Hi,
I think you could also use this way (via array, see
http://r.789695.n4.nabble.com/apply-over-list-of-data-frames-td3057968.html)
b <- list()
b[[1]] = matrix(1:4, 2, 2)
b[[2]] = matrix(10:13, 2, 2)
b[[3]] = matrix(20:23, 2, 2)
b.a <- array(unlist(b), dim=c(2, 2, 3))
(b.mean <- apply(X = b.
On Mar 24, 2011, at 8:35 PM, barbara costa wrote:
Hi to all,
Does anybody knows why this is giving an error?
data(ToothGrowth)
# Two-way design with options
bargraph.CI(dose, len, group = supp, data = ToothGrowth,
xlab = "Dose", ylab = "Growth", cex.lab = 1.5, x.leg = 1,
Hello
I simply want to plot two variables against one 'year' variable in
qplot.
Is any way of doing this without reshaping data in long format and using
facet function afterwards?
Thank you
Denis
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R-help@r-project.org mailing list
https://stat.ethz.c
Hi Barbara,
Works just fine for me:
> require(sciplot)
Loading required package: sciplot
> data(ToothGrowth)
# Two-way design with options
bargraph.CI(dose, len, group = supp, data = ToothGrowth,
xlab = "Dose", ylab = "Growth", cex.lab = 1.5, x.leg = 1,
col = "black", angle
Hi Everybody,
I have a data set in which each observation has a pair of students,
with each kid id'd by a 4 digit number:
> head(PAIRS)
student1 student2
2 2213 2200
4 2198 2195
5 2199 2191
6 2229 2221
7 2247 2249
8 2250 2263
There is no significa
Trying to send to the list, take 3 (for some reason my mails get
bounced when I send to r-help@r-project.org, and not when I send to
r-h...@stat.math.ethz.ch)
On Thu, Mar 24, 2011 at 8:34 PM, Steve Lianoglou
wrote:
> Hi Jaimin,
>
> Please include r-help in your reply -- it's not included by defau
Hi to all,
Does anybody knows why this is giving an error?
data(ToothGrowth)
# Two-way design with options
bargraph.CI(dose, len, group = supp, data = ToothGrowth,
xlab = "Dose", ylab = "Growth", cex.lab = 1.5, x.leg = 1,
col = "black", angle = 45, cex.names = 1.25,
Worked just fine. I had been incorrectly trying
tapply(x, l.c.1, quantile(probs=0.75))
rather than
tapply(x, l.c.1, quantile, probs=0.75)
Thanks for the help -
SR
Steven H. Ranney
On Thu, Mar 24, 2011 at 6:03 PM, Peter Alspach
wrote:
> Tena koe Steven
>
> The ... argument of the apply se
Hi Jaimin,
Please include r-help in your reply -- it's not included by default,
so you'll have to use your email client's "reply all" functionality
when responding to posts coming from the mailing list.
This way (i) more people will be able to help you than just me; and
(ii) if someone runs into
Hi Steven,
See the prob argument under ?quantile. The following should be what you
want:
tapply(x, l.c.1, quantile, prob = 0.75)
HTH,
Jorge
*
*
On Thu, Mar 24, 2011 at 7:18 PM, Steven Ranney <> wrote:
> All -
>
> I have an example data frame
>
> x l.c.1
> 43.38812035 085
> 47.5571
Just have a look at ?quantile and the probs argument.
tapply(x, l.c.1, quantile,probs=0.75)
Anyway, quantiles and quartiles are not the same. I guess you meant the
3rd quartile.
> All -
>
> I have an example data frame
>
> x l.c.1
> 43.38812035 085
> 47.55710661 085
> 47.55710661 085
Tena koe Steven
The ... argument of the apply series of functions allows one to pass arguments
to the called function. So:
tapply(x, l.c.1, quantile, probs=0.75)
should work (although I haven't tested it).
HTH .
Peter Alspach
> -Original Message-
> From: r-help-boun...@r-projec
All -
I have an example data frame
x l.c.1
43.38812035 085
47.55710661 085
47.55710661 085
51.99211429 085
51.99211429 095
54.78449958 095
54.78449958 095
56.70201864 095
56.70201864 105
59.66361903 105
61.69573564 105
61.69573564 105
63.77469
On 11-03-24 4:08 PM, Thomas Lumley wrote:
On Fri, Mar 25, 2011 at 5:11 AM, David.Epstein
wrote:
Hello, I don't want to find out how to make packages unless that becomes
necessary. Also, I don't want to clog up the computer memory with functions
that I'm not using. (It would be great if someone
On Thu, Mar 24, 2011 at 3:25 PM, Pamela Allen wrote:
>
>
>
> Hi All,
>
>
>
> I have a data set of daily measurements of river flow.
