Thank you for the explanation, and the fortune-ish quote,
“As the documentation for substitute() says, there is no guarantee
that the result makes sense.”
Best,
baptiste
On 19 May 2010 02:59, Duncan Murdoch wrote:
> On 18/05/2010 4:36 PM, baptiste auguie wrote:
>>
>> Dear list,
>>
>> I am puzz
On Wed, May 19, 2010 at 7:20 AM, milton ruser wrote:
> *but* going back to my question, is not true
> that already installed packages will not be
> reinstaled. I ran:
>
> install.packages(c("ggplot2"),dependencies=T)
> install.packages(c("mgcv","bbmle","akima","drc","sensitivity","tgp"),dependenci
*but* going back to my question, is not true
that already installed packages will not be
reinstaled. I ran:
install.packages(c("ggplot2"),dependencies=T)
install.packages(c("mgcv","bbmle","akima","drc","sensitivity","tgp"),dependencies=T)
and several packages installed during the first install.pa
Look at the grconvertX and grconvertY functions for a built in solution with
much more flexibility.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help
?grconvertX
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Oliver
> Sent: Monday, May 17, 2010 8:24 AM
> To
Thank you it worked perfectly. I just needed to close the window that was
the problem. Do you know how to close the window automatically and why does
that matter? Thanks!
Amitoj
On Tue, May 18, 2010 at 12:10 PM, Amitoj Chopra wrote:
> I tried doing that and this is what I go:
>
>
> dlg <-
Hello,
Thanks for your help so far, I am still having the same problem but I
think I am getting closer. Just to recap, since setting up R on my
server, so that everyone shares a common Library, I have had issues
getting packages to load. Here is an example with the package
survival. I have the sur
Thanks Ista..
I will take your suggestion.
Regards
Vijayan Padmanabhan
"What is expressed without proof can be denied
without proof" - Euclide.
Ista
Zahn
Thanks Joshua..
It really helped in polishing my coding essentials
in R.
Thanks & Regards
Vijayan Padmanabhan
"What is expressed without proof can be denied
without proof" - Euclide.
Joshua
On 05/18/2010 10:41 PM, Greg Snow wrote:
Have you read the BoxCox paper? It has the theory in there for dealing with an
offset parameter (though I don't know of any existing functions that help in
estimating both lambdas at the same time). Though another important point (in
the paper as well
Have you read the BoxCox paper? It has the theory in there for dealing with an
offset parameter (though I don't know of any existing functions that help in
estimating both lambdas at the same time). Though another important point (in
the paper as well) is that the lambda values used should be
An option similar to Bert's but looking more like a standard hypothesis test
output is to use the function:
SnowsCorrectlySizedButOtherwiseUselessTestOfAnything from the TeachingDemos
package.
If you want a more useful result then you will need to be less general and more
specific.
--
Gregory
The clip function will limit the area that other functions (including abline)
plot to, so that is one option.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mai
Hi Taki,
This should be doable with "gnls" by properly specifying the `weights'
argument, although I cannot figure out how to do it without spending much time
(someone like Doug Bates would know for sure).
But let me ask you: did you try the straightforward nonlinear optimization
(e.g. optim)
In this case, Ben's approach is the way to go.
I'm curious how you fit nls without knowing the model formula beforehand?
...Tao
- Original Message
> From: array chip
> To: r-help@r-project.org; TaoShi
> Sent: Tue, May 18, 2010 5:22:47 PM
> Subject: Re: [R] automate curve drawing
On 18/05/2010 4:36 PM, baptiste auguie wrote:
Dear list,
I am puzzled by this,
substitute(expression(x), list(x = factor(letters[1:2])))
# expression(1:2)
Why do I get back the factor levels inside the expression and not the
labels?
As the documentation for substitute() says, there is no gu
Sorry that I forgot to attach the data file.
Indeed, this is a much simpler way to do this.
