Many thanks for your helpful suggestions and
the detailed feedback!
We will have a look at your suggestions before buying
the (quite expensive) PASS software.
-Udo
Quoting Tobias Verbeke :
Frank E Harrell Jr wrote:
Greg Snow wrote:
I don't know of a single package that is comparable to PASS
milton ruser wrote:
Hi there,
try this:
setwd("/home/mcr")
dir(path=".", pattern=".txt$")
OR
dir(path="/home/mcr/.", pattern=".txt$")
$ means finished with ".txt"
A word of warning. The pattern is a regular expression, so the above
means that the name ends with an arbitrary character foll
Hi Ravi,
See if the following helps:
=
Lines <- "Date1 Date2
29-Dec-06 25-Jan-07
29-Dec-06 25-Jan-07
29-Dec-06 25-Jan-07
2-Jan-07 25-Jan-07
2-Jan-07 25-Jan-07
2-Jan-07 25-Jan-07"
DF <- read.table(con<- textConnection(Lines), skip = 1)
close(con);
names(DF) <- scan(textCo
Dear R users,
I have two nonlinear equations, f1(x1,x2)=0 and f2(x1,x2)=0. I try to use optim
command by minimize f1^2+f2^2 to find x1 and x2. I found the optimal solution
changes when I change initial values. How to solve this?
BTW, I also try to use grid searching. But I have no information
Hi R,
I have two columns of date in a CSV file in the below format
29-Dec-06 25-Jan-07
29-Dec-06 25-Jan-07
29-Dec-06 25-Jan-07
2-Jan-0725-Jan-07
2-Jan-0725-Jan-07
2-Jan-0725-Jan-07
I read in R using dat<-read.csv("Z:\\data.csv").
> class(dat[,1])
[1] "
Hi there,
try this:
setwd("/home/mcr")
dir(path=".", pattern=".txt$")
OR
dir(path="/home/mcr/.", pattern=".txt$")
$ means finished with ".txt"
good luck,
milton
brazil=toronto
On Fri, Jul 17, 2009 at 12:08 AM, torpedo fisken wrote:
> Hi
> I got a script that works on file with the extensio
Kyle Werner wrote:
Does anyone know how to get the C-index from a logistic model - not using
the dataset that was used to train the model, but instead using a fresh
dataset on the same model?
I have a dataset of 400 points that I've split into two halves, one for
training the logistic model, and
Does anyone know how to get the C-index from a logistic model - not
using the dataset that was used to train the model, but instead using
a fresh dataset on the same model?
I have a dataset of 400 points that I've split into two halves, one
for training the logistic model, and the other for evalua
Hi
I got a script that works on file with the extension ".qed" that are
al located in a folder '~/works/'
Is there are R function that will fetch all the filenames from the works subdir,
similar to
'ls ~/works/*.qed'
Thanks in advance
__
R-help@r-proj
Thanks, 'cut()' was indeed what I was looking for
thanks
2009/7/16 :
> Hi: I'm not sure if I understand what you want but below might help you ?
> see cut for more details by doing ?cut.
>
>
> probs<-c(0.001,0.5,0.02,.05,0.12,0.23,0.5,0.6,0.59,0.7)
> probs
>
> temp <- cut(probs,breaks=seq(0,1,le
Does anyone know how to get the C-index from a logistic model - not using
the dataset that was used to train the model, but instead using a fresh
dataset on the same model?
I have a dataset of 400 points that I've split into two halves, one for
training the logistic model, and the other for evalua
Wouldn't the "response" of a survival model be survival times? You
may want to look at survfit and survest.
--
DW
On Jul 16, 2009, at 9:19 PM, Sean Brummel wrote:
> With type="response" in the predict funtion, I was expecting an
> expected survival time given covariates( in my dataset I h
when you post this can you please remove the tiff file? my supervisor
doesn't want me to let it out. sorry!
On Thu, Jul 16, 2009 at 4:02 PM, Mehdi Khan wrote:
> I imported the attached tiff file and converted the coordinate system to
> long lat and graphed it:
>
> californiatiff<- readGDAL("cal
Don MacQueen wrote:
>
> When I try this:
>
>> tmp <- randz<-matrix(rnorm(200),2000,1)
>> dim(tmp)
> [1] 2000 1
>
> It gives a result in less than one second. Very quick. And the
> resulting matrix is not alarmingly big.