> I would like to create a “ts” object from this data.
>
>
>
> Here’s a sample data set:
>
> date <- as.Date(c(1:300), format="%Y")
>
> year=as.numeric(format(date, f
Hello,
Im new to R and I just successfully installed Jri. However I dont quite get
the rtest examples. Is it working?
One the console:
Creating Rengine (with arguments)
Rengine created, waiting for R
re-routing stdout/err into R console
rBusy(0)
And the java window is asking me:
to save the wor
On Thu, Mar 24, 2011 at 2:24 PM, Kevin Burnham wrote:
> I have a data file with indicates pretest scores for a linguistics
> experiment. The data are in long form so for each of 33 subjects there are
> 400 rows, one for each item on the test, and there is a column called
> ‘Correct’ that shows ‘C
On Mar 24, 2011, at 1:37 PM, JP wrote:
Using Trellis, am successfully setting up a number of panels (25) in
which I
have two box and violin plots.
I would like to colour - one plot as RED and the other as BLUE (in
each
panel). I can do that with the box plots, but the violin density
are
It's likely that the loss function has a log() or 1/x and the finite difference
approximations to gradients have added / subtracted a small number and caused a
singularity. Unfortunately, you'll need to dig into the fGarch code or write
your own
(ouch!). Or perhaps the fGarch package maintainer wi
On 25/03/11 09:08, Thomas Lumley wrote:
On Fri, Mar 25, 2011 at 5:11 AM, David.Epstein
wrote:
Hello, I don't want to find out how to make packages unless that becomes
necessary. Also, I don't want to clog up the computer memory with functions
that I'm not using. (It would be great if someone
On Mar 24, 2011, at 1:15 PM, jessiecy2001 wrote:
Hi all,
I am a new user for R.
I faced a questions about subscript out of bounds.
Although I checked some other helps before I sent this email, I
still don't know
how to deal with this problems.
I want to use "domain" function to run specie
On Mar 24, 2011, at 1:13 PM, Madaliso Mulaisho wrote:
I am wondering if there is a good way to work with data that is
indexed in
time, via timestamps with a resolution in milliseconds. As I
understand it,
the POSIX classes have a resolution i n terms of seconds, and will not
process fracti
On 24 March 2011 at 13:13, Madaliso Mulaisho wrote:
| I am wondering if there is a good way to work with data that is indexed in
| time, via timestamps with a resolution in milliseconds. As I understand it,
| the POSIX classes have a resolution i n terms of seconds, and will not
| process fractio
Hi Kevin,
Try this (untested):
sapply(split(pretestdata, Subject), function(l) with(l$Correct == "C"))
HTH,
Jorge
On Thu, Mar 24, 2011 at 3:24 PM, Kevin Burnham <> wrote:
> I have a data file with indicates pretest scores for a linguistics
> experiment. The data are in long form so for each
Try this:
tapply(mydata$Correct, mydata$Subject, function(x)sum(x == "C"))
Sarah
On Thu, Mar 24, 2011 at 3:24 PM, Kevin Burnham wrote:
> I have a data file with indicates pretest scores for a linguistics
> experiment. The data are in long form so for each of 33 subjects there are
> 400 rows, o
Hi Hui,
Ssee ?Reduce for more details. You might try something along the lines of
> mymean <- function(x) Reduce("+", x)/length(x)
> add(b)
[,1] [,2]
[1,] 10.3 12.3
[2,] 11.3 13.3
HTH,
Jorge
On Thu, Mar 24, 2011 at 11:07 AM, Hui Du <> wrote:
> Hi All,
>
>
Sorry, the above should have been
> mymean <- function(x) Reduce("+", x)/length(x)
> mymean(b)
[,1] [,2]
[1,] 10.3 12.3
[2,] 11.3 13.3
Apologies for the noise.