Thanks
--- On Tue, 5/18/10, Ben Bolker wrote:
> From: Ben Bolker
> Subject: Re: [R] automate curve drawing on nls() object
> To: r-h...@stat.math.ethz.ch
> Date: Tuesday, May 18, 2010, 7:33 PM
> arra
well, this is not going automate enough because you have to know how the model
(formula) looks like, and how many parameters there are in the model beforehand
to do what you are suggesting.
Thanks
--- On Tue, 5/18/10, Shi, Tao wrote:
> From: Shi, Tao
> Subject: Re: [R] automate curve drawin
I can't directly answer your question regarding 'expression', but can you just
replace b, c,d, and e with coef(obj)[1], coef(obj)[2], ... etc. You still can
automate the whole process this way, right?
- Original Message
> From: array chip
> To: r-help@r-project.org
> Sent: Tue, M
It depends on what you mean by display values. In addition to the other
suggestions also look at the identify function as well as TkIdentify and
HTKidentify in the TeachingDemos package.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.40
array chip yahoo.com> writes:
> Hi, I would like to use the curve() function to draw the
> predicted curve from an nls() object. for example:
>
Is there a reason you don't want to use plot() and predict() ???
> dd<-read.table("dd.txt",sep='\t',header=T,row.names=1)
note that this is no
> precip.1 <- subset(DF, precipitation!="NA")
> b <- ddply(precip.1$precipitation, .(precip.1$gauge_name), cumsum)
> DF.precip <- precip.1
> DF.precip$precipitation <- b$.data
I suspect what you want here is
ddply(precip.1, "gauge_name", transform, precipitation = cumsum(precipitation))
Hadley
You have run out of memory. What OS are you using, how much physical
memory do you have? how large are the objects you already have in your
data space? Have you removed all extraneous objects and done 'gc()'?
The solution is to get better control of your data space and
understand what is using i
Hi, I would like to use the curve() function to draw the predicted curve from
an nls() object. for example:
dd<-read.table("dd.txt",sep='\t',header=T,row.names=1)
obj<-nls(y~c+(d-c)/(1+(x/e)^b),data=dd,start=list(b=-1, c=0, d=100, e=150))
coef(obj)
b c d e
This sounds like a perfect use of "list" and the 'sapply/lapply' functions:
> # list with 100 dataframes
> df.list <- lapply(1:100,function(x) data.frame(a=1:10, b=1:10))
> # create a new list with a random value of 'a'
> new.list <- sapply(df.list, function(x) sample(x$a, 1))
>
>
> new.list
[1]
Hello!
I'm having trouble figuring out how to apply a function to a set of vectors
or data frames, when that set is over 1000 large.
I've created 1891 vectors named as follows:
vector.names <- paste("vector",1:1891,sep="")
These can be referred to easily by using get as follows:
vector <
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Jan Schubert
> Sent: Tuesday, May 18, 2010 12:51 PM
> To: r-help@r-project.org
> Subject: [R] sem error "no variance or error-variance parameter"
>
>
> Hi,
> I am sorry to post
Dear colleagues,
thank you so much for your help.
Hans, I think the Remez algorithm is what I need. I will brush up on
fortran language.
Ravi, thanks anyway, I appreciated.
All the best,
Patrizio
On Tue, May 18, 2010 at 12:10 PM, Hans W Borchers
wrote:
>
> I guess you may be looking for the R
On 2010-05-18 11:00, John Kane wrote:
I don't think you can do this
precipitation!="NA")
Actually, that will work here, although it should always be avoided.
Do use is.na().
The main problem seems to be that the ddply() call doesn't work.
I would just use tapply() and unlist():
b <- with(pr
Hello,
Can you run str() on your data and report the results?
str(en.id.pr)
I suspect that one of your variables that you are trying to sum over
is a factor.
HTH,
Josh
2010/5/18 Changbin Du :
>> head(en.id.pr)
> valid.gene_id b.pred rf.pred svm.pred
> 1521 2500151211 0 0
Thanks, David!
Yes, I found it just as you said. It works now after change to numeric.
