Hi,
I recently added another 1,000 observations to a data frame I have for
a total of 10,861 observations. Observations are by row and the frame
has 15 columns. R now keeps converting certain columns to factors
whenever I run regressions and can't seem to figure out the correct
scaling
Hi all,
I'm quite new to R and have a very basic question regarding how one gets
the standard error of the mean for factor levels under aov. I was able to
get the factor level means using:
summary(print(model.tables(rawfixtimedata.aov,"means"),digits=3)),
where rawfixtimedata.aov is my aov m
could someone explain why the name of FPVAL gets " .value" concatenated
onto it when the code below is run and temp is returned.
I've been trying to figure this out for too long. It doesn't matter when
I put the FPVAL in the return statement. It happens regardless of
whether it's first or last.
Hi, John,
Thanks very much. I have another question about gformlayout().
addSettingGUI <- function(container){
Setting <- list(type = "ggroup",
horizontal = FALSE,
children = list(
list(type="fieldset",
columns = 1,
ronggui gmail.com> writes:
>
> I want to write a function mimic the function of select.list(), here
> is my preliminary version.
>
> select <- function(x,multiple=TRUE,...){
> ans<-new.env()
> g <- gwindow(title=title,wid=200,heigh=500)
> x1<-ggroup(FALSE,con=g)
> x2<-gtable(x,multiple=multiple
Could the R Windows FAQ 2.16 be relevant?
http://cran.r-project.org/bin/windows/base/rw-FAQ.html#R-can_0027t-find-my-file
--
D Winsemius
On Dec 12, 2008, at 11:34 PM, Larry Ma wrote:
Dear R Users,
I have just installed R 2.8 today and also updated the Hmisc package
to the latest 3.4-4 v
Dear R Users,
I have just installed R 2.8 today and also updated the Hmisc package to the
latest 3.4-4 version. However, the sas.get function no longer works in the new
Hmsic 3.4-4. Below is a sample code and output from the old and new, but the
old one worked.
Can someone tell me how sas.ge
Define nm as the names to be multiplied and then use mapply
on m[-1] and d1[nm] so that the latter's columns are rearranged:
> nm <- names(m)[-1]
> data.frame(m[1], mapply("*", m[-1], d1[nm]))
class feather jet
1 birds 20 20
2 planes 60 40
On Fri, Dec 12, 2008 at 8:02 PM, Bill Mc
My apology to the list, as it turns out the moment after I hit the send
button, I found that the first element in the list was a null. The error
message reflects that.
> _
> From: Barker, Chris
> Sent: Friday, December 12, 2008 6:30
> To:
The answer I have seen before on this list (but cannot find at the
moment) was a suggestion to solve for one of the variables in terms of
the other two and then plot with standard 3d plotting routines. I also
seem to remember that some of the examples in an RGL vignette had
examples of grap
Barker, Chris wrote:
>
>
> I'd appreciate some tips, I'm making a simple mistake trying to
> extract elements from a list of fit objects.
>
> The following command works, coef( cph.list[[1]] ) , coef(
> cph.list[[3]] ) and so forth. These extract regression coefficients
> from
Good day all,
I am having seom trouble building a simple vector. Below my sample code
shows what ime trying to do and i have pointed out where the issue is. What
happens not is that a single "TAZDetermine_FEET" is selected by i need
multiple values, as many as there are "cands". I am think
I'd appreciate some tips, I'm making a simple mistake trying to
extract elements from a list of fit objects.
The following command works, coef( cph.list[[1]] ) , coef(
cph.list[[3]] ) and so forth. These extract regression coefficients
from the appropriate list element.
When I
I am trying to make estimates of the predictive power of ARIMA models estimated
by the auto.arima() function.
I am looping through a large number of time seiries and fitting ARIMA models
with the following code.
data1 <- read.csv(file = "case.csv", header = T)
data <- data1
output = c(1:leng
to the person who just asked about the feather and jet problem: i lost
your email but below works. there should be a shorter way by
avoiding the last statement and doing everything in the lapply but I've
had enough of feathers and jets tonight.
m <- data.frame(class = c("birds", "planes"), feat
disregard my last question. i forgot that d1 was a dataframe and was
thinkiing it was a vector. i think i can fix what i was doing now.