>
> There must be some other problem. Perhaps your compu
On Jul 16, 2009, at 10:17 PM, Kum-Hoe Hwang wrote:
I work for a research institute. I have used R for several years.
I think there are some good and bad sides followings:
Good sides are: I can use new statistical methods from R. no license
fee..
Bad sides are : physical memory in PC is an
I work for a research institute. I have used R for several years.
I think there are some good and bad sides followings:
Good sides are: I can use new statistical methods from R. no license fee..
Bad sides are : physical memory in PC is an obstacle (max. 3GB), some
package of R is still being deve
Kel Lam wrote:
My institute has been heavily dependent on SAS for the past while, and
SAS is starting to charge us a very deep amount for license renewal.
Since we are a non-profit organization that is definitely not
sustainable. The team is brainstorming possibility of switching to R,
at least
With type="response" in the predict funtion, I was expecting an expected
survival time given covariates( in my dataset I have a few covariates but
not in the example)... a natural predictor. The prediction function is
clearly not returning a probability of survival at a given time since my
example
I realize function write FASTA expects a list with two items, respectively,
description and sequence.
However, just passing a list won't work (please, see code at the bottom of this
message)
I saw there is the helper function CharacterToFASTArecords(x) that presumably
generates the right input
On Jul 16, 2009, at 8:19 PM, Sean Brummel wrote:
> Thanks for the help but...
>
> I did the required transformations at the end of the code. The
> thing that I dont understand is: Why is the predicted value (from
> the predict function) not either the mean or median. Sorry I was not
> clea
Thanks for the quick reply.
I found that page and do have the survival package installed.
My problem is that I don't understand a few things:
1) What do I input for the "formula" field? (In RapidMiner, the
functions don't have this input.)
2) How do I indicate which field is the "grou
I have a relatively simple finance application of GMM. Given the moment
condition:
E[m*R]=0
where m=m[theta]
I would like to constrain m>0. Any ideas?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.eth
*hum*
How much of the "employes" of your institute use statistical softwares? 1-5?
5-20? 20-50?
I think a brief description could help on the discussion.
cheers
milton
On Thu, Jul 16, 2009 at 5:40 PM, Kel Lam wrote:
> My institute has been heavily dependent on SAS for the past while, and
> S
Thanks for the help but...
I did the required transformations at the end of the code. The thing that I
dont understand is: Why is the predicted value (from the predict function)
not either the mean or median. Sorry I was not clear in my explanation.
Thanks,
Sean
On 7/16/09, David Winsemius w
just got my westgrid account, and finished downloading R-2.9.1 onto
Glacier(.westgrid.ca). how do i run my files in R on westgrid? I have
several ".r" and ".txt" files that run from a main driver program - so all i
need to run is the driver file (which is .txt). Are their any good sites
that ha
On Jul 16, 2009, at 7:41 PM, Sean Brummel wrote:
I am trying to generate predictions from a weibull survival curve
but it
seems that the predictions assume that the shape(scale for
survfit) parameter is one(Exponential but with a strange rate
estimate?).
Here is an examle of the problem, th
Hi, Steve!
Thanks for reply!
Acutally, that's what I thought too. But I believe some special control
terms are needed. Do you know how could I find some examples?
Thanks again.
Popo
2009/7/16 Steve Lianoglou
> Hi,
>
>
> On Jul 16, 2009, at 2:03 PM, popo UBC wrote:
>
> Hi there!
>>
>> S provi
I am trying to generate predictions from a weibull survival curve but it
seems that the predictions assume that the shape(scale for
survfit) parameter is one(Exponential but with a strange rate estimate?).
Here is an examle of the problem, the smaller the shape is the worse the
discrepancy.
###
On Jul 16, 2009, at 6:41 PM, Ross Boylan wrote:
If m is a matrix and s is a logical matrix of the same dimensions, I'd
like to be able to update m with
m[s] <- 0
If m2 is another matrix, I'd also like to be able to do
m[s] <- m2
where elements of m for which s is TRUE get the corresponding
el
I imported the attached tiff file and converted the coordinate system to
long lat and graphed it:
californiatiff<- readGDAL("california1.tif")
proj4string(californiatiff)
> rasterprojection <- spTransform(californiatiff, CRS("+proj=longlat")
however, when using the plot command for rasterprojecti
My institute has been heavily dependent on SAS for the past while, and
SAS is starting to charge us a very deep amount for license renewal.