On Tue, May 18, 2010 at 1:53 PM, David Winsemius wrote:
>
> On May 18, 2010, at 4:32 PM, Changbin Du wrote:
>
> head(en.id.pr)
>>>
>>valid.gene_id b.pred rf.pred svm.pred
>> 15212500151211 0
Dear R users,
A parallel implementation of the np package titled `npRmpi' is now available on
CRAN. This package can take advantage of multiple core computing environments
to reduce the run time associated with the methods contained in the np package.
Kindly see the vignette for details and ex
On May 18, 2010, at 4:32 PM, Changbin Du wrote:
head(en.id.pr)
valid.gene_id b.pred rf.pred svm.pred
15212500151211 0 00
366 639679745 0 00
19652502081603 1 11
1420 644148030 1 11
1565250062648
Hi,
I am having a problem getting character (string) variables into a
dataframe. I seem to have no problem with numeric fields, but character
fields are appearing as a "1".
In my problem I need to retrieve some information from an SQL database
using RODBC one line at a time and then update
Ivan Allaman yahoo.com.br> writes:
>
>
> Hi Ben!
>
> First I thank you for your attention. Unfortunately, the ANOVA does not work
with vglm. In
> another email, Rafael warned me that actually a lot of zeros does not
> necessarily imply a distribution of zeros binomail inflated. So how could I
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Martin Maechler
> Sent: Tuesday, May 18, 2010 1:25 PM
> To: milton ruser
> Cc: r-help@r-project.org
> Subject: Re: [R] avoiding reinstall already installed *package*
>
> On Tue,
Dear list,
I am puzzled by this,
substitute(expression(x), list(x = factor(letters[1:2])))
# expression(1:2)
Why do I get back the factor levels inside the expression and not the
labels? The following work as I expected,
substitute(expression(x), list(x = letters[1:2]))
# expression(c("a", "b")
> head(en.id.pr)
valid.gene_id b.pred rf.pred svm.pred
15212500151211 0 00
366 639679745 0 00
19652502081603 1 11
1420 644148030 1 11
15652500626489 1 11
18162501711016
On Tue, May 18, 2010 at 18:06, milton ruser wrote:
> Hi Martin,
>
> thanks for your reply, and very thanks for your kind tips about "package"
> and "library"
> So, I was trying to understand *why* we load packages using library().
>
I've started to use and suggest using require(.) instead
{as
Hi,
On Tue, May 18, 2010 at 2:49 PM, Godavarthi, Murali
wrote:
> Hi,
>
> I'm new to 'R' and need some help on the "Load" command. Any responses
> will be highly appreciated. Thanks in advance!
>
> As per manuals, the "Load" command expects a binary file input that is
> saved using a "save" comman
Hi,
I am sorry to post the message again but I really need some advise on that.
I am using the R version 2.11.0 and the version of sem package: sem_0.9-20
under Windows XP.
I read the questions:
http://r.789695.n4.nabble.com/computationally-singular-and-lack-of-variance-parameters-in-SEM-td891081.
Thank you Adrian,
its working fine.
Knut
__
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, rep
I tried doing that and this is what I go:
dlg <- aDialog(items=list(
ProtienCode=stringItem("")
),
OK_handler=function(.) { # . is reference to dlg object
values <- .$to_R()
f <- function(ProtienCode)
pdb <- read.pdb(.$get_ProteinCode())
#cat("ProteinCode is",ProtienCode,"\n")
do.call(f, values)
Hi,
I'm new to 'R' and need some help on the "Load" command. Any responses
will be highly appreciated. Thanks in advance!
