Dear Eric
This test is unfortunately AFAIK not implemented in R. You can
nevertheless use the similar Zivot and Andrews test in package urca.
library(urca)
?ur.za
Best
Matthieu
Le 06. 06. 10 12:00, r-help-requ...@r-project.org a écrit :
Message: 24
Date: Sat, 5 Jun 2010 15:45:29 +0100
From:
On Sun, Jun 6, 2010 at 11:53 PM, Daisy Englert Duursma
wrote:
> Hello r-help.
>
> I have been working on making a graph and have several solutions but
> they are tedious at best.
>
> Here is an example dataset:
>
> catg<-(c(1,2,3,2,4,3,2,1,4,3,1))
> min<-(c(1,2,3,3,4,5,6,6,3,2,1))
> max <-(c(10,6,
Actually the 'apply' is not necessary.
The original poster has stocks as rows
rather than the customary columns, so
the following should suffice:
retmat <- diff(log(t(pricemat)))
Questions that are specifically financial
should be sent to r-sig-finance (you need
to subscribe before posting).
O
Hi R experts.
how can I average two sets of columns?
dd <- data.frame(b = c("chr2", "chr1", "chrY", "chr13", "chrX"),
w=11:15, x = 1:5, y = c(8, 3, 9, 9,7),
z = c(1, 1, 1, 2, 8))
expected result for mean(w,x) and mean(y,z) is
1 chr2 6 4.5
2 chr1 7 2
3 chrY 8 5
4 chr13 9 5.5
Hey Everyone,
I have been stumped by this all day.
Basically, I have a data.frame of multiple columns. Of concern are "id" &
"date"
For some reason, oftentimes there are duplicates of data with the same date.
I would like to remove the duplicates per different id (removing duplicate
dates for
Hello,
I have a couple of questions about the ols function in Frank Harrell's rms
package.
Is there any way to specify variables by their column number in the data
frame rather than by the variable name?
For example,
library(rms)
x1 <- rnorm(100, 0, 1)
x2 <- rnorm(100, 0, 1)
x3 <- rnorm(100, 0,
I'm sorry for not clearer describing my motive with this--
So this is what I'm trying to do- Take a survival object and utilize
it in ggplot.
ggkm<-function(time,event,stratum) {
m2s<-Surv(time,as.numeric(event))
fit <- survfit(m2s ~ stratum)
f$time<-fit$time
f$surv<-fit$surv
Hi:
Using the plyr package,
summarise(dd, b = b, wx = (w + x)/2, yz = (y + z)/2)
b wx yz
1 chr2 6 4.5
2 chr1 7 2.0
3 chrY 8 5.0
4 chr13 9 5.5
5 chrX 10 7.5
HTH,
Dennis
On Sun, Jun 6, 2010 at 9:53 PM, Yuan Jian wrote:
> Hi R experts.
>
> how can I average two sets of columns?
>
You can use diff.zoo like this:
library(zoo)
z <- zoo(matrix(1:24, 6))
z
diff(log(z))
# also try
diff(z, arith = FALSE) - 1
See ?diff.zoo
and read the three zoo vignettes (pdf documents):
vignette(package = "zoo") # lists them
vignette("zoo")
etc.
On Sun, Jun 6, 2010 at 11:11 PM, Anyi Zhu wr
Joris Meys gmail.com> writes:
>
> There might be a way doing that in R, but if you really need to take
> measures of a picture in some standardized way, I suggest you take a
> look at ImageJ. More than R that seems the right tool for the job.
> Some people have been experimenting with connecting
The package EBImage provides general purpose functionality for the
reading, writing, analysis and displaying of images.
> library(EBImage)
> x = readImage('lena.gif')
> display(x)
ImageJ is OK to carry out some image processing operations. But if you
need to perform quantitative image analysis
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Uwe Ligges
On 07.06.2010 09:01, Jeff08 wrote:
Hey Everyone,
I have been stumped by this all day.
Basically, I have a data.frame of multiple columns.
It helps if you provide sample data, so we can try some things out
ourselves. Right now, I can't run your function without guessing how
your data might look like.