Since we are a non-profit organization that is definitely not
sustainable. The team is brainstorming possibility of switching to R,
at least gradually. I am
On Thu, Jul 16, 2009 at 4:44 PM, Erik Iverson wrote:
> I have a list of logical expressions, and I would really like it if the
> "names" of the components of the list were identical to the corresponding
> logical expression.
>
> So, as an example:
>
> df.example <- data.frame(a = 1:10, b = rnorm(
On Jul 16, 2009, at 5:53 PM, KARSTEN G HOLMQUIST wrote:
Hello,
I have a series of questions that I hope will be simple to answer.
Basically I would like a code to do the following so that I can
compute the distribution free test for the slope of a postulated
regression line (Theil test)
If m is a matrix and s is a logical matrix of the same dimensions, I'd
like to be able to update m with
m[s] <- 0
If m2 is another matrix, I'd also like to be able to do
m[s] <- m2
where elements of m for which s is TRUE get the corresponding element of
m2.
However, this doesn't work in R 2.7.1.
Dear R-friends,
I removed the directory, untarzed the source files and
now I can compile the R.
Thanks for those suggested it off-list.
Bests
milton
On Thu, Jul 16, 2009 at 6:08 PM, milton ruser wrote:
> Dear all,
>
> I'm trying to compile R 2.9.1 on my home directory under debian,
> and as I
Could you please give a reproducible example?
Thanks,
Hadley
On Thu, Jul 16, 2009 at 5:41 PM, Alex Brown wrote:
> Hi there,
>
> I'm trying to find out the command to stop clipping to plot region in
> ggplot.
>
> I have a bar chart (axis flipped) with labels on the bars, but the
> lab
cddesjar wrote:
>
> Hi,
> I am trying to do run the following model saved in "C:/bugs/sus.bug"
>
> model {
>for (i in 1:n){
> y[i] ~ dpois(lamdba[i])
> log(lambda[i]) <- mu+bmale[male[i]]+bschn[schn[i]]+epsilon[i] #
> epsilon[i] ~ dnorm(0,tau.epsilon)
>}
>mu ~ dnorm(
Dear all,
I'm trying to compile R 2.9.1 on my home directory under debian,
and as I need to play with RPy, I tryed compile R using:
./configure --enable-R-shlib
But, during the compilation, I get the error below:
-
using as R_SHELL for scripts ... /bin/sh
conf
When I try this:
tmp <- randz<-matrix(rnorm(200),2000,1)
dim(tmp)
[1] 2000 1
It gives a result in less than one second. Very quick. And the
resulting matrix is not alarmingly big.
There must be some other problem. Perhaps your computer has only a
very small amount of memory?
Steve Lianoglou-6 wrote:
>
> Hi,
>
> On Jul 16, 2009, at 4:54 PM, Jose Narillos de Santos wrote:
>
>> Hi I want to simulate random numbers normal distributed with this size
>> (2000,1).
>>
>> I tried this but my computer exhaust there is a fast way to make it?
>>
>>
>> randz<-matrix(rnorm
I suppose it's conceivable that your object named "object" has some
non-standard character(s) in it that cause the code underlying cat()
to do something weird. For example,
cat('abcdef','\r','\n')
bcdef
Appears to truncate the first character.
Basic debugging suggests breaking down the pr
Hello,
I have a series of questions that I hope will be simple to answer. Basically I
would like a code to do the following so that I can compute the distribution
free test for the slope of a postulated regression line (Theil test). As I am
testing the null hypothesis that slope = 0 against t
As others noted, you can use the built in function colSums, but you
said you're writing your own. Given what you've got so far, that
makes the issue one of structuring the output.
Try
csum <- function(m)
{
a = data.frame(m)
s = lapply(a,sum)
unlist(s)
}
lapply() return
That is what you have. It just prints to screen that way.
If you want you could put it into a one column data.frame, that is,
mydata <- data.frame(myvector) which may give you the layout you want.