As per manuals, the "Load" command expects a binary file input that is
saved using a "save" command. However it is required that we need to
call the 'R' program from
J
On Tue, 18 May 2010, Vinh Nguyen wrote:
Binder's estimating equations are the usual way of applying weights to a Cox
model, so nothing special is done apart from calling coxph(). To quote the
author of the survival package, Terry Therneau, "Other formulae change in
the obvious way, eg, the weigh
On 05/18/2010 11:52 AM, Chuck Cleland wrote:
On 5/18/2010 12:38 PM, Praveen Surendran wrote:
Dear all,
I am trying to read an SPSS file into a data frame in R using method
read.spss(),
sample<- read.spss(file.name,to.data.frame=TRUE)
But dates in the data.frame 'sample' are coming as inte
Dear R Help,
I am trying to fit a nonlinear model for a mean function $\mu(Data_i,
\beta)$ for a fixed covariance matrix where $\beta$ and $\mu$ are low-
dimensional. More specifically, for fixed variance-covariance matrices
$\Sigma_{z=0}$ and $\Sigma_{z=1}$ (according to a binary covariate $
John Lande writes:
> dear all,
>
> we am trying to improve the performance of my R code, with the implentation
> of some function with custom C code.
> we found difficult to import and export/import data structure such us
> matrices or data.frame into the external C functions.
Please give a *ver
Hi Ben!
First I thank you for your attention. Unfortunately, the ANOVA does not work
with vglm. In
another email, Rafael warned me that actually a lot of zeros does not
necessarily imply a distribution of zeros binomail inflated. So how could I
test if my variable is or not a binomial zero infl
The "glassolist" function says that if rholist is set to NULL, then "10
values in a (hopefully reasonable) range are used". I'm trying to figure out
how these 10 values are chosen, I can't find this in any documentation.
Thanks for your help!
--
View this message in context:
http://r.789695.n4.n
On May 18, 2010, at 12:07 PM, Arantzazu Blanco Bernardeau wrote:
Hello
Well, the problem is, that arcilla is the percentage of clay in the
soil sample. So, for linear model, I need to work with that number
or value. Now, R thinks that arcilla (arcilla means clay in
spanish), is a factor,
Under UNIX I usually write something like
system(paste("echo '",...,"'", sep=sep))
where I replace the '...' with whatever I need to show.
This would need some adjustments for non-scalars, though.
Benno
Am 18.Mai.2010 um 19:10 schrieb Jimmy Söderly:
> Dear R users,
>
> I am using th
Hi Jimmy,
You can use message() instead of cat():
if (i%%100==0) message(i)
Best,
Ista
On Tuesday 18 May 2010 1:10:51 pm Jimmy Söderly wrote:
> Dear R users,
>
> I am using the Sweave package and I am doing some MCMC. I have a loop
> function for my MCMC. Every 100 iterations, I want the number
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of tetonedge
> Sent: Tuesday, May 18, 2010 9:14 AM
> To: r-help@r-project.org
> Subject: [R] unsigned 4 byte number
>
>
> Does anybody know how to read in a unsigned 4 byte number
On Mon, May 17, 2010 at 7:42 PM, Vinh Nguyen wrote:
> Dear R-help,
>
> I would like to compute the variance for the proportion of treatment
> effect by a surrogate in a survival model (Lin, Fleming, and De
> Gruttola 1997 in Statistics in Medicine). The paper mentioned that
> the covariance matri
Dear R users,
I am using the Sweave package and I am doing some MCMC. I have a loop
function for my MCMC. Every 100 iterations, I want the number of iterations
already done to appear on my screen (but not on the final document). Is that
possible ?