Yet, try something like this :
> ts <- list(a=1:5)
> names(ts$a) <- letters[1:5]
> n <- length(names(ts$a))
> tmp <- sapply(1:n,func
Dear Alex,
after installation of ROptRegTS there should be a folder "scripts" in
the package directory of ROptRegTS which includes some further examples.
If you are assuming normal errors you should switch to package RobRex
which is "optimized" for normal errors. There you will also find a
fol
On 06/07/2010 02:49 AM, Iasonas Lamprianou wrote:
Hi all,
does anyone have any practical examples of how this command can be used in
prettyR?
add.value.labels(x,value.labels)
I mean, can we have the SPSS style of using numbers and/or labels if we want to?
Thank you for your time
Hi Jason,
Th
HI all.,
I have some univariate time series that need to be prewhitened. HOw this can
be performed in R.
I am thinking of to fit an ARIMA model and substract this from the original
series. Is this the correct way
THanks in advance
nuncio
--
Nuncio.M
Research Scientist
National Center for Antarcti
Thank you,
I'll have a go and if I have more questions I'll probably come back to you...
Dr. Iasonas Lamprianou
Assistant Professor (Educational Research and Evaluation)
Department of Education Sciences
European University-Cyprus
P.O. Box 22006
1516 Nicosia
Cyprus
Tel.: +357-22-713178
Fax:
Thank you,
I'll have a go and if I have more questions I'll probably come back to you...
Dr. Iasonas Lamprianou
Assistant Professor (Educational Research and Evaluation)
Department of Education Sciences
European University-Cyprus
P.O. Box 22006
1516 Nicosia
Cyprus
Tel.: +357-22-713178
Fax: +3
On 06/07/2010 04:53 PM, Daisy Englert Duursma wrote:
Hello r-help.
I have been working on making a graph and have several solutions but
they are tedious at best.
Here is an example dataset:
catg<-(c(1,2,3,2,4,3,2,1,4,3,1))
min<-(c(1,2,3,3,4,5,6,6,3,2,1))
max<-(c(10,6,8,6,7,3,10,9,10,8,9))
cbin
On Mon, 7 Jun 2010, nuncio m wrote:
HI all.,
I have some univariate time series that need to be prewhitened. HOw this can
be performed in R.
That depends on what exactly you want to do for prewhtening. For example,
if you want to prewhiten using an AR(1) model, you could apply ar() and
then
Dear Daisy,
This is easy woth ggplot2. Use geom_crossbar() and set the middle bar at
the minimum of maximum.
library(ggplot2)
dataset <- data.frame(catg = (c(1,2,3,2,4,3,2,1,4,3,1)),
min = (c(1,2,3,3,4,5,6,6,3,2,1)),
max = (c(10,6,8,6,7,3,10,9,10,8,9)))
ggplot(dataset, aes(x = catg, y = min, ym
The ?plotmath page and the example " ## How to combine "math" and numeric
variables :" should be helpful...?
>>> Thomas Bschorr 05/06/2010 10:36:42 >>>
Hi,
I desperately try to do s.th. like
m=1.23455
sig=0.84321
plot(1,1)
text(0.8,1,sprintf("=%1.2f±%1.2f",m,sig))
where actually the gree
I did not realise making a zoo object is that convenient. Thanks a lot
Gabor.
-Original Message-
From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Sent: June-07-10 5:20 AM
To: anyi@gmail.com
Cc: r-help@r-project.org
Subject: Re: [R] Computing day-over-day log return for a mat
Greetings to you all.
I have two datasets - Time and magnitude. For a particular location, the
magnitude of the parameter varies with time. I wish to obtain a polar
coordinate distribution of time (0-24h) and magnitudes so as to visualize
how magnitude varies with different times of the day (e.g.,
Err, sorry, forget my previous message. It has nothing to do with
Tinn-R or Eclipse. If you save that code in a file and you use the
source command in the console, you get the same output. Tinn-R and
Eclipse both source the code to R instead of sending it line by line.
