/docs.html) has LU. Here is an
example adapted from armadillo's documentation using Rcpp/RcppArmadillo
and inline:
require( inline )
require( RcppArmadillo )
fx <- cxxfunction( signature( A_ = "matrix"), '
using namespace arma ;
mat A = as(A_);
mat
Donald,
Mixed types are handled in flexmix and in depmixS4, not sure about ordinal
in flexmix (depmixS4 does not handle ordinal but does handle multinomial,
constraints may be an
option to deal with ordinal); both have glm distributions, ie gaussian,
binary and many
others.
Best, Ingmar
On Sat, Ju
The solution to (most R problems) is as follows:
1.if asking for help include reproducible examples including parts of
your data otherwise we can just guess what kind of data you have.
2. In general, refer to the help pages of the functions you use (
help(read.csv),help(data.frame) ) (
i ha
Dear all,
Sandy Weisberg and I would like to announce version 2.0-0 of the car
package, now on CRAN. We've released this major revision of the package in
anticipation of the publication of An R Companion to Applied Regression,
Second Edition (Sage, in press), co-authored by us, which should be
ava
Dear R People:
Could someone recommend a good c/c++ code (or Fortran) for LU
decomposition, please?
Sorry to bother about this. I'm trying to do some "non-R" work that
requires a matrix inversion.
Thanks,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Scienc
Hello,
(ccing Rcpp-devel too because this is relevant)
This comes up every now and again on packages that are completely
unrelated to Rcpp. We don't know yet why or what to do to fix the issue.
I believe (but this might not be the case) that this is due to packages
that do use Rcpp and fail
Try this:
> mytable
id col1 col2 col3
1 10048
2 11122
3 12083
4 13055
> colName <- 'col3'
> mytable[order(mytable[[colName]]),]
id col1 col2 col3
2 11122
3 12083
4 13055
1 10048
> colName <- 'id'
>
What about these two options?
#One way
ifelse(equated > 120, 120, equated)
#Another way
equated[equated > 120] <- 120
HTH,
Josh
On Mon, Jul 26, 2010 at 10:26 AM, ying_chen wang
wrote:
> I am new to R. Used to use FORTRAN. R is so different from FORTRAN. The
> following codes would work in FOT
On Mon, Jul 26, 2010 at 2:06 PM, ying_chen wang
wrote:
> Thanks, it works. I found out the solution a moment ago. The 2nd one works.
>
> But, the weird thing is that if I use 'x', it works. If I use 'equated', it
> didn't work. Not sure why.
What is 'x' ?
>
> Thanks again.
>
> G
>
> On Mon, Jul
Hi I am new to R.
I am having this problem
t1 <- read.csv("myfile.csv")
t2 <- data.frame(t1)
which have 10 row and 10 columns
t2[1,1] does not give the first element but it gives the levels, how can I
fix it.
I will be thankful to community.
--
View this message in context:
http://r.789695.n
verry sorry for posting the mwssage below from wrong email account.
please reject it. sorry for the inconvenience...
mirek
On 07/26/2010 08:54 PM, r-help-boun...@r-project.org wrote:
Your mail to 'R-help' with the subject
using string variable as order() function argument
Is being held u
hi, this is more a statistical question than a R question. but I do want to
know how to implement this in R.
I have 10,000 data points. Is there any way to generate a empirical
probablity distribution from it (the problem is that I do not know what
exactly this distribution follows, normal, beta?
Hi,
Is this what you want?
##
mytable <- read.table(textConnection("
id col1 col2 col3
1004 8
1112 2
1208 3
1305 5
"), header = TRUE)
mytable
columnname <- "col3"
mytable[order(mytable[, columnname]), ]
###
Josh
Dear All,
The following code should return 1, but it returns 0:
source("http://r-bc.googlecode.com/svn/trunk/R/bc.R";)
bc("9 % 2")
Do you confirm this bug?
Paul
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE
Is this what you want:
> equated<-c(111.0,112.06, 112.9, 113.8, 115.0, 116.2, 117.0, 118.0, 120.5,
+ 120.5, 120.5)
> equated[equated > 120] <- 120
> equated
[1] 111.00 112.06 112.90 113.80 115.00 116.20 117.00 118.00 120.00
120.00 120.00
>
You should read up on 'indexing' in the R Intro paper.