Usually I can rely on
> if (i%%100==0) cat(i, "\
On 18-May-10 16:42:40, Peter Ehlers wrote:
> On 2010-05-18 10:05, (Ted Harding) wrote:
>> On 18-May-10 15:49:37, Martin Maechler wrote:
>>> { I've modified the subject; I can't stand it hitting square into
>>>my face ... }
"mr" == milton ruser
on Tue, 18 May 2010 12:36:23
I don't think you can do this
precipitation!="NA")
have a look at ?is.na
--- On Tue, 5/18/10, stephen sefick wrote:
> From: stephen sefick
> Subject: [R] Function that is giving me a headache- any help appreciated
> (automatic read )
> To: r-help@r-project.org
> Received: Tuesday, May 18, 20
On 5/18/2010 12:38 PM, Praveen Surendran wrote:
> Dear all,
>
>
>
> I am trying to read an SPSS file into a data frame in R using method
> read.spss(),
>
> sample <- read.spss(file.name,to.data.frame=TRUE)
>
>
>
> But dates in the data.frame 'sample' are coming as integers and not in the
>
On Tue, May 18, 2010 at 8:50 AM, Thomas Lumley wrote:
>
> > I
>>
>> don't believe so since svycoxph() calls coxph() of the survival
>> package and weights are applied once in the estimating equation. If
>> the weights are implemented in the ratio, could you point me to where
>> in the code this
On 2010-05-18 10:05, (Ted Harding) wrote:
On 18-May-10 15:49:37, Martin Maechler wrote:
{ I've modified the subject; I can't stand it hitting square into
my face ... }
"mr" == milton ruser
on Tue, 18 May 2010 12:36:23 -0300 writes:
mr> Dear R-experts,
mr> I am installing ne
Dear all,
I am trying to read an SPSS file into a data frame in R using method
read.spss(),
sample <- read.spss(file.name,to.data.frame=TRUE)
But dates in the data.frame 'sample' are coming as integers and not in the
actual date format given in the SPSS file.
Appreciate if anyone can help
Hello everybody
the problem has been solved. It was my mistake not to be ensured that any N/A
had dissapeared. I do apologize for the inconveniences caused ;)
friendly greeting sfrom Spain!
Arantzazu Blanco Bernardeau
Dpto de Química Agrícola, Geología y Edafología
Universidad de Murcia-Campu
note: whole function is below- I am sure I am doing something silly.
when I use it like USGS(input="precipitation") it is choking on the
precip.1 <- subset(DF, precipitation!="NA")
b <- ddply(precip.1$precipitation, .(precip.1$gauge_name), cumsum)
DF.precip <- precip.1
DF.precip$precipitation <-
Ivan Allaman yahoo.com.br> writes:
>
>
> I'm trying to use the inflated binomial distribution of zeros (since 75% of
> the values are zeros) in a randomized block experiment with four
> quantitative treatments (0, 0.5, 1, 1.5), but I'm finding it difficult,
> since the examples available in VGA
Arantzazu,
Your problem is that the data were probably imported
from Excel where you had at least one cell containing "#N/A".
You need to replace those cases in your dataframe with NA.
Then you should be able to do as.numeric(as.character(arcilla)).
-Peter Ehlers
On 2010-05-18 10:07, Arantzazu
Does anybody know how to read in a unsigned 4 byte number from a binary file?
According to the help for readBin, the signed argument only applies to
size=1 or 2. But if you declare any size larger it assumes it is signed?
Thanks
--
View this message in context:
http://r.789695.n4.nabble.com/unsi
One last thing:
before you take my advice on how to recode your nominal/categorical
"clay" variable for your regression model, take some time to see how
other people talk about this and do some searching on phrases like
"regression model with nominal variables" (that's just the one I
used).
You'l
Hi again,
If you used the function read.table() to read from a csv file into a
data.frame, it is weird that numeric data are converted into factors.
I would check in the original data that you don't have a typo somewhere.
I don't know all the possibilities, but a special character can
definite
Hi Martin,
thanks for your reply, and very thanks for your kind tips about "package"
and "library"
So, I was trying to understand *why* we load packages using library().
I suggest that developers killl the problem on its root, deleting library
function :-)
Good to know already installed packages w
Dear all,
Just one last question. There seems to be no problem in writing
> z = system.time(y <- f(x))
or
> z <- system.time(y <- f(x))
Then z contains the named vector of the elapsed times, and y the value of
the function f(x).
Am I right ?
Thank you very much,
Gustave
2010/5/17 Alexander
Hello
Well, the problem is, that arcilla is the percentage of clay in the soil
sample. So, for linear model, I need to work with that number or value. Now, R
thinks that arcilla (arcilla means clay in spanish), is a factor, and gives me
the value as a factor, so the output of the linear model i
If I perceive the issue:
Day <- c(rep(97, each = 11), rep(98:278, each = 12))
-
A R learner.
--
View this message in context:
http://r.789695.n4.nabble.com/Repeating-Name-for-Rows-tp2221457p2221492.html
Sent from the R help mailing list archive at Nabble.com.