--- On Thu, 7/16/09, Andriy Fetsun wrote:
> From: Andriy Fetsun
> Subject: [R] Transformation
Hello,
I'm having a problem in R. I'm getting an error message that reads, "Too
many open files". I'm opening files and closing them (and unlinking
them), but when I go through that process 509 times, the program halts
and I get this error message: "cannot open the connection" with warning
me
Hi,
On Jul 16, 2009, at 4:54 PM, Jose Narillos de Santos wrote:
Hi I want to simulate random numbers normal distributed with this size
(2000,1).
I tried this but my computer exhaust there is a fast way to make it?
randz<-matrix(rnorm(200),2000,1)
I'll refrain from asking if yo
I'm guessing you want to perform some sort of transformation on all the
elements in the matrix you've posted and that you've only presented
those 3 elements as an example of how the transformation will affect
those 3 elements. Is that right?
--
David
-Original Message-
From: r-help-bo
This is awesome!
Total continue to be amazed.
Thanks again!
--- On Thu, 7/16/09, Gabor Grothendieck wrote:
> From: Gabor Grothendieck
> Subject: Re: [R] Best way to replace :SS with :00
> To: "Jason Rupert"
> Cc: R-help@r-project.org
> Date: Thursday, July 16, 2009, 3:45 PM
> Try the follow
Hi I want to simulate random numbers normal distributed with this size
(2000,1).
I tried this but my computer exhaust there is a fast way to make it?
randz<-matrix(rnorm(200),2000,1)
[[alternative HTML version deleted]]
__
R-help@
Try this:
> s <- c("df.example$a > 7", "df.example$b < 4")
> sapply(s, function(x) eval(parse(text = x)))
df.example$a > 7 df.example$b < 4
[1,]FALSE TRUE
[2,]FALSE TRUE
[3,]FALSEFALSE
[4,]FALSE
On 17/07/2009, at 8:09 AM, Andriy Fetsun wrote:
Hi Colleagues,
Could you please help?
I get as the output of my calculations following
[1] 0.00e+00 1.89e-04 3.933000e-05 1.701501e-04
2.040456e-04
[6] 3.119242e-04 2.545665e-04 1.893930e-03 1.303112e-03
9.880183e-04
Try the following which replaces : followed by any two
characters followed by space with the required string:
> x <- c("HH:MM:SS AM", "HH:MM:SS PM")
> sub(":.. ", ":00 ", x)
[1] "HH:MM:00 AM" "HH:MM:00 PM"
On Thu, Jul 16, 2009 at 4:20 PM, Jason Rupert wrote:
>
> Dang it. I forgot to mention the
Dear R-help,
I am having quite a difficult time coming up with what I want to do involving
named lists.
I have a list of logical expressions, and I would really like it if the "names"
of the components of the list were identical to the corresponding logical
expression.
So, as an example:
On Thu, Jul 16, 2009 at 10:09 PM, Andriy Fetsun wrote:
> [1] 0.00e+00 1.89e-04 3.933000e-05 1.701501e-04 2.040456e-04
> [6] 3.119242e-04 2.545665e-04 1.893930e-03 1.303112e-03 9.880183e-04
> [11] 1.504378e-03 1.549246e-03 5.877690e-04 4.771359e-04 8.528219e-04
That it
Hi Colleagues,
Could you please help?
I get as the output of my calculations following
[1] 0.00e+00 1.89e-04 3.933000e-05 1.701501e-04 2.040456e-04
[6] 3.119242e-04 2.545665e-04 1.893930e-03 1.303112e-03 9.880183e-04
[11] 1.504378e-03 1.549246e-03 5.877690e-04 4.771
Dear sir,
I have 2 questions:
Question 1:
suppose that we have the following:
a<-matrix(c(0,2,0,4,0,6,5,8,0),nrow=3)
colnames(a)<-c("F1","F2","F3")
rownames(a)<-c("A1","A2","A3")
a
ind <- which(a == 0, arr = TRUE)
mapply("[", dimnames(a), as.data.frame(ind))
I want to add a column to the resu
You can still use my suggestion (sent offline)
if you have a vector of strings, or just one
> my.str <- "HH:MM:SS AM"
> substr(my.str,7,8) <- "00"
> my.str
[1] "HH:MM:00 AM"
Doesn't matter if it's AM or PM because the function is just acting
on the 7th and 8th characters.