Best,
Jorge
On Thu, Mar 24, 2011 at 3:54 PM, Jorge Ivan Velez <> wrote:
> Hi Hui,
>
> Ssee ?Reduce for m
?strptime or ?options
> d1 <- as.POSIXct("2009-09-30 10:00:00.543")
> d1
[1] "2009-09-30 10:00:00 CDT"
> # Oh, yeah, I forgot
> op <- options(digits.secs=6) # or 3
> d1
[1] "2009-09-30 10:00:00.543 CDT"
> # The accuracy was there already
HTH,
David L. Reiner, PhD
Head Quant
XR Trading LLC
-
> I want to use "domain" function to run species distribution model in R.
> I used "sp.occ.do<-domain(env.pre, pred_train, factors=c('continent')) &
> sp.occ.pre.pro<-predict(env.pre, sp.occ.do)" to predict species distribution.
> BUT, I got "Error in domdist(object, ln[j], vals[, ln[j]]) : subscr
Are you taking the same online course as Barth Riley, who posed a very
similar question only this morning?
On Mar 24, 2011, at 11:07 AM, Hui Du wrote:
Hi All,
Suppose I have data like
b[[1]] = matrix(1:4, 2, 2)
b[[2]] = matrix(10:13, 2, 2)
b[[3]] = matrix(20:23, 2, 2)
[[1]]
package msm has some examples with this type of type, ie modeling disease
state transitions in continuous time, using multi-state markov models.
hth, Ingmar
On Thu, Mar 24, 2011 at 6:22 PM, Rasanga Ruwanthi wrote:
> Dear List,
>
> I have some longitudinal data, each patient was followed at times
On Fri, Mar 25, 2011 at 5:11 AM, David.Epstein
wrote:
> Hello, I don't want to find out how to make packages unless that becomes
> necessary. Also, I don't want to clog up the computer memory with functions
> that I'm not using. (It would be great if someone in this forum would
> explain how memor
Peter,
Regarding 1) I do not agree. See the following, simplified example:
x <- data.frame(ID=rep(1:2, each=4), Visit=rep(c(1:4), 2),
ptA=c(7,8,9,10,17,18,19,20), ptB=c(5,6,7,8,21,20,19,18))
In this data frame you have only 2 patients with 4 visits each, but the
correlation of ptA and ptB is
I have a data file with indicates pretest scores for a linguistics
experiment. The data are in long form so for each of 33 subjects there are
400 rows, one for each item on the test, and there is a column called
Correct that shows C for a correct response and E for an incorrect
response. I
Hi All,
I have a data set of daily measurements of river flow. I would like to
create a "ts" object from this data.
Here's a sample data set:
date <- as.Date(c(1:300), format="%Y")
year=as.numeric(format(date, format = "%Y"))
month=as.numeric(format(date, format = "%m"))
julianday=
Hi there,
I was doing great with R, ggplot and tikzDevice, when suddenly it stopped
working.
I can do a plot just fine on the R online graphics device. I can also open a
tikzdevice via tikz(). However, the plot (both ggplot and R graphics plot)
call leads to the error:
"createLockFile(lockname):
Using Trellis, am successfully setting up a number of panels (25) in which I
have two box and violin plots.
I would like to colour - one plot as RED and the other as BLUE (in each
panel). I can do that with the box plots, but the violin density areas just
take on one colour.
My basic call is as
Dear List,
I have some longitudinal data, each patient was followed at times 0, 12, 16, 24
weeks and measure severity of a illness (0-worse, 1-same, 2-better). So,
longitudinal response is categorical. I was wondering whether lmer in R can
fit a model for this type of data. If so, how we code
Hi all,
I am a new user for R.
I faced a questions about subscript out of bounds.
Although IÂ checked some other helps before I sent this email, I still don't
know
how to deal with this problems.
I want to use "domain" function to run species distribution model in R.
I used "sp.occ.do<-domain
On Thu, Mar 24, 2011 at 12:36 PM, Ian Schiller
wrote:
> Thanks for your prompt answer Mr. Ripley. It appears the version of tar I'm
> using is exactly the same as the one included in the latest Rtools. I
> noticed that the Rtools/bin directory was not first in path, so I corrected
> this. Ho
Yes, yes statisticians. But just remember that this is a plague that
we have visited on ourselves. And continue to -- just peruse
practically any statistical journal.