On 18-May-10 15:49:37, Martin Maechler wrote:
> { I've modified the subject; I can't stand it hitting square into
> my face ... }
>> "mr" == milton ruser
>> on Tue, 18 May 2010 12:36:23 -0300 writes:
> mr> Dear R-experts,
> mr> I am installing new libraries using
> mr> in
Hello
so, here you have the output of the data frame. The data frame comes from a csv
file.
I could take Gr_2 instead of arcilla, because it is the same value... but
curiously, it is a factor as well.
> str(caperf)
'data.frame': 556 obs. of 38 variables:
$ Hoja : int 818 818 818 8
Hi,
Sorry, I'm not really getting what going on here ... perhaps having
more domain knowledge would help me make better sense of our question.
In particular:
On Tue, May 18, 2010 at 11:35 AM, Arantzazu Blanco Bernardeau
wrote:
>
> Hello
> I have a data array with soil variables (caperf), in whi
On Mon, 17 May 2010, Vinh Nguyen wrote:
Dear R-help,
Let me know if I should email r-devel instead of this list. This
message is addressed to Professor Lumley or anyone familiar with the
survey package.
Does svycoxph() implement the method outlined in Binder 1992 as
referenced in the help fil
Hi,
I think that providing the output from str(data array or whatever you
have) would help.
Because, for now, we don't have much idea of what you really have.
Moreover, some sample data is always welcomed (using the function dput
for example)
Ivan
Le 5/18/2010 17:36, Arantzazu Blanco Berna
On 17.05.2010 18:46, Kevin E. Thorpe wrote:
Hello.
In this post:
http://finzi.psych.upenn.edu/Rhelp10/2010-March/233815.html
Uwe Ligges suggests using BRugs rather than R2WinBUGS under windows. He
also notes that it is not in the main CRAN repository, but it is in
"extras" which is a default
{ I've modified the subject; I can't stand it hitting square into
my face ... }
> "mr" == milton ruser
> on Tue, 18 May 2010 12:36:23 -0300 writes:
mr> Dear R-experts,
mr> I am installing new libraries using
mr> install.packages("ggplot2",dependencies=T).
mr> But I
Try
arcilla<-as.numeric(as.character(clay))
best
milton
On Tue, May 18, 2010 at 12:36 PM, Arantzazu Blanco Bernardeau <
aramu...@hotmail.com> wrote:
>
>
> sorry I had a mistake sending my question without a subject. I do resend
> again. Please excuse me.
> > Hello
> > I have a data array with
Hi,
On Tue, May 18, 2010 at 11:36 AM, milton ruser wrote:
> Dear R-experts,
>
> I am installing new libraries using
> install.packages("ggplot2",dependencies=T).
> But I perceive that many dependencies are already installed. As I am using
> a low-band internet, how can avoid reinstall installed l
sorry I had a mistake sending my question without a subject. I do resend again.
Please excuse me.
> Hello
> I have a data array with soil variables (caperf), in which the variable
> "clay" is factor (as I see entering str(caperf)) . I need to do a regression
> model, so I need to have arcilla
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of M.Ribeiro
> Sent: Tuesday, May 18, 2010 7:13 AM
> To: r-help@r-project.org
> Subject: [R] Counting Frequencies in Data Frame
>
>
> Hi,
> I am sure there is an easy way to do it,
Dear R-experts,
I am installing new libraries using
install.packages("ggplot2",dependencies=T).
But I perceive that many dependencies are already installed. As I am using
a low-band internet, how can avoid reinstall installed libraries?
cheers
milton
[[alternative HTML version deleted]]
Hello
I have a data array with soil variables (caperf), in which the variable "clay"
is factor (as I see entering str(caperf)) . I need to do a regression model, so
I need to have arcilla (=clay) as a numeric variable. For that I have entered
as.numeric(as.character(arcilla))
and even enterin
Hi,
Others will have fancier solutions, but is the way I would do it:
dat <- read.table(textConnection("1 2 3
1 aa ab ab
2 ab ab ab
3 aa aa aa
4 bb bb bb"), header=TRUE)
closeAllConnections()
countAB <- function(x) {
aa <- length(which(x == "aa"))
ab <- length(which(x == "ab"))
bb <- leng
Hi,
On Tue, May 18, 2010 at 10:51 AM, wrote:
> Hi, all.