On Jul 16, 2009,
Dang it. I forgot to mention the actual format of the time is the following:
"HH:MM:SS AM" or "HH:MM:SS PM"
And I would still hope for them to end up with the following format:
"HH:MM:00 AM" or "HH:MM:00 PM"
How would you propose handling that condition?
I tried to use strsplit with items to
probably :
a <- rep("HH:MM:SS", 5)
substring(a, 7) = "00"
a
Not sure if there is an R way to do this or a regular express way, but here
is what I am trying to do.
I've got lots of data where the format is HH:MM:SS, but I need to format it
like HH:MM:00, i.e. round the second down to zero.
What
It is because the nesting structure perfectly explains the data (i.e., there
is only one observation and, therefore, no variation for each Ind in each
Treatment).
e=rnorm(90,0,1)
x=rep(1:3,30)
y=rep(1:30,each=3)
z=x+y+e
ano=aov(z~factor(y)/factor(x))
ano
residuals(ano)
Best,
Daniel
---
Dear R-Users,
i want to use the function mice of the mice package with data, that
contains dates, but this gives an error
x <- Sys.time()
dat <- data.frame(dates=(1:10)*60*60*24+x, size=rnorm(10))
dat$dates[c(3,7)]<-NA
mice(dat)
Fehler in check.imputationMethod(imputationMethod,
defaultImpu
Hint: You have 27 observations fit by 27 fixed effects (including the mean).
You need to consult a statistician, as you seem not to have the basic
statistical understanding required. Person should be a random effect. Was
this a homework problem, perchance?
Bert Gunter
Genentech Nonclinical Biost
It your times are chron objects then
trunc(tt, "minutes")
where tt are your times may do what you want. It truncated but that seems to
be what you want.
--- On Thu, 7/16/09, Jason Rupert wrote:
> From: Jason Rupert
> Subject: [R] Best way to replace :SS with :00
> To: R-help@r-project.or
Hi,
Not sure if there is an R way to do this or a regular express way,
but here is what I am trying to do.
I've got lots of data where the format is HH:MM:SS, but I need to
format it like HH:MM:00, i.e. round the second down to zero.
What is the best way to do this?
Probably not the be
Have you looked at the chron package? It has a trunc.times function:
??"times"# would have shown this to you
And the help page appears to provide exactly what was requested.
-puzzlement follows
I did get somewhat unexpected results when I applied what seems to be
the obvi
On 16-Jul-09 19:40:20, Noah Silverman wrote:
> Hello,
> I'm brand new to using R. (I've been using Rapid Miner, but would
> like to move over to R since it gives me much more functionality.)
>
> I'm trying to learn how to do a conditional logit model.
>
> My data has one dependent variable, 2 in
Hello,
I'm brand new to using R. (I've been using Rapid Miner, but would like
to move over to R since it gives me much more functionality.)
I'm trying to learn how to do a conditional logit model.
My data has one dependent variable, 2 independent variables and a
"group" variable.
example:
c
I am having trouble setting up a nested anova model. Here is a
truncated version of my data:
> data
Ind Treatment PC1
1 PER14SC 1.14105282
2 PER14SH 1.45348615
3 PER14AC 2.45096904
4 PER25SC 1.23687887
5 PER25SH 4.54797450
6 PER2
Not sure if there is an R way to do this or a regular express way, but here is
what I am trying to do.
I've got lots of data where the format is HH:MM:SS, but I need to format it
like HH:MM:00, i.e. round the second down to zero.
What is the best way to do this?
Thanks again.
Jason
_
What is dataset?
What is this supposed to be doing?
newsample<-dataset[sample(m,replace=T),]
--- On Thu, 7/16/09, MarcioRibeiro wrote:
> From: MarcioRibeiro
> Subject: [R] Sample Function
> To: r-help@r-project.org
> Received: Thursday, July 16, 2009, 11:55 AM
>
> Hi listers,
> Suppose I
Dear voidobscura,
Try either:
colSums(mdat)
# or
apply(mdat, 2, sum)
See ?colSums and ?apply for more details.