I'm running a logistic regression model in R. I've used both the Zelig and Car
packages. However, I'm wondering if there is a simple way to get the model fit
statistics for the model. (pseudo R-square, chi-square, log liklihood,etc)
Thanks
___
I am new to R. Used to use FORTRAN. R is so different from FORTRAN. The
following codes would work in FOTRAN. I am trying to put an upper limit at
120. If the score is > 120, it is assigned 120. Or else, keep the original
values.
version 1:
equated<-11
result<-11
equated<-c(111.0,112.06, 112.9, 1
Hello,
In my script I would like to use a loop, which sorts the dataframe
according to different columns, pointed by the string variable.
id col1 col2 col3
1 1004 8
2 1112 2
3 1208 3
4 1305 5
Usually the order() function can be
Dear friends,
I have just gotten a strange error message back from Uwe saying
that the most recent version of psych failed to pass R CMD check for
Windows.
The error message was less than helpful, in that it seems to have
failed when trying to include the Rcpp library, which I do not
dire
The obvious:
Take a small sample, say 25-50. Get an estimate of your distribution
from that. Then use this to determine how many more (if any)
additional samples you need for desired precision. This latter can
probably easily be done via simulation/bootstrap if you don't want to
specify a paramet
Basically, we have a population of 4,392 documents and we want to find out
the number of patents per document. We don’t want to go through all 4,392
documents, but want a reliable sample size from which to draw inferences. I
feel like this count data will not follow a normal distribution, but more
okay, thanks I will try the R-devel list :)
BR
Roland
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-c
On 26/07/10 19:01, Michael Lachmann wrote:
Hi,
R seems to have a feature that isn't used much, which I don't really
now how to call. But, the dimnames function, can in addition to giving
names to rows/columns/dim 3 rows/dim 4 rows... can also give labels to
the dimensions themselves.
Thus,
On 26/07/2010 3:09 PM, rrich...@fh-lausitz.de wrote:
I'm build with the usage of the tcltk/tcltk2 package a gui which is
started with rscript.exe. At this gui the user is able to open and close
plot windows.
Now I'm looking for a way that a r function is called when a
grDevices::windows win
Dear all,
I'm doing a survival analysis with time-dependent covariates. Until now, I
have used a simple Cox model for this, specifically the coxph function from
the survival library. Now, I would like to try out an accelerated failure
time model with a parametric specification as implemented for e
I'm build with the usage of the tcltk/tcltk2 package a gui which is
started with rscript.exe. At this gui the user is able to open and close
plot windows.
Now I'm looking for a way that a r function is called when a
grDevices::windows window is closed.
Is there a way to attach a callback func
On Mon, Jul 26, 2010 at 2:24 PM, Erin Hodgess wrote:
> Whoops...sorry about that. Here we go:
>
>> dput(xgh)
> structure(c(5, 6, 1, 5, 0, 0, 13, 9, 6, 4, 6, 0, 0, 9, 15, 10,
> 5, 6, 0, 0, 0, 12, 8, 3, 6, 0), index = structure(c(14775, 14776,
> 14777, 14778, 14779, 14780, 14781, 14782, 14783, 1478
On 26/07/2010 2:56 PM, mirek wrote:
Hello,
In my script I would like to use a loop, which sorts the dataframe
according to different columns, pointed by the string variable.
id col1 col2 col3
1 1004 8
2 1112 2
3 1208 3
4 1305
Hello,
In my script I would like to use a loop, which sorts the dataframe
according to different columns, pointed by the string variable.
id col1 col2 col3
1 1004 8
2 1112 2
3 1208 3
4 1305 5
Usually the order() function can be
Whoops...sorry about that. Here we go:
> dput(xgh)
structure(c(5, 6, 1, 5, 0, 0, 13, 9, 6, 4, 6, 0, 0, 9, 15, 10,
5, 6, 0, 0, 0, 12, 8, 3, 6, 0), index = structure(c(14775, 14776,
14777, 14778, 14779, 14780, 14781, 14782, 14783, 14784, 14785,
14786, 14787, 14788, 14789, 14790, 14791, 14792, 14793
On Mon, Jul 26, 2010 at 2:17 PM, Erin Hodgess wrote:
> Dear R People:
>
> I would like to combine a zoo object with some observations at the end.
>
> Here is the set up:
>
>> xgh
> 2010-06-15 2010-06-16 2010-06-17 2010-06-18 2010-06-19 2010-06-20 2010-06-21
> 5 6 1
I doubt that there are any.
For a "bimodal" distrbution, I think one would have to specify in
detail the nature of the distribution and then define what one means
by an "outlier" (a slippery, sinister notion, at best and a flimsy
cloak for skulduggery at worst) .
As has been said her frquently be
Dear R People:
I would like to combine a zoo object with some observations at the end.