On Fri, Dec 12, 2008 at 8:52 PM, markle...@verizon.net wrote:
could someone explain what is happening below ? I was trying to solve
a related question on
could someone explain what is happening below ? I was trying to solve a
related question on the list and then , as I was solving it,
I was getting strange answers and then I noticed below. It's obviously
not a bug but I don't get it. Thanks.
m <- data.frame(class = c("birds", "planes"), feather
On Dec 10, 2008, at 8:03 AM, Frank E Harrell Jr wrote:
Maithili Shiva wrote:
Hi,
I have the response variable 'Y' and four predictors say X1, X2, X3
and X4. Assuming all the assmptions like Y follows normal
distribution etc. hold and I want to run linear multiple
regression. How do I run
Thanks for the help. Data frames are gradually starting to make sense to me.
Here's a related question. I have two data frames, one looks like this:
> m <- data.frame(class = c("birds", "planes"), feather = c(1,3), jet = c(2,4))
> m
class feather jet
1 birds 1 2
2 planes 3 4
Hello,
I have a POSIXct column in a dataframe, but when I change the value
assigment (for instance, VencimientosxDia$fe_ini_ope <-
as.Date(fecha) ) automatically change the format attribute with "2008-12-01
06:00:00".
How can I drop the hh:mm:ss from format attribute?
> str(VencimientosxDia)
'da
You might try something like
do.can <- candisc(do.mod, data=do)
plot(do.can)
But, in your example there is only one canonical dimension there are
only two levels of the
factor.
In any case, do.can$scores has the scores, so you can try to plot them
however you like.
For a 1-D plot in a case
Dear Celine
If you are considering using support vector machines in R,
you might find it worthwhile to read the paper:
http://www.jstatsoft.org/v15/i09/paper
Best, Niels
Celine Carret wrote:
Dear All,
Apologies for sending this email to both list, but at this point I'm not
sure which one co
Hello,
I am using RODBC and some times when I use a sqlSave I recive this message:
Microsft Visual C++ Runtime Library
RunTime Error!
Program C:\Archivo de programa\R\R-2.8.0\bin\Rgui.exe
This application has requested the Runtime to terminate it in an
unusual way.
P
On Dec 10, 2008, at 10:36 AM, Fazekas, Jacob wrote:
Good morning,
Is there anyway to do Mixed Exponential Distribution in R?
require(fortunes)
fortune(109)
I am trying
to load some lag-weighted empirical survival distribution into R and
run
a mixed exponential on that data.
You mig
Greg Snow wrote:
> Where do you get "should" and "expect" from? All the regular expression
> tools that I am familiar with only match non-overlapping patterns unless you
> do extra to specify otherwise. One of the standard references for regular
> expressions if you really want to understand w
On Fri, Dec 12, 2008 at 3:08 PM, Vitalie Spinu wrote:
> On Fri, 12 Dec 2008 18:27:02 +0100, hadley wickham
> wrote:
>
>
>>> or may be just
>>> mtcars[cyl>3&last(20)]
>>>
>>> or this is already too far?
>>
>> This would be a considerable extension because then the selector would
>> need to know ab
aggregate(x, list(x$y, x$a), FUN=median, na.rm=TRUE)
On Fri, Dec 12, 2008 at 2:13 PM, wrote:
> Hello R-experts,
>
> I have a simple question: how do I aggregate data using the median
> function with na.rm = TRUE, i.e.
>
> aggregate(x, list(x$y, x$z), FUN = '')
>
> I have tried 'median(x, na.
Hello R-experts,
I have a simple question: how do I aggregate data using the median
function with na.rm = TRUE, i.e.
aggregate(x, list(x$y, x$z), FUN = '')
I have tried 'median(x, na.rm = TRUE)' but that doesn't help.
Any suggestions are very much appreciated,
Thorsten
__
Dear R users,
Using R lme function, I found that both fixed and random effects of variable
A on variable B are significant. Now, I'd like to analyze what variables
are predicting differences in the slope. In other words, I'd like to know
what variables (e.g., variable C) are predicting individua
Dear R-experts,
I want to produce a very small png file (35 x 18 px) that contains a
histogram without a figure region or margins, only the pure heights.