("People who live in grass houses shouldn't stow thrones")
Cheers,
Bert
(To non-native English speakers: my apologies for the bad
On Thu, Mar 24, 2011 at 1:28 PM, Ian Schiller
wrote:
> Ok guys. First thanks again for your help. It is very appreciated. Now
> here's the thing. I downloaded the Rcmd.bat and included in the path. I've
> also changed the location of the package source and ran R CMD build mypackage
> (Rcmd
Hello, I don't want to find out how to make packages unless that becomes
necessary. Also, I don't want to clog up the computer memory with functions
that I'm not using. (It would be great if someone in this forum would
explain how memory is used when I type library(MASS) and then use only one
funct
I am wondering if there is a good way to work with data that is indexed in
time, via timestamps with a resolution in milliseconds. As I understand it,
the POSIX classes have a resolution i n terms of seconds, and will not
process fractional seconds from a string. Is this correct. I realize that
On 24/03/2011 12:36 PM, Ian Schiller wrote:
Thanks for your prompt answer Mr. Ripley.
Although Brian Ripley is faster at most things than I am, in this case I
was more prompt than him.
It appears the version of tar I'm using is exactly the same as the one
included in the latest Rtools.
Hi All,
Suppose I have data like
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide
Hi All,
Suppose I have data like
b[[1]] = matrix(1:4, 2, 2)
b[[2]] = matrix(10:13, 2, 2)
b[[3]] = matrix(20:23, 2, 2)
[[1]]
[,1] [,2]
[1,]13
[2,]24
[[2]]
[,1] [,2]
[1,] 10 12
[2,] 11 13
[[3]]
[,1] [,2]
[1,] 20 22
[2,] 21 23
On Thu, 24 Mar 2011, jdanielnd wrote:
Hello!
I am not familiar to deal with S4 objects in R, so this question can be
stupid, but I hope I can get an answer. :P
I'm trying to extract what are the response and explanatory variables from a
Binary Tree and Random Forest.
I could already extract t
Hi All,
I have a data set of daily measurements of river flow.
I would like to create a ts object from this data.
Heres a sample data set:
date <- as.Date(c(1:300), format="%Y")
year=as.numeric(format(date, format = "%Y"))
month=as.numeric(format(date, format = "%m"))
juliand
For the record, you can avoid all the Cygwin tar problems in 2.13.0
by setting the environment variable TAR to 'internal': you most likely
do not even need Rtools.
You may get other issues, though (especially from Cygwin and
permissions), which is why it is not the default yet on Windows.
On
> (Rcmd.bat is not working so I assume there is something I'm doing wrong)
I'd just like to let everyone knows that Rcmd.bat IS working properly after
all. I had messed with the PATH "environmental variable". In fact, like Gabor
Grothendieck pointed out to me, one can just copy the Rcmd.bat fi
> You likely have the Hide extensions options enabled on your Windows system.
Yes you're right! Guys, I thank you all for your precious help!
Ian
**
IAN SCHILLER, M.Sc.
S
Ok guys. First thanks again for your help. It is very appreciated. Now
here's the thing. I downloaded the Rcmd.bat and included in the path. I've
also changed the location of the package source and ran R CMD build mypackage
(Rcmd.bat is not working so I assume there is something I'm doing w
> A couple of things to try:
- try re-installing Rtools and R
- put Rcmd.bat from http://batchfiles.googlecode.com on your path and
then try re-building
Rcmd.bat build mypackage
Rcmd.bat will automatically find R and certain other files from the
registry and temporarily set certain environment va
> Although Brian Ripley is faster at most things than I am, in this case I
was more prompt than him.
My mistake. I'm sorry!
> I don't know the CRAN policy on this, but I think it would be worthwhile
fixing the problem you can see, in case there are other problems that
you can't see.
Agree.
Oh Sorry Mr Murdoch! I read a post where both you and Brian Ripley were
involved and the confusion probably arose from that!
Ok thanks Mr Ligges.
Ian
**
IAN SCHILLER, M
On 24.03.2011 17:36, Ian Schiller wrote:
Thanks for your prompt answer Mr. Ripley. It appears the version of tar I'm
using is exactly the same as the one included in the latest Rtools. I noticed
that the Rtools/bin directory was not first in path, so I corrected this.
However, I'm still n
Thanks for your prompt answer Mr. Ripley. It appears the version of tar I'm
using is exactly the same as the one included in the latest Rtools. I noticed
that the Rtools/bin directory was not first in path, so I corrected this.
However, I'm still not able to create a tar.gz file with the R CM
Thanks a lot,
Arnau.
El 24/03/2011, a las 16:46, Bernd Weiss escribió:
> Am 24.03.2011 11:28, schrieb Arnau Mir:
>> Hello.