>
> I have R installed in my computer. I guess I did something in my previous
> session, and now every time I start R, I find the following message:
>
> "Fatal error: unable to restore saved data in .RData"
>
> I uninstalled R and installe
On 18/05/2010 10:51 AM, gbre...@ssc.wisc.edu wrote:
Hi, all.
I have R installed in my computer. I guess I did something in my previous
session, and now every time I start R, I find the following message:
"Fatal error: unable to restore saved data in .RData"
I uninstalled R and installed it ag
Hello,
I have a large data frame (47:2186), where i want to label every 12th row.
This command works,
Day <-rep(97:278, each = 12)
However i need 97 to only labeled 11 rows and then from 98:278 can be
labeled every 12 times.
Thanks!
__
R-help@r-proj
Hi,
On Tue, May 18, 2010 at 10:50 AM, John Lande wrote:
> dear all,
>
> we am trying to improve the performance of my R code, with the implentation
> of some function with custom C code.
> we found difficult to import and export/import data structure such us
> matrices or data.frame into the exte
Hi, all.
I have R installed in my computer. I guess I did something in my previous
session, and now every time I start R, I find the following message:
"Fatal error: unable to restore saved data in .RData"
I uninstalled R and installed it again and I'm still getting this message.
Can anyone he
dear all,
we am trying to improve the performance of my R code, with the implentation
of some function with custom C code.
we found difficult to import and export/import data structure such us
matrices or data.frame into the external C functions.
we already tried the solution from "Writing R Exte
Henrique Dallazuanna schrieb:
Try this also:
pmax(test - 1, 1)
O
test <- c(1,3,5,7,9,11,12,13,14)
test
test <- pmax(test - 1, 1)
test
This works for 1
what about if I would dec 11: to 14 to close the gap between 9 and 10 ?
I did not find the answer with the help file
Thank you Knut
_
Hi,
I am sure there is an easy way to do it, but I can't find it.
I have a data frame that has 15 columns and 7000 rows.
The only values inside the data.frame are "aa", "ab", "bb" as you can see an
example bellow.
1 2 3
1 aa ab ab
2 ab ab ab
3 aa aa aa
4 bb bb bb
What I would like to do, i
Sorry, I made two mistakes. The first was matching the female with the male.
The second was 2 variables should be selected randomly every time.
Followed is a revised copy:
## Import data.
moms <- read.delim("females.txt", sep =" ", stringsAsFactors = FALSE, header
= TRUE)
dads <- read.delim("mal
Thank you this was helpful
Chris Anderson
Data Analyst
Medical Affairs
wk: 925-677-4870
cell: 707-315-8486
Fax:925-677-4670
-Original Message-
From: foolish.andr...@gmail.com [mailto:foolish.andr...@gmail.com] On Behalf Of
Felix Andrews
Sent: Monday, May 17, 2010 5:55 PM
To: Anderson, C
Thanks for the replies! If anybody encounters a similar problem, the function
that now does what I wanted is attached below.
Best
Jannis
trnsf.coords = function(array_x,array_y)
# This function transfers relative coordinates between 0 and 1 for two arrays
with x
# and y values into the coo
Dears,
a way to define x and y positions in plots in relative numbers (e.g in
fractions between 0 and 1 referring to relative positions inside the plot
region) would really help me. One example I would need this to would be to add
text via text() to a plot always at a defined spot, e.g the up
Hi,
For 2., I don't know if it's possible to retrieve the axis limits, but
you can surely specify them in your call to plot (with arguments xlim
and ylim).
That's a cheap solution and others probably have better ones.
Ivan
Le 5/18/2010 16:23, Jannis a écrit :
Dears,
a way to define x and
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