To get what you want, you ne
Hello Ben,
Does the workaround pointed out later in the thread you're
responding to (from the last paragraph of a very long 'Note'
section of ?pdf) help?
Well, I did not try to edit my fonts.conf but I feel this is not a PDF
issue. I have no problem to have greek letters correctly rendered i
Dear list,
I am getting weired with fitting data with a 1/x-polynomial. Suggest I have
the following data:
x <- c(1,2,3,4,5,6,7)
y <- c(100,20,4,2,1,.3,.1)
I may fit this with a linear model
fit1 = lm(y ~ I(x))
Getting plot out of this model I applied
library(polynom)
pol1 = polynomial(fit1$co
Hi
I am new in using R. I am facing problem getting the raw data like the mean
value, standard deviation value form the .CEL file.
Please help me, how can I get these raw information from a .CEL file into an
exel file?
Thanks
--
View this message in context:
http://r.789695.n4.nabble.com/raw
Thanks a lot Sayan, I'll give it a try and let you know how it goes.
From: sayan dasgupta [mailto:kitt...@gmail.com]
Sent: June-07-10 2:13 AM
To: anyi@gmail.com
Cc: r-help@r-project.org
Subject: Re: [R] Computing day-over-day log return for a matrix containing
multiple time series
Hi
some example would be helpful.
r-help-boun...@r-project.org napsal dne 07.06.2010 09:01:27:
>
> Hey Everyone,
>
> I have been stumped by this all day.
>
> Basically, I have a data.frame of multiple columns. Of concern are "id"
&
> "date"
>
> For some reason, oftentimes there are duplicat
subset(x, !duplicated(paste(x$id, x$date))
On Mon, Jun 7, 2010 at 3:01 AM, Jeff08 wrote:
>
> Hey Everyone,
>
> I have been stumped by this all day.
>
> Basically, I have a data.frame of multiple columns. Of concern are "id" &
> "date"
>
> For some reason, oftentimes there are duplicates of data w
Hi
r-help-boun...@r-project.org napsal dne 07.06.2010 14:28:58:
> Dear list,
>
> I am getting weired with fitting data with a 1/x-polynomial. Suggest I
have
> the following data:
>
> x <- c(1,2,3,4,5,6,7)
> y <- c(100,20,4,2,1,.3,.1)
>
> I may fit this with a linear model
>
> fit1 = lm(y ~ I
All questions about manipulations of sequence data (guess that's the
CEL file you're talking about...) can better be asked on the specific
mailing list of BioConductor :
http://www.bioconductor.org/docs/mailList.html
And please, don't swear in church any more. We don't put things in
Excel files.
Xiongqing Zhang yahoo.com> writes:
>
>
> Dear Mr. or Ms.,
>
> I used the R-software to run the zero-inflatoin negative binomial model, but I
could not get the results.
> And the error message is "solve.default(as.matrix(fit$gaussian)) ". In the
model, I introduced 3 dummy
> variables. I do no
Hello,
I'm running R 2.10.1 on Windows XP.
I'd like to read some files from a ftp site, but I don't how to do it.
I've tried the functions url, download.file but each time I have a
message "cannot open URL 'ftp://ftp.cea.fr'" .. and I don't kwow why ..
Thanks for the help
Benoit
--
Benoit Wast
On 06/06/2010 10:49 PM, Mark Seeto wrote:
Hello,
I have a couple of questions about the ols function in Frank Harrell's rms
package.
Is there any way to specify variables by their column number in the data
frame rather than by the variable name?
For example,
library(rms)
x1<- rnorm(100, 0, 1)
Hi Hadley and everyone,
here's a patch for ggplot2 that fixes the behavior of
opts(legend.position={left,top,bottom}). If you try the following code
in an unmodified ggplot2
options(warn = -1)
suppressPackageStartupMessages(library("ggplot2"))
data <- data.frame(
x = c(1, 2, 3, 4, 5, 6),
Greetings Moderators!