>>
>> Somebody knows how to write in R all the permutations of a given set?
>> For example, is the set is {a,b,c} I want the following:
>> a,b,c
>> a,c,b
>> b,c,a
>> b,a,c
>
On Thu, Mar 24, 2011 at 9:29 AM, Michael Bach wrote:
> Dear R users,
>
> Given this data:
>
> x <- seq(1,100,1)
> dx <- as.POSIXct(x*900, origin="2007-06-01 00:00:00")
> dfx <- data.frame(dx)
>
> Now to play around for example:
>
> subset(dfx, dx > as.POSIXct("2007-06-01 16:00:00"))
>
> Ok. Now fo
Dear Orvalho Augusto,
Thanks for your response. But we met a new problem. We set as waht you said
in the last email and run installing agian. But we got the error:
rJava.h:22:17: error: jni.h: No such file or directory.
Actually we had found "jni.h" under {JAVA_HOME}/../include and the path had
be
Well this starts veering towards fortune(194) and fortune(218).
Though I did one time receive a response from an editor and reviewers asking
for fewer p-values in favor of more confidence intervals. I was excited that
someone was willing to move in that direction (unfortunately, that particular
Am 24.03.2011 11:28, schrieb Arnau Mir:
Hello.
Somebody knows how to write in R all the permutations of a given set?
For example, is the set is {a,b,c} I want the following:
a,b,c
a,c,b
b,c,a
b,a,c
c,a,b
c,b,a
library(gregmisc)
permutations(3, 3, letters[1:3])
gives
[,1] [,2] [,3]
[1,]
On Mar 24, 2011, at 10:02 AM, Michael Bach wrote:
> Hi David,
>
> your approach selects the datapoints between "2007-06-01 10:00:00"
> and "2007-06-01 14:00:00" true enough. However, my real dataset is
> for several months and years. So I need data points between 10:00:00
> and 14:00:00 onl
Hello.
Somebody knows how to write in R all the permutations of a given set?
For example, is the set is {a,b,c} I want the following:
a,b,c
a,c,b
b,c,a
b,a,c
c,a,b
c,b,a
Thanks,
Arnau.
Arnau Mir Torres
Edifici A. Turmeda
Campus UIB
Ctr
>
> I wanted to point out that this is a little easier to express with the
> lubridate package:
>
> subset(dfx, dx > ymd("2007-06-01") & hour(dx) > 14 & hour(x) < 10)
>
Thanks, will have a look at it..
>
> but I presume you meant:
>
> subset(dfx, dx > ymd("2007-06-01") & hour(dx) > 10 & hour(x)
@Kenn: Yeah I have a way of loosing concentration when fiddling around and
trying to get it work. Thanks for pointing out that I supplied a third
argument when I wanted an additional test in the second argument instead.
Also my selection choices indeed do not make any sense, no wonder your
"parser"
On Thu, Mar 24, 2011 at 8:29 AM, Michael Bach wrote:
> Dear R users,
>
> Given this data:
>
> x <- seq(1,100,1)
> dx <- as.POSIXct(x*900, origin="2007-06-01 00:00:00")
> dfx <- data.frame(dx)
>
> Now to play around for example:
>
> subset(dfx, dx > as.POSIXct("2007-06-01 16:00:00"))
>
> Ok. Now fo
Make your own copy of predict.rpart:
mypredict <- rpart:::predict.rpart
(Even better to grab the source code version, then you can start with a
copy that has comments in it!)
Edit it to add type='where': about 1/3 of the way down you will see that
the routine has computed a variable "where"; th
On 24/03/2011 10:36 AM, Ian Schiller wrote:
Dear all,
I'm trying to create my own package by following for instance
http://epub.ub.uni-muenchen.de/6175/2/tr036.pdf In step 6 (page 19) it says that the
command 'R CMD build' should result in a tar.gz file. However,
after running the command I
On 03/24/2011 06:29 AM, Michael Bach wrote:
Dear R users,
Given this data:
x<- seq(1,100,1)
dx<- as.POSIXct(x*900, origin="2007-06-01 00:00:00")
dfx<- data.frame(dx)
Now to play around for example:
subset(dfx, dx> as.POSIXct("2007-06-01 16:00:00"))
Ok. Now for some reason I want to extract
Dear all,
I'm trying to create my own package by following for instance
http://epub.ub.uni-muenchen.de/6175/2/tr036.pdf In step 6 (page 19) it says
that the command 'R CMD build ' should result in a tar.gz file.