In the base graphic this was easy:
png(filename = "hist.png", res = 72, width=35, height=18)
par(mar=c(0,0,0,0), oma=c(0,0,0,0))
Hi Ista,
one way could be:
ncharacters<-unlist(lapply(x,function(x)nchar(gsub(' ','',x
ncharacters
From: Ista Zahn
To: r-help@r-project.org
Sent: Friday, December 12, 2008 10:31:10 AM
Subject: [R] character count
Dear list,
I have a variable that consists
Hello List and thanks in advance for all of your help,
I am trying implement a permutation test of a multinomial logistic
regression ('multinom' within the nnet package). In the end I want to
compare the parameter estimate from my data to the distribution of
randomized parameter estimates.
I ha
I am using R as a data manipulation tool for a SQL database. So in some of
my R scripts I use the RODBC package to retreive data, then run analysis,
and use the sqlSave function in the RODBC package to store the results in a
database.
There are two problems I want to avoid, and they are highly re
Hi Sebpe,
the analysis of the data that you describe could be a complex and
lengthy process, in which decisions that you are confronted by are
affected by previous decisions that you have made. I recommend
obtaining the assistance of a statistician, preferably a local one
whose door you can knock
On Fri, 12 Dec 2008 18:27:02 +0100, hadley wickham wrote:
or may be just
mtcars[cyl>3&last(20)]
or this is already too far?
This would be a considerable extension because then the selector would
need to know about all other variables in the dataset, and you'd need
someway of combining selec
John M. Chambers Statistical Software Award - 2009
Statistical Computing Section
American Statistical Association
The Statistical Computing Section of the American Statistical
Association announces the competition for the John M. Chambers
Statistical Software Award. In 1998 the Association for Co
Thanks Jorge,
I am aware of the method you suggest, however, it's not quite what I was
asking for.
I was attempting to do this change on 56 dataframes, so I attempted to
do so with a list of all rather than each individually.
Thanks, DaveT.
From: Jorg
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Fernando Bizuet
> Sent: Friday, December 12, 2008 12:55 PM
> To: r-help@r-project.org
> Subject: [R] loop with dates
>
> Hello,
>
> I am trying to do a loop with dates, but w
You have not told us your OS (despite the posting guide).
For Windows this is covered in the rw-FAQ. Similar advice would apply on
a Unix-alike.
For Mac OS X, see the recent archives of the R-sig-mac list. E.g. the
strangely titled threads (look in a threaded view) around
https://stat.ethz
Hi everybody,
I would like to estimate prevalence ratio and confidence intervals.
I tried to do a log-binomial regression, but there was a failure of
convergence.
Now, I would like to learn how to do a poisson regression with robust
variance.
I am trying to estimate coefficients with poisson r
Ok, I knew somebody come with regex solution. My regex skills are limited
so I do not use it very often.
Regards
Petr
r-help-boun...@r-project.org napsal dne 12.12.2008 17:03:38:
> On Fri, Dec 12, 2008 at 11:00 AM, Petr PIKAL
wrote:
> > Hi
> >
> >
> > r-help-boun...@r-project.org napsal dne 1
Try iterating over the index rather than the value of
each component:
s <- seq(fini,to = ffin, by='days')
for (i in seq_along(s)) print(s[[i]])
On Fri, Dec 12, 2008 at 1:55 PM, Fernando Bizuet wrote:
> Hello,
>
> I am trying to do a loop with dates, but when I try to use the index is not
> a da
Try this:
weekdays(seq(fini,to = ffin, by='days'))
or in a loop:
sapply(as.character(seq(fini,to = ffin, by='days')),
function(d)weekdays(as.Date(d)))
On Fri, Dec 12, 2008 at 4:55 PM, Fernando Bizuet wrote:
> Hello,
>
> I am trying to do a loop with dates, but when I try to use the inde
> My end would be the output of your end(). If there are no args and no local
> context, I don't see the need for it to be a function call. It would just
> be defined as something like
>
> end <- structure( function(n) c(rep(FALSE, n-1), TRUE), class="selector")
Oh, I see what you mean.
> I'm n
Hello,
I am trying to do a loop with dates, but when I try to use the index is not
a date.