I moderated the message below just now (one of two identical
test messages). I rejected it, with a covering note to the
author (John Munroe , who does not appear
to be subscribed to the list) that I was doing so because
approving it would serve no useful purpose, and pointing
Apologoes for sending a message to R-help that should have gone
to the moderators' mailing-list. I had overlooked that the "From:"
in the MIME-attached messsage was different from the "From:" in
the covering message, and I replied to the wrong address.
Ted.
On 07-Jun-10 14:00:02, Ted Harding wrote
Dear R People:
I found the following code on the Internet:
> x <- co2
> m <- HoltWinters(x, alpha=.1, beta=0, gamma=0)
> p <- predict(m, n.ahead=240, prediction.interval=T)
> plot(m, predicted.values=p)
>
and it works fine.
My question: is there an equivalent for ARIMA objects, please? When
I
Dear Ms and Mr R,
I am seeking a visually appealing presentation of 3 variables (~30
datapoints) that are correlated and that should to into the same direction.
The objective is to show that they do go in the same direction and what
differences there might be while also giving an idea of the v
reproducible example needed.
Georg Ehret wrote:
Dear Ms and Mr R,
I am seeking a visually appealing presentation of 3 variables (~30
datapoints) that are correlated and that should to into the same direction.
The objective is to show that they do go in the same direction and what
difference
Dear Anonymous
I suggest you to have a look on the code in package ccgarch, which might
help you in seeing how this have been done. A useful information would
be to know what kind of optimization algorithm the authors use.
best
Matthieu
Le 07. 06. 10 12:00, r-help-requ...@r-project.org a écr
Hi Georg,
Take a look at ?pairs as well as [1] and [2].
HTH,
Jorge
[1] http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=159
[2] http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=149
On Mon, Jun 7, 2010 at 11:16 AM, Georg Ehret <> wrote:
> Dear Ms and Mr R,
> I am
On Mon, 7 Jun 2010, ogbos okike wrote:
Greetings to you all.
I have two datasets - Time and magnitude. For a particular location, the
magnitude of the parameter varies with time. I wish to obtain a polar
coordinate distribution of time (0-24h) and magnitudes so as to visualize
how magnitude vari
Dear R People:
A very fine solution exists in the "forecast" package, by using plot.forecast.
Thanks though!
Sincerely,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com
_
See ?plot.igraph, example:
plot(g, layout=mlayout,
vertex.label.color = "red",
vertex.color = "green",
vertex.frame.color = "blue",
edge.color = "orange",
edge.width = 1 + (1:18)/2
)
Uwe Ligges
On 06.06.2010 17:58, Etienne Cuvelier wrote:
Hello,
I am using the igraph pack
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of sayan dasgupta
> Sent: Sunday, June 06, 2010 11:13 PM
> To: anyi@gmail.com
> Cc: r-help@r-project.org
> Subject: Re: [R] Computing day-over-day log return for a
> matrix cont
Hi,
I have the following two dataframes (Df1 and Df2):
Df1 has a layout like
ID Location Var1 Var2 Var3 . Var20
A Loc2 1 11. 1
A Loc1 0 10. 1
B Loc1 0 00. 0
A Loc3
Dear Bert,
The easiest thing would be to merge both datasets and then multiply the
corresponding columns.
both <- merge(df1, df2)
both[, 3:22] * both[, 23:42]
HTH,
Thierry
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org namens Bert Jacobs
Verzonden: ma 7-6-2010 20:29
Aan: r
Hi everyone,
I am using bio3d package for some sequence analysis because of its nice
integrated MUSCLE execution. However, I have been encountering some errors
while I was running seqaln function, which calls for installed MUSCLE program.
In my case, I stored the MUSCLE program on the desktop
Does the collapse argument to the paste function do what you want? Possibly
nested inside another paste.
--
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:
Hi,
I tried on this, but couldn't figure it out:
Suppose I have a dataframe as follows:
dat <- data.frame(Day=rep(rep(c(1,2), each=4),2), Animal = rep(c(1,2),
each=8), Count=c(rnorm(8, 100), rnorm(8,90)))
2 animals are being examined on 2 different days. Count is the result.