Here is the set up:
> xgh
2010-06-15 2010-06-16 2010-06-17 2010-06-18 2010-06-19 2010-06-20 2010-06-21
5 6 1 5 0 0 13
2010-06-22 2010-06-23 2010-06
Hi,
R seems to have a feature that isn't used much, which I don't really
now how to call. But, the dimnames function, can in addition to giving
names to rows/columns/dim 3 rows/dim 4 rows... can also give labels to
the dimensions themselves.
Thus, I can do:
A = matrix(1:9,3,3)
dimnames(A
Hi!
I moved the definition of the vlay function before the grid.newpage call,
and now it works! This is weird, I don't get it what was wrong in the first
place, if someone can enlighten me, I would feel better.
Sebastian
To make it clear, this works:
<>=
vlay <- function(x,y) viewport(layout.p
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of jd6688
> Sent: Monday, July 26, 2010 9:23 AM
> To: r-help@r-project.org
> Subject: [R] gapped sequence data summary
>
>
>Id cat1 location item_values p-values sequence
> a1
Hi all,
I have no idea if this question is to easy to be answered, but I´m starting
with R. So, here we go.
I have a large dataset with a lot of steps a judicial case. A sample is
attached.
I´d like to do a cluster analysis to try to understand with one is the most
usual path followed by this le
Thanks a lot!
This solves my problem!
Joh
On Monday 26 July 2010 17:06:37 Joshua Wiley wrote:
> Hi,
>
> Here is another option if you already have a list you want to convert.
> This will handle different elements of the list being different
> lengths.
>
> #Using your example data
> mydata <-
Id cat1 location item_values p-values sequence
a111 1 3002737 100 0.01 1
a112 1 3017821 102 0.05 2
a113 2 3027730 103 0.02 3
a114 2 3036220 104 0.04 4
a115 1 3053984 105 0.03 5
a
Hello,
Does anybody know if the OOP module (Chambers & Temple Lang) is going to
replace the the S4 (and the S3) class system?
http://www.omegahat.org/OOP/oop.pdf
Cheers!!
Albert-Jan
~~
All right, but apart from the sanitation
Dear Ravi - I echo everything you wrote, useR2010 was an amazing experience
(for me, and for many others with whom I have spoken about it).
Many thanks should go to the wonderful people who put their efforts into
making this conference a reality (and Kate is certainly one of them).
Thank you for ex
You can also automate it with this:
do.multirow<-function(df, which=1:ncol(df)){
for(c in which){
runs <- rle(as.character(df[,c]))
if(all(runs$lengths>1)){
tmp <- rep("", nrow(df))
tmp[c(1, 1+head(cumsum(runs$lengths),-
1))] <-
paste
Hi Uli,
I am not sure if this is the problem that you really want to solve. The
answer is the solution to the equation y = x * T^(x-1), provided a solution
exists. There is no optimization involved here. What is the real problem
that you are trying to solve?
If you want to solve a more meaning
Thanks Jim and Enrique, that should work since I am only trying to show
the month along my X axis it regardless of what year it is.
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish & Wildlife Service
California, USA
- Original Message
> From: jim hol
I just run the code below with sweave and works fine
It looks like you might be missing the sequence of vplay
<>=
library(ggplot2)
vplay<- function(x, y)
viewport(layout.pos.row=x, layout.pos.col=y)
grid.newpage()
p <- ggplot(diamonds, aes(x=carat, y=..density..)) +
geom_histogram(binwidth=0.
Hi,
Here is another option if you already have a list you want to convert.
This will handle different elements of the list being different
lengths.
#Using your example data
mydata <- list(c(1,2,3),c(4,5,6))
data.frame(
OriginalListIndex = rep(x = seq_along(mydata),
times = unlist(lapply(myd
Thank you very much for your quick help.
In fact, I was trying to show confidence intervals from two different
methods (for various groups in panels) that had some overlaps and some
exclusive areas. And using "fill" hides the area of the one underneath.
Hatching may have been very good for this a
It works, thanks a lot!