Fcorte <- as.Date('2008/11/30',format = "%Y/%m/%d")
fini <- Fcorte + 1
ffin <- seq(fini,by='months',length=2)[2] - 1
for (i in seq(fini,to = ffin, by='days'))
print (weekdays(i)) # i does
That depends on what you want evens() & last(5) to mean. Does that mean the
last 5 evens (returning 5 values) or the values in the last 5 that are also
even items (returning either 2 or 3 values depending on if the structure has an
odd or even number of elements). It could be interpreted eithe
Roy Robertson wrote:
> After installing a new version of R, how do I make the packages that I
> have already installed and use on the old version available to the new
> version?
Here is my solution. It has worked over many releases of R,
though it will probably require one initial re-installat
> evens() & last(5)
wouldn't x[evens()][last(5)] do the & already?
or is different, though.
Claudia
--
Claudia Beleites
Dipartimento dei Materiali e delle Risorse Naturali
Università degli Studi di Trieste
Via Alfonso Valerio 6/a
I-34127 Trieste
phone: +39 (0 40) 5 58-34 47
email: cbelei...@un
On 12/12/2008 1:06 PM, hadley wickham wrote:
Oh yes, that's a good point. But wouldn't the following do the job?
"&.selector" <- function(a, b) {
function(n) a(n) & b(n)
}
or
"&.selector" <- function(a, b) {
function(n) intersect(a(n), b(n))
}
depending on whether selectors return logical
Dear Michael,
You haven't told us what your data is, and we can only surmise -- not very
helpful for you and annoying for those who try to help.
Apologies, I am brand new to R and this mailing list. Will try to be more
concise.
Here is my data a NEW verion of my data:
Curvature Diameter Qu
Have a look at this
http://thread.gmane.org/gmane.comp.lang.r.general/87014/focus=133050
it is safer to have all you packages reinstalled once you have updated R.
HTH.
On Fri, 12 Dec 2008 18:47:33 +0100, Roy Robertson
wrote:
After installing a new version of R, how do I make the packages th
>> Oh yes, that's a good point. But wouldn't the following do the job?
>>
>> "&.selector" <- function(a, b) {
>> function(n) a(n) & b(n)
>> }
>>
>> or
>>
>> "&.selector" <- function(a, b) {
>> function(n) intersect(a(n), b(n))
>> }
>>
>> depending on whether selectors return logical or numeric v
After installing a new version of R, how do I make the packages that I
have already installed and use on the old version available to the new
version?
Thank you,
Roy Robertson
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinf
On 12/12/2008 12:23 PM, hadley wickham wrote:
On Fri, Dec 12, 2008 at 11:18 AM, Duncan Murdoch wrote:
On 12/12/2008 11:38 AM, hadley wickham wrote:
On Fri, Dec 12, 2008 at 8:41 AM, Duncan Murdoch
wrote:
On 12/12/2008 8:25 AM, hadley wickham wrote:
From which you might conclude that I don
On Fri, Dec 12, 2008 at 12:11 PM, Vitalie Spinu wrote:
> On Fri, 12 Dec 2008 17:38:13 +0100, hadley wickham
> wrote:
>
>> You could also imagine similar iterators for random sampling, like
>> samp(0.2) to choose 20% of the indices, or boot(0.8) to choose 80%
>> with replacement. first(n) could a
Erin Hodgess gmail.com> writes:
>
> Dear R People:
>
> Are there functions for spatial AR and ARIMA models in R, please?
Would looking at the "Spatial" task view on your CRAN mirror help? If by AR
you mean simultaneous or conditional autoregressive models using spatial
weights (or moving ave
On Fri, Dec 12, 2008 at 11:11 AM, Vitalie Spinu wrote:
> On Fri, 12 Dec 2008 17:38:13 +0100, hadley wickham
> wrote:
>
>> You could also imagine similar iterators for random sampling, like
>> samp(0.2) to choose 20% of the indices, or boot(0.8) to choose 80%
>> with replacement. first(n) could a
On Fri, Dec 12, 2008 at 11:18 AM, Duncan Murdoch wrote:
> On 12/12/2008 11:38 AM, hadley wickham wrote:
>>
>> On Fri, Dec 12, 2008 at 8:41 AM, Duncan Murdoch
>> wrote:
>>>
>>> On 12/12/2008 8:25 AM, hadley wickham wrote:
>
> From which you might conclude that I don't like the design of su
On 12/12/2008 11:38 AM, hadley wickham wrote:
On Fri, Dec 12, 2008 at 8:41 AM, Duncan Murdoch wrote:
On 12/12/2008 8:25 AM, hadley wickham wrote:
From which you might conclude that I don't like the design of subset, and
you'd be right. However, I don't think this is a counterexample to my
ge
I want to write a function mimic the function of select.list(), here
is my preliminary version.