HTH,
Jorge
On Thu, Jul 16, 2009 at 2:25 PM, voidobscura wrote:
>
> Alright, so I am trying to write my own function to calculate column sums
> in
> a matrix. I want the result a
The easiest way is to just do something like this:
> mdat <- matrix(c(4,2,3, 11,12,13), nrow = 2, ncol=3)
> mdat
[,1] [,2] [,3]
[1,]43 12
[2,]2 11 13
> as.vector ( colSums ( mdat ) )
[1] 6 14 25
>
HTH
--
David
-Original Message-
From: r-help-boun...@r-project.o
Alright, so I am trying to write my own function to calculate column sums in
a matrix. I want the result as a single list with the values.
So far I have:
csum<-function(m)
{
a = data.frame(m)
s = lapply(a,sum)
return(s)
}
What is the easiest way to have it return in a f
It sounds like you might want to draw the convex hull for each group
of points. There is a package called "chplot" which appears to
do this, though I haven't used it...
albyn
On Thu, Jul 16, 2009 at 06:23:54PM +0100, Ahmed, Sadia wrote:
> Hi,
>
> I'll try to keep my question brief and to the p
Look at the chull function and its examples.
--
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 Ahmed, Sadia
> S
This is just a variation of David's solution below:
intersect(X,Y)
# [1] "red" "blue" "green"
HTH,
Jorge
On Thu, Jul 16, 2009 at 12:07 PM, David Huffer wrote:
> Or...
>
> > X = c ("red", "blue", "green", "black" ) ; Y = c("red", "blue",
> "green", "magenta", "cyan")
> > unique ( c ( X [X %
Hi,
I'll try to keep my question brief and to the point, to anyone who helps 'THANK
YOU!'
I want to get a circle/ring/buffer or some other form of drawn line around
certain points on a x,y plot, the points usually don't form a circle, often
they form a 'wobbly' shape, so ideally I'd like some
Hi,
On Jul 16, 2009, at 2:03 PM, popo UBC wrote:
Hi there!
S provide some functions, such as "qq.weibull", to produce various
qqplot
for model checking. But I can't find the corresponding version in R.
Does
any R package available to produce these qqplots?
Many thanks in advance!!
Per
Hi there!
S provide some functions, such as "qq.weibull", to produce various qqplot
for model checking. But I can't find the corresponding version in R. Does
any R package available to produce these qqplots?
Many thanks in advance!!
Cheers~~~
Popo
[[alternative HTML version del
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of MarcioRibeiro
> Sent: Thursday, July 16, 2009 8:56 AM
> To: r-help@r-project.org
> Subject: [R] Sample Function
>
>
> Hi listers,
> Suppose I have a dataset with n=10 observation
Kurt Hornik is currently out of office and will be back soon.
Moreover, CRAN was also unmanned in repspect to all other issues since
all other CRAN maintainers have been attending useR! and/or DSC conferences.
Just a bit more patience, CRAN will be in regular duty soon.
Best,
Uwe Ligges
J
Hi,
I am trying to do run the following model saved in "C:/bugs/sus.bug"
model {
for (i in 1:n){
y[i] ~ dpois(lamdba[i])
log(lambda[i]) <- mu+bmale[male[i]]+bschn[schn[i]]+epsilon[i] #
epsilon[i] ~ dnorm(0,tau.epsilon)
}
mu ~ dnorm(0,.0001)
bmale ~ dnorm(0,.0001)
tau.epsilon
have a look at the help page for ?"["; try also this
a[[2]]
a[[2]][2]
a[[3]][3]
Best,
Dimitris
voidobscura wrote:
a
[[1]]
[1] 1 2 3
[[2]]
[1] 4 5 6
[[3]]
[1] 7 8 9
I need to access individual elements, such as the 5 or the 9. Can anyone
please tell me the syntax to do this?
tia
--
Di
Like,
> a = list (1:3,4:6,7:9)
> a
[[1]]
[1] 1 2 3
[[2]]
[1] 4 5 6
[[3]]
[1] 7 8 9
> a [[2]] [2]
[1] 5
> a [[3]] [3]
[1] 9
>
HTH
--
David
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of voidobscura
Sent: Thursday, July 16,
> a
[[1]]
[1] 1 2 3
[[2]]
[1] 4 5 6
[[3]]
[1] 7 8 9
I need to access individual elements, such as the 5 or the 9. Can anyone
please tell me the syntax to do this?
tia
--
View this message in context:
http://www.nabble.com/Problems-with-lists...-tp24519517p24519517.html
Sent from the R help
Hi listers,
Suppose I have a dataset with n=10 observations and I want to sample with
replacement.