Cheers
-Original Message-
From: jim holtman [mailto:jholt...@gmail.com]
Sent: Monday, July 26, 2010 4:04 PM
To: Olivier Coupiac
Cc: r-help@r-project.org
Subject: Re: [R] search and replace in a list of strings
The most "usable" form is to convert it to POSIXct si
You can also automate it with this:
do.multirow<-function(df, which=1:ncol(df)){
for(c in which){
runs <- rle(as.character(df[,c]))
if(all(runs$lengths>1)){
tmp <- rep("", nrow(df))
tmp[c(1, 1+head(cumsum(runs$lengths),-1))] <-
paste(
Dear David, Petr, and Alain,
thank you very much for your fast responses. It's a typical
"handbook-not-read-error" at my side. I will dig deeper into the
plot-functions and the assignment of data. I was not aware of that the
vector "a" is handled as a vector of factors with 10 levels. Thanks f
Try this:
stack(data.frame(list('A' = c(1,2,3), 'B' = c(4,5,6
On Mon, Jul 26, 2010 at 11:46 AM, Johannes Graumann <
johannes_graum...@web.de> wrote:
> Hi,
>
> Any ideas on how to efficiently convert
>
> > list(c(1,2,3),c(4,5,6))
>
> to
>
> > data.frame(OriginalListIndex=c(1,1,1,2,2,2),Item=c
Hi!
I am troubled by Sweave which I want to use in order to plot graphics which
I build up successively by the use of grid.layout. Here is the code:
<>=
## combined plot via grid viewports
grid.newpage()
pushViewport(viewport(layout=grid.layout(2,1)))
vlay <- function(x,y) viewport(layout.pos.ro
Hi,
Any ideas on how to efficiently convert
> list(c(1,2,3),c(4,5,6))
to
> data.frame(OriginalListIndex=c(1,1,1,2,2,2),Item=c(1,2,3,4,5,6))
Thanks for any hints,
Joh
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Hi,
assign(paste(c("tmax.", 1950, 12), collapse="") ,1:10) does what you want.
Alain
On 26-Jul-10 16:23, Panos Hadjinicolaou wrote:
Thanks for the reply. Indeed the paste function results in concatenation:
> paste(c("tmax.", 1950, 12), collapse="")
[1] "tmax.195012"
but I am looking
have a look at assign() -- Best, Dimitris
On 7/26/2010 4:23 PM, Panos Hadjinicolaou wrote:
Thanks for the reply. Indeed the paste function results in concatenation:
> paste(c("tmax.", 1950, 12), collapse="")
[1] "tmax.195012"
but I am looking for a way to subsequently get rid of the - "
Thanks for the reply. Indeed the paste function results in concatenation:
> paste(c("tmax.", 1950, 12), collapse="")
[1] "tmax.195012"
but I am looking for a way to subsequently get rid of the - " - in order to
use tmax.195012 as an object (e.g. to define a vector with that name). Any
ideas
Dear all,
I'm looking for a way to solve a simple optimization problem with a
nonlinear constraint. An example would be
max x s.t. y = x * T ^(x-1)
where y and T are known values.
optim() and constrOptim() do only allow for box or linear constraints,
so I did not succedd here. I als
Hi,
Just as an example, here are three threads that discuss it.
http://www.mail-archive.com/r-help@r-project.org/msg16881.html
http://r.789695.n4.nabble.com/advice-opinion-on-vs-in-teaching-R-td1014502.html#a1014502
http://www.mail-archive.com/r-help@r-project.org/msg100034.html
Cheers,
Josh
Check the mail archieve on this; there has been a long discussion.
To avoid trouble in the future, use "<-" as the assignment operator.
On Mon, Jul 26, 2010 at 9:51 AM, Alaios wrote:
> Hello
> I notice that in Linux the "=" operator works like the "<-" operator
> So a=3 is similar to a<-3.
> Co
The most "usable" form is to convert it to POSIXct since most of the
plotting routines know how to handle date/time. You could do:
myDate <- as.POSIXct(paste(WD$date, WD$time), format='%m.%d.%Y %H:%M:%S')
e.g.,
> as.POSIXct('07.07.2010 12:34:15', format='%m.%d.%Y %H:%M:%S')
[1] "2010-07-07 12:3
Dear all,
i´m search for informations referring to R and Seaside respectively
VisualWorks or other cross-platforms for the implementation of Smalltalk.
Do you know a interface or a bridge,which enables work with such a
R-Smalltalk connection?
Best Regards
Nero
--
View this message in context:
Hi everybody,
I have a 4*4 matrix WD of wind data of the following form:
> WD
$date
[1] "07.07.2010" "07.07.2010" "07.07.2010" "07.07.2010"
$time
[1] "00:00:00" "00:10:00" "00:20:00" "00:30:00"
$CH1Avg
[1] 3.02 3.04 2.94 2.71
I would like to transform the date and time strings in usable number
Hello
I notice that in Linux the "=" operator works like the "<-" operator
So a=3 is similar to a<-3.
Could you please verify me that is correct? I would like to use "=" operator.
Do
you think that might be a problem in the future?
Best Regards
Alex
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