Now I need to point
Hi,
Is there a way to recursively change the variables imported with
read.table(..) - and then attached with attach(..) - from as.factor to
as.integer?
I have about 9 variables that are now recognized as factors but I need to do
statistics on them so I need to change them in as.integer. It works,
I'm not sure if this is the right mailing list. My apologies if I'm
sending the question to the wrong mailing list.
(The r-packages list moderator suggested using the R-help list.)
My question is about the "chickwts" data in the datasets package.
According to help(chickwts), the source of the data
Dear all,
I have a problem when using some classification functions (Kmeans, PAM,
FANNY...) with a distance matrix, and i would to understand how it
proceeds for the positioning of centroids after one execution step.
In fact, in the classical formulation of the algorithm, after each step,
t
hi, everyone:
i have a data frame with one single character column as followings:
DS_xxx_yyy_24hr_zzz_1
DS_xxx_yyy_24hr_zzz_2
DS_xxx_yyy_24hr_zzz_3
DS_xxx_yyy_24hr_zzz_4
DS_xxx_yyy_24hr_zzz_1
DS_xxx_yyy_24hr_zzz_2
DS_xxx_yyy_24hr_zzz_3
DS_xxx_yyy_24hr_zzz_4
DS_xxx_yyy_24hr_zzz_1
i need to parse o
Hello Sir,
I am using mgcv package for my data.
My model is y~x1+f(x2),I want to find out the function f(x2) .
Following is the code.
sm1=gam(y~x1+s(x2),family=binomial, f)
summary(sm1)
plot(sm1,residuals=TRUE, xlab="AGE",pch=20)
In this plot I am getting S(x2,1.93) on y axixs
How shou
Once you have your image displayed in a plot, look at the updateusr function in
the TeachingDemos package for a way to modify the coordinate system to what you
want so that using locator() gives the results that you want.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Hea
Hello,
Sorry for the very Off TOpic post, but I need some data on past football
(soccer) world cups. I'd like to find (or calculate) the time to the
first goal of each match (where a goal was scored).
I''ve looked at the uefa website and can't find what I want, maybe
someone here can help ?
Hi all,
Consider the following.
a<-list(c("MHsub","MHsub","SFD","Duplex"))
b<-list(c("MHsub","MHsub","SFD"))
c<-list(c("MHpark","SFD","SFD"))
d<-list(c("MultiFam","MultiFam","MultiFam","Duplex"))
all<-list(a,b,c,d)
names(all)<-c("5","6","7","8")
all2<-list()
for(i in 1:length(all)){
all2[[i
Try this:
gsub(".*_(.*)_.*_.*", "\\1", s)
On Mon, Jun 7, 2010 at 2:21 PM, xin wei wrote:
>
> hi, everyone:
> i have a data frame with one single character column as followings:
> DS_xxx_yyy_24hr_zzz_1
> DS_xxx_yyy_24hr_zzz_2
> DS_xxx_yyy_24hr_zzz_3
> DS_xxx_yyy_24hr_zzz_4
> DS_xxx_yyy_24hr_zz
Do you just want a rectangle with different vertical color bands? Then look at
the image function. Do you want a small rectangle that goes the length of the
x-axis with these color bands, but gives room for another plot above that?
Then look at the subplot function (TeachingDemos package) alo
1) Give us a reproducible example.
2) You probably don't want to use attach.
3) Yes there is a way, but without an example, it's hard to say what
will work. Assume a data.frame called df, maybe something like the
following, which is untested...
df[is.factor(df)] <- lapply(df[is.factor(df)],
Are you interested in only those 35 animals (not every going to look at any
other animals other than those 35, but you want to predict what will happen for
those 35)? Or are the 35 animals a sample of a larger population of animals?
If the later (seems the most likely case) then you probably wan
On Mon, Jun 7, 2010 at 9:05 AM, sidahmed BENABDERRAHMANE
wrote:
> Dear all,
>
> I have a problem when using some classification functions (Kmeans, PAM,
> FANNY...) with a distance matrix, and i would to understand how it proceeds
> for the positioning of centroids after one execution step.