However, after running the command I end up with a tar file only (no gz
extensio
Hello,
I drew a simple area plot using ggplot2 using
set <- read.table(file="http://www.jovian.nl/set.csv";, head=1, sep=",")
library(ggplot2)
ggplot() +
layer(
data = set, mapping = aes(x = time, y = hours),
geom = "area", stat="smooth", color="red"
) +
layer(
data = set, mapping = aes(x
On Thu, Mar 24, 2011 at 1:29 PM, Michael Bach wrote:
> Dear R users,
>
> Given this data:
>
> x <- seq(1,100,1)
> dx <- as.POSIXct(x*900, origin="2007-06-01 00:00:00")
> dfx <- data.frame(dx)
>
> Now to play around for example:
>
> subset(dfx, dx > as.POSIXct("2007-06-01 16:00:00"))
>
> Ok. Now fo
Hello!
I am not familiar to deal with S4 objects in R, so this question can be
stupid, but I hope I can get an answer. :P
I'm trying to extract what are the response and explanatory variables from a
Binary Tree and Random Forest.
I could already extract the response variable from a Binary Tree u
Thomas,
You need to specify both the labels and the desired location:
plot(1:10, xaxt = "n")
axis(1, xaxp=c(2, 9, 7))
axis(4, labels=c("A", "B", "C", "D", "E"), at=seq(2, 10, by=2))
You want the labels at 2,4,6,8,10 so you need to specify that,
not 1:5.
Sarah
On Thu, Mar 24, 2011 at 9:55 AM, T
Hi David,
your approach selects the datapoints between "2007-06-01 10:00:00"
and "2007-06-01 14:00:00" true enough. However, my real dataset is for
several months and years. So I need data points between 10:00:00 and
14:00:00 only - independent of the date. I thought the name for that is
"aggregat
I have looked at many examples and tried many different combinations
of doing this, but with no luck. I have something like this:
plot(1:10, xaxt = "n")
axis(1, xaxp=c(2, 9, 7))
axis(4)
but, what I need is to have different labels for axis-4 than those for
axis-2 (the vertical axes) — that is
On Mar 24, 2011, at 9:29 AM, Michael Bach wrote:
Dear R users,
Given this data:
x <- seq(1,100,1)
dx <- as.POSIXct(x*900, origin="2007-06-01 00:00:00")
dfx <- data.frame(dx)
Now to play around for example:
subset(dfx, dx > as.POSIXct("2007-06-01 16:00:00"))
Ok. Now for some reason I want t
thx , i have solved this by sep=","
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
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Dear R users,
Given this data:
x <- seq(1,100,1)
dx <- as.POSIXct(x*900, origin="2007-06-01 00:00:00")
dfx <- data.frame(dx)
Now to play around for example:
subset(dfx, dx > as.POSIXct("2007-06-01 16:00:00"))
Ok. Now for some reason I want to extract the datapoints between hours
10:00:00 and 1
Hi Barth,
Use rowMeans() like this:
mat <- matrix(rnorm(100), ncol=10)
cbind(sum1.5 = rowSums(mat[,1:5]), sum2.6 = rowSums(mat[,2:6]))
Best,
Ista
On Thu, Mar 24, 2011 at 12:10 PM, Barth B. Riley wrote:
> Dear list
>
> I want to know if there is a function in R that enables one to create a new
>
An improvement on my suggestion using codetools is to use
codetools::findGlobals(f, merge=FALSE)
which separates the functions and variables:
$functions
[1] "*" "cos"
$variables
[1] "omega" "rho"
This is important, because R distinguishes between functions and
variables by usage when it is
On 3/23/2011 10:27 AM, Grossmann Patrick wrote:
Dear R-Project helpdesk member,
I am currently working on some basic correspondence analysis using the R packages
'ca'. As I compute the CA by example<- ca(data), and plotting example via
plot(example), I am trying to figure out, how to get
2011/3/24 Javier López-de-Lacalle :
> Hi everybody:
>
> I need to get the names of the arguments in an object of class "expression".
> I've got the following expression:
>
>> x <- expression(rho * cos(omega))
>
> Is there any function or procedure that returns the names of the arguments
> (in the e
On Mar 23, 2011, at 5:35 PM, mipplor wrote:
exactly, i dont know why the header are separated by invisible
periods.
and it only happen to the header.