select <- function(x,multiple=TRUE,...){
ans<-new.env()
g <- gwindow(title=title,wid=200,heigh=500)
x1<-ggroup(FALSE,con=g)
x2<-gtable(x,multiple=multiple,con=x1,expand=TRUE)
gbutton("OK",con=x1,handler
Hi,
We are trying to classify atmospheric patterns using PCA and kmeans. In
T-mode, the (non-normalized) loadings of the PCs are used as input
variables for the k-means classification.
As first step we perform PCA using "princomp" but the computed loadings
produced by princomp are normalized
On Fri, 12 Dec 2008 17:38:13 +0100, hadley wickham wrote:
You could also imagine similar iterators for random sampling, like
samp(0.2) to choose 20% of the indices, or boot(0.8) to choose 80%
with replacement. first(n) could also be useful, selecting the first
min(n, length(vector)) observatio
I got the whole thing to work, with Hadley Wickham's help.
is.POSIXt <- function(x) inherits(x, "POSIXt")
dates <- sapply(df, is.POSIXt)
df[dates] <-lapply(df[dates], as.Date)
Farrel Buchinsky
On Fri, Dec 12, 2008 at 11:26, Farrel Buchinsky wrote:
>
> Something strange. The lapply only proce
On Fri, Dec 12, 2008 at 8:41 AM, Duncan Murdoch wrote:
> On 12/12/2008 8:25 AM, hadley wickham wrote:
>>>
>>> From which you might conclude that I don't like the design of subset, and
>>> you'd be right. However, I don't think this is a counterexample to my
>>> general rule. In the subset functi
Just to muddy the waters a bit further. Currently we can do things like:
> pascal.tri <- numeric(0)
> class(pascal.tri) <- 'pasctri'
>
> `[.pasctri` <- function(x, ...) {
+ dots <- list(...)
+ n <- dots[[1]]
+ row <- choose(n, 0:n)
+ if(length(dots) > 1) {
+ row <- row[ dots[[2]] ]
+ }
+ row
+ }
Something strange. The lapply only processed the first row and then wrote
that value to every row of the original dataframe. It is as if the lapply is
indeed processing every item on the list, namely each column, but the ifelse
or the as.Date is getting messed up. Not only is it only processing the
On Fri, Dec 12, 2008 at 11:05 AM, Ista Zahn wrote:
> Thanks to everyone who responded. This turns out to be amazingly easy.
> To count characters including spaces:
> nchar(x)
> To count characters excluding spaces:
> nchar(gsub(" *","",x))
The * is unnecessary.
>
> Thanks!
>
> On Fri, Dec 12, 20
Hi,
I'm working on leaf characteristics of trees. Each tree is characterised by
about 10 leaf traits.
The trees were sampled at 9 different locations (about 20 to 30
trees/location, NOT balanced), grouped in 3 different climatic zones
(Sahelian, Soudanian and Guinean) (NOT balanced).
Further, eac
Hello all,
Does anyone know if there exists any function in R that resembles the
"lincom" and "nlcom" of STATA?. These functions computes point
estimates, standard errors, significance levels, confidence intervals,
etc. for linear and non linear combinations of previous estimated
parameters. Down
Besides the impute package, are there others that have alternative impute
approaches ? I hope to compare their performances.
Thanks
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide ht
Thanks to everyone who responded. This turns out to be amazingly easy.
To count characters including spaces:
nchar(x)
To count characters excluding spaces:
nchar(gsub(" *","",x))
Thanks!
On Fri, Dec 12, 2008 at 11:00 AM, Petr PIKAL wrote:
> Hi
>
>
> r-help-boun...@r-project.org napsal dne 12.12.