>
> In
Usually when read.table converts numbers to factors it means that there is a
problem in the data file, an extra character somewhere. It is best to fix the
problem in the source data (or a copy of the source data) so that the data
imports properly rather than try to fix it post hoc. If that is
Look at the strapply function in the gsubfn package. It may do what you want.
--
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.
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Chirok Han
> Sent: Monday, June 07, 2010 9:51 AM
> To: r-help
> Subject: [R] Source of chickwts data
>
> I'm not sure if this is the right mailing list. My apologies if I'm
> sen
predict(sm1,type="terms",terms="s(x2)")
see ?predict.gam
Cheers
Joris
On Mon, Jun 7, 2010 at 7:25 PM, Dipa Hari wrote:
>
>
> Hello Sir,
> I am using mgcv package for my data.
> My model is y~x1+f(x2),I want to find out the function f(x2) .
> Following is the code.
>
> sm1=gam(y~x1+s(x2),family=
Try this:
`colnames<-`(do.call(rbind, lapply(all2, '[', nm <-
unique(unlist(sapply(all2, names), nm)
On Mon, Jun 7, 2010 at 4:20 PM, LCOG1 wrote:
>
> Hi all,
> Consider the following.
>
>
> a<-list(c("MHsub","MHsub","SFD","Duplex"))
> b<-list(c("MHsub","MHsub","SFD"))
> c<-list(c("MHpark"
Sorry for bothering all of you. In the end it turned out to be much simpler
than I thought. Takes a while to get used to the vectorizing idea.
require(survival)
require(ggplot2)
ggkm<-function(time,event,stratum) {
stratum<-as.factor(stratum)
m2s<-Surv(time,as.numeric
Paul
The FIFA database doesn't have times that goals are scored either.
The best I have found is at http://www.worldcup-history.com/, but you
have to check individual match reports for the times that goals are
scored.
albyn
On Mon, Jun 07, 2010 at 09:50:34PM +0100, Paul wrote:
> Hello,
>
> Sorry
Hi, I am trying to do some shading in R and am using the polygon() setting.
I want the border on the top and bottom of the polygon, but not on the left
and right. Is there a way to do this in R? Thanks in advance!
beloitstudent
--
View this message in context:
http://r.789695.n4.nabble.com/B
these websites seem to have the data. though I have not checked for
completeness. the rsssf in particular is seems to be concerned with
collecting and archiving these kinds of football data:
http://www.rsssf.com/tablesw/worldcup.html
http://wapedia.mobi/en/1930_FIFA_World_Cup_Group_1
hope that
Many thanks. That's the right one.
It would be better corrected in the next version of R.
Thanks again.
Chirok
On Tue, Jun 8, 2010 at 6:01 AM, William Dunlap wrote:
>> -Original Message-
>> From: r-help-boun...@r-project.org
>> [mailto:r-help-boun...@r-project.org] On Behalf Of Chirok Ha
Henrique,
The last message gave me exactly what i needed. Im still not sure whats
going on, i will need to reference the do.call function some more. Thanks for
your help.
Josh Roll
Lane Council of Governments
Transportation Planner
Office: 541-682-2454
Cell: 406-671-5069
jr...@lcog.org
www
Hi,
I have used DeSolve package for my ODE problem regarding infectious disease
transmission and currently am trying to pass lots (roughly a thousand) of model
parameters to the C compiled model (I have to use C compiled code instead of R
code purely because of the speed).
I can't go define it
On Mon, Jun 7, 2010 at 6:55 PM, beloitstudent wrote:
> [...] I am trying to do some shading in R and am using the polygon() setting.
> I want the border on the top and bottom of the polygon, but not on the left
> and right. [...]
One way would be to add the borders after as lines. For example:
pl
Thanks to all,
I did eventually come up with a solution which involved sub-setting
the data first into categories, making an empty plot, and then
plotting rectangles. I really like the ggplot2 and I will use that
next time.