If it continues as a problem, it because you have not followed my
advice from yesterday.
--
View this message in context:
http://r.78
I tried this as an exercise and here's what I arrived to:
collector <- function(expr){
RES <- list()
foo <- function(x) unlist(lapply(x, as.list))
EXPR <- foo(expr)
while(length(EXPR) >0){
if(is.symbol(EXPR[[1]])){
RES <- c(RES, EXPR[[1]])
EXPR <- EXPR[-1
Hi,
I'm trying to create a distance matrix. And it works out somewhat ok. However,
I suspect that there are
some efficiency issues with my efforts. Plz have a look at this:
donor <- matrix(c(3,1,2,3,3,1,4,3,5,1,3,2), ncol=4)
receiver <-
matrix(c(1,4,3,2,4,3,1,5,1,3,2,1,4,5,3,5,1,3,2,4,5,1,2,3
Dear list
I want to know if there is a function in R that enables one to create a new
matrix based on values in an existing matrix, whereby each column in the new
matrix is a row-specific summary (e.g., sum, mean) of multiple columns in the
original matrix. For instance, to take a 12-column mat
On 11-03-24 5:03 AM, Javier López-de-Lacalle wrote:
Hi everybody:
I need to get the names of the arguments in an object of class "expression".
I've got the following expression:
x<- expression(rho * cos(omega))
Is there any function or procedure that returns the names of the arguments
(in th
How about simply
> df <- data.frame(id = 1:6,
+ xout = c(12.34, 21.34, 2.34, 4.56, 3.24, 3.45),
+ xin = c(NA, 34,67,87,34, NA))
>
> with(df, c(NA, xin[-1]/xout[-length(xout)]))
[1]NA 2.755267 3.139644 37.179487 7.456140NA
>
BTW You seem to
> Date: Thu, 24 Mar 2011 03:35:31 -0700
> From: kont...@alexanderbachmann.de
> To: r-help@r-project.org
> Subject: [R] Problem with Snowball & RWeka
>
> Dear Forum,
>
> when I try to use SnowballStemmer() I get the following error message:
>
> "Could not i
Dear r-help,
I'm having this DF
df <- data.frame(id = 1:6,
xout = c(1234, 2134, 234, 456, 324, 345),
xin= c(NA, 34,67,87,34, NA))
and would like to calculate the fraction (xin_t / xout_t-1)
The result should be:
# NA, 2.76, 3.14, 37.18, 7.46, NA
I am sure ther
Hi Alex
On 24 March 2011 04:47, Alex Olssen wrote:
> I am looking for some recommended reading.
>
> I want to read up on the estimation systems of linear equations using
> maximum likelihood?
For a comprehensive review of estimating systems of linear equations, see:
http://www.jstatsoft.org/v
Dear Forum,
when I try to use SnowballStemmer() I get the following error message:
"Could not initialize the GenericPropertiesCreator. This exception was
produced: java.lang.NullPointerException"
It seems to have something to do with either Snowball or RWeka, however I
can't figure out, what to
Hi everybody:
I need to get the names of the arguments in an object of class "expression".
I've got the following expression:
> x <- expression(rho * cos(omega))
Is there any function or procedure that returns the names of the arguments
(in the example: "rho" and "omega")?
I tried a rough appro
Dear All
I don't know is this right place to ask this question. I made one program in
R environment and now I need to call that program from the from the java
application using JRI package. I tried in both eclipse and java compiler but
the Rengine re = new Rengine(args, false, new TextConsole());
On 03/24/2011 06:30 PM, Dieter Menne wrote:
...
Nobody ever got fired for producing a p-value :-(
Oh, fortune, fortune! Or an R-squared!
Jim
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting
Hello. I am using fGarch to estimate the following model:
Call:
garchFit(formula = fmla, data = X[, i], trace = F)
Mean and Variance Equation:
data ~ arma(1, 1) + garch(1, 1)
Conditional Distribution:
norm
Coefficient(s):
mu ar1 ma1 omegaalpha1 beta1
-0.94934
Rob,
If 'sex' is indeed an exogenous variable (ie. predictor only), you can
simply code it as (1=male, 2=female) and use it as a covariate in any sem
model. In lavaan, you can explicitly use the argument 'fixed.x=TRUE', which
will regard all exogenous covariates as fixed variables. Their
means/var
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