Or :
nchar(gsub("[[:space:]]", "", x))
On Fri, Dec 12, 2008 at 2:00 PM, Petr PIKAL wrote:
> Hi
>
>
> r-help-boun...@r-project.org napsal dne 12.12.2008 16:31:10:
>
> > Dear list,
> > I have a variable that consists of typed responses. I wish to compute
> > a variable equal to the number of cha
On Fri, Dec 12, 2008 at 11:00 AM, Petr PIKAL wrote:
> Hi
>
>
> r-help-boun...@r-project.org napsal dne 12.12.2008 16:31:10:
>
>> Dear list,
>> I have a variable that consists of typed responses. I wish to compute
>> a variable equal to the number of characters in the original variable.
>> For exam
Hi
r-help-boun...@r-project.org napsal dne 12.12.2008 16:31:10:
> Dear list,
> I have a variable that consists of typed responses. I wish to compute
> a variable equal to the number of characters in the original variable.
> For example:
>
> > x <- c("convert this to 32 because it has 32 charact
Claudia Beleites wrote:
>>> Wacek:
>>>
x[3:]
instead of
x[3:length(x)]
x[3:end]
>>> I don't think that would help:
>>> what to use for end - 3 within the convention that negative values mean
>>> exclusion?
>>>
>> might seem tricky, but not impossible:
nchar (c("convert this to 47 because it has 47 characters", "this one has 26
characters", "13 characters"))
HTH Claudia
--
Claudia Beleites
Dipartimento dei Materiali e delle Risorse Naturali
Università degli Studi di Trieste
Via Alfonso Valerio 6/a
I-34127 Trieste
phone: +39 (0 40) 5 58-34 47
On Fri, 12 Dec 2008 12:43:17 +0100, Liviu Andronic
wrote:
I'm not sure whether this can (partially) fill the gap, but Quick-R
[1] has a decent overview of data manipulation, and is essentially a
reference card. The maintainer of the Web site is open to
improvements, so it might make sense to
I just realized that my idea of doing something without going into the
extraction functions itself won't work
:-( it was a nice dream, though.
The reason is that there is no general way to find out what the needed length
is: At least I'm just writing a class where 2 kinds of columns are involve
Dear list,
I have a variable that consists of typed responses. I wish to compute
a variable equal to the number of characters in the original variable.
For example:
> x <- c("convert this to 32 because it has 32 characters", "this one has 22
> characters", "12 characters")
[Some magic function h
Here is how to emulate matlab end in R in the case of matrices.
Rather than redefine the matrix class (which would be a bit intrusive) we
just define a subclass of matrix called matrix2. Note in the examples that
matrix2 survives some operations such as + but not others such as crossprod
so in th
Is try() what you're looking for?
Sarah
On Fri, Dec 12, 2008 at 5:43 AM, wrote:
> Is there any way to handle errors retuened by R functions ... something like
> the exception handling in C++ ?
>
> Thank you so much,
> Maura
>
> Alice Messenger ;-) chatti anche con gli amici di Windows Live Mes
> > Wacek:
> >> x[3:]
> >> instead of
> >> x[3:length(x)]
> >> x[3:end]
> >
> > I don't think that would help:
> > what to use for end - 3 within the convention that negative values mean
> > exclusion?
>
> might seem tricky, but not impossible:
>
> x[-2]
> # could mean 'all except for 2nd', as it i
Re. Workshop on Directions in Statistical Computing, Copenhagen 13-14
July 2009.
http://www.r-project.org/dsc-2009
The web interface for registration and submission of abstracts is now open.
(The site is still under construction in a number of respects. In
particular, we haven't negotiated disco
paste()
?paste
- Original Message -
From: "Philip Whittall"
To:
Sent: Friday, December 12, 2008 1:30 PM
Subject: Re: [R] Extracting the name of an object into a character string
and vice versa
I am still struggling to map a character string to an object name and
vice versa in R
I have managed to get rid the two warning messages simply by uninstalling and
installing again R. Very likely that I disrupted the installation of a large
package.
> Date: Fri, 12 Dec 2008 15:07:38 +0100
> From: lig...@statistik.tu-dortmund.de
> To: dare...@hotmail.com
> CC: r-h...@stat.math.e
Dear Dr. Harrell,
Thanks for response so quickly. Here is example code to illustrate the
problem.
> n <- 1000
> set.seed(731)
> age <- 50 + 12*rnorm(n)
> label(age) <- "Age"
> sex <- factor(sample(c('Male','Female'), n,
+ rep=TRUE, prob=c(.6, .4)))
> cens <- 15*runif(n)
> h <- .02*e
Dear All,
Apologies for sending this email to both list, but at this point I'm not
sure which one could help me the most.