My new sample is of size m=9.
So, I am using the following code...
newsample<-dataset[sample(m,replace=T),]
The problem is that generates the new sample and the last observation of my
data set is nev
Hi All,
This week SPSS Inc. announced version 18, which has some very
interesting enhancements regarding R. It will now ship with R on the
DVD, along with the PASW Statistics-R Integration Package. Previously
this required a download of R, and of the Integration Package. This
may be the first time
Hi there,
I'm trying to find out the command to stop clipping to plot region in
ggplot.
I have a bar chart (axis flipped) with labels on the bars, but the
labels are clipped at the plot region box.
I know it's possible to turn this off for base and lattice, but how
about ggplot?
Hi listers,
I am
--
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Sent from the R help mailing list archive at Nabble.com.
__
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All,
I've been using AQUAMACS Emacs to develop R code and run/test the code
by clicking "eval line and step" from within the editor which opens R
in a new tab and displays the output. This calls X11 when a display
window opens up. This was all working fine weeks ago but now when I
try to get a
It has to be related to 'cat' because the output of 'cat' is truncated. I am
just tyring to find out some possible reasons as to why it is truncated. I have
been unable to form an array like is in the test program. Do you think there is
something else that is gobbling up the output from cat that
Or...
> X = c ("red", "blue", "green", "black" ) ; Y = c("red", "blue",
"green", "magenta", "cyan")
> unique ( c ( X [X %in% Y] , Y [Y %in% X] ) )
[1] "red" "blue" "green"
>
David
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf
Leave the mgp setting alone and use xlab='' when you create the original plot.
This will put the y-label in the usual place and not create the x-label. Then
use the "title" or "mtext" function to place the x label where you want it.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistic
Is there a function in R to do a-b where a and b are two non-numeric
sets
(or intersection complement of these two sets).
I think you're looking for "setdiff"
-steve
--
Steve Lianoglou
Graduate Student: Physiology, Biophysics and Systems Biology
Weill Medical College of Cornell University
C
Hi,
Is there a function in R to do a-b where a and b are two non-numeric sets
(or intersection complement of these two sets).
Kind Regards,
Praveen.
[[alternative HTML version deleted]]
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https:
Try this:
> Lines <- "19:21:59.855
+ 19:21:59.905
+ 19:21:59.955
+ 19:22:0.5
+ 19:22:0.55
+ 19:22:0.105
+ 19:22:0.155
+ 19:22:0.205
+ 19:22:0.255
+ 19:22:0.305
+ 19:22:0.355
+ 19:22:0.405"
>
> DF <- read.table(textConnection(Lines))
> library(chron)
> tt <- times(DF[[1]])
> diff(tt)
[1
Hello,
I have been attempting to write a script that automatically calculates time
intervals from a list of hh.mm.ss.ms timestamps. Should be easy, but I keep
running into problems. Any help would be greatly appreciated!!
Problems:
1. If I parse the timestamp, it doesn't always return the digits
By the way, note that read.zoo passes the ... arguments to read.table
and so can use the same skip= and nrows= arguments that read.table
uses. These can be used to read in a subset of the rows.
On Thu, Jul 16, 2009 at 10:35 AM, Gabor
Grothendieck wrote:
> There is no such limitation. There is l
Hi,
The following works fine:
> f
[1] 0.08 0.03 0.04
> A2
[,1] [,2] [,3]
[1,] 0.00e+00 0.00 0.00e+00
[2,] 0.00e+00 0.00 0.00e+00
[3,] 2.999651e-03 0.0009094342 1.945708e-03
[4,] 4.124431e-05 0.0001360390 1.203345e-05
[5,] 3.027932e-04 0
Thxs to all of you
Alberto
On Thu, Jul 16, 2009 at 7:03 AM, Gabor Grothendieck wrote:
> Here is a variation on the solution below (first line
> is the same but second differs):
>
> > ind <- which(a == 0, arr = TRUE)
> > mapply("[", dimnames(a), as.data.frame(ind))
> [,1] [,2]
> [1,] "A1" "F1
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