I have 4 sets of data, 1 test and 3 different sets of controls.
The measurements are binary, with a matrix of 0 and 1
I'm measuring across time (rows, ~815) the behaviour of
Is there any way to handle errors retuened by R functions ... something like
the exception handling in C++ ?
Thank you so much,
Maura
Alice Messenger ;-) chatti anche con gli amici di Windows Live Messenger e
tutti i telefonini TIM!
Vai su http://maileservizi.alice.it/alice_messenger/index.htm
Bonjour!
Je suis en train de faire un projet utilisant Affymatrix mais j'ai un problème à
télécharger un package depuis R: une fois que je télécharge le package et je
demande de faire "rma" ca sort l'erreur suivante et je n'ai aucune idée de
comment je pourrais faire.
> library(preprocessCore)
>
Thanks very much for the replies.
Both suggestions worked perfectly.
That's one more step towards understanding R.
Thanks again.
--
View this message in context:
http://www.nabble.com/Row-order-in-plot-tp20962774p20974800.html
Sent from the R help mailing list archive at Nabble.com.
__
The following error occurs every now and then by calling a function of wmTSA
package:
Error in `row.names<-.data.frame`(`*tmp*`, value = c("1", "0")) :
invalid 'row.names' length
I would greatly appreciate some guidelines about how to catch such an error
upon its occurrence and have it handl
Please, can you tell me if you know how can i use the library Igraph in C#
?
Thanks
Nadifi Taoufik
*
This message and any attachments (the "message") are con...{{dropped:15}}
__
R-
>> Wacek:
>>
>>
>>> x[3:]
>>> instead of
>>> x[3:length(x)]
>>> x[3:end]
>>>
>>>
>> I don't think that would help:
>> what to use for end - 3 within the convention that negative values mean
>> exclusion?
>>
>>
>>
>>
>
> might seem tricky, but not impossible:
>
> x[-2]
Any tips?
DaveT.
>-Original Message-
>From: Thompson, David (MNR)
>Sent: December 9, 2008 04:03 PM
>To: 'r-help@r-project.org'
>Subject: assign()ing within apply
>
>Hello,
>
>I'm trying to convert a character column in several dataframes
>to lower case.
>
>###
>#
># Sample data and 'spp
On 12/12/2008 8:25 AM, hadley wickham wrote:
From which you might conclude that I don't like the design of subset, and
you'd be right. However, I don't think this is a counterexample to my
general rule. In the subset function, the select argument is treated as an
unevaluated expression, and the
I have asked the maintainer for a fix and didn't get a reply, which is why I
posted to the list to see if there were any other ideas.
Sorry to bother. . . with my warning message.
Patrick
-Original Message-
From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
Sent: Friday, December 1
Claudia Beleites wrote:
> Am Freitag 12 Dezember 2008 13:10:20 schrieb Patrick Burns:
>
>> How about:
>>
>> x[, -seq(to=ncol(x), length=n)]
>>
> Doing it is not my problem. I just agree with Mike in that I would like if I
> could do shorter than:
>
> x[, 1 : (ncol(x) - n)]
>
> which I bt
And the *Error* message was (that was a warning)?
The warning comes from some contributed code, probably in Tinn-R. Please
find out where it is and ask the maintainer for a fix. But it looks
harmless enough.
On Fri, 12 Dec 2008, Richardson, Patrick wrote:
When R starts I get the following
When R starts I get the following error message
Warning message:
In grep(paste("[{]TclEval ", topic, "[}]", sep = ""), tclvalue(.Tcl("dde
services TclEval {}")), :
argument 'useBytes = TRUE' will be ignored
Has anyone else run into this issue and is there any way to can fix this?
Thanks,
Patri
-- Weitergeleitete Nachricht --
Betreff: Re: [R] The end of Matlab
Datum: Freitag 12 Dezember 2008
Von: Claudia Beleites
An: r-help@r-project.org
Am Freitag 12 Dezember 2008 13:10:20 schrieb Patrick Burns:
> How about:
>
> x[, -seq(to=ncol(x), length=n)]
Doing it is not my prob
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