My ugly script is below
Thanks,
Daisy
plot(1, xlim=c(5,45),ylim=c(0,1
Hello,
I searched the archives but found no answers.
How to modify the hisgram color of function pairs.panels of Psych package ?
I tried col() but it was the line color modified.
Thanks.
Elaine
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R-help@
Hi Joris,
Thanks very much. I think the %in% ls() was the key. I just needed to know
how to select elements that existed in ls().
Scott
On Sat, Jun 5, 2010 at 4:22 PM, Joris Meys wrote:
> Hi,
>
> One possible way to get around it is using following idea :
> X1 <- rnorm(10)
> X2 <- rnorm(10)
>
I have a matrix of, say, M and N dimensions:
my_matrix=matrix(c(1:60),nrow=6,ncol=10)
I have two "id" vectors corresponding to the rows and columns, e.g.:
id_m=seq(10,60,by=10)
id_n=seq(100,1000,by=100)
I would like to create a "proper" database (let's say a data.frame for
this example -- i'm g
Hi,
I just need to know where the R was developed, for a journal article citation.
Thanks
Ronjon
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PLEASE do read the posting g
Sample Data.Frame format
Name is Returns.nodup
X id ticker date_ adjClose totret RankStk
427225 427225 00174410AHS 2001-11-1321.661001235
"id" uniquely defines a row
What I am trying to do is filter out id's that have less than 1500 data
points (by date
Hi,
I had a questions about doing ridge regression in R.
Why is it that when I try this on datasets with more predictors than samples
(p>n) using lambda=0, it still finds coefficients for all predictors? I thought
when lambda=0, it should be like ordinary regression and therefore not find
coe
Sample Data.Frame format
Name is Returns.names
X id ticker date_ adjClose totret RankStk
427225 427225 00174410AHS 2001-11-1321.661001235
"id" uniquely defines a row
What I am trying to do is add missing data for each ID.
Important Information: Date is
Datta, Ronjon wrote:
Hi,
I just need to know where the R was developed, for a journal article citation.
There's a function for that!
citation()
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PLEASE do read the post
Jeff08 wrote:
Sample Data.Frame format
Name is Returns.nodup
X id ticker date_ adjClose totret RankStk
427225 427225 00174410AHS 2001-11-1321.661001235
"id" uniquely defines a row
What I am trying to do is filter out id's that have less than 1500 data
Hi Ronjon,
>From citation():
> citation()
To cite R in publications use:
R Development Core Team (2010). R: A language and environment for
statistical computing.
R Foundation for Statistical Computing, Vienna, Austria. ISBN
3-900051-07-0, URL
http://www.R-project.org.
A BibTeX entry for
Try this:
> mm <- matrix(1:6, 3, dimnames = list(LETTERS[1:3], letters[1:2]))
> mm
a b
A 1 4
B 2 5
C 3 6
> library(reshape)
> melt(mm)
X1 X2 value
1 A a 1
2 B a 2
3 C a 3
4 A b 4
5 B b 5
6 C b 6
On Mon, Jun 7, 2010 at 10:33 PM, Jonathan Greenberg
wrote:
>
Jeff08 wrote:
Sample Data.Frame format
Name is Returns.names
X id ticker date_ adjClose totret RankStk
427225 427225 00174410AHS 2001-11-1321.661001235
"id" uniquely defines a row
What I am trying to do is add missing data for each ID.
Important Infor
Hi Jonathan,
Following Gabor Grothendieck's advice, try also:
mm <- matrix(1:6, 3, dimnames = list(LETTERS[1:3], letters[1:2]))
as.data.frame.table(mm)
HTH,
Jorge
On Mon, Jun 7, 2010 at 10:33 PM, Jonathan Greenberg <> wrote:
> I have a matrix of, say, M and N dimensions:
>
> my_matrix=matrix(
I think what you are groping for is something like this
my_matrix <- matrix(1:60, nrow = 6)
id_a <- seq(10,60,by=10)
id_b <- seq(100,1000,by=100)
my_database <- cbind(
expand.grid(id_a = id_a, id_b = id_b),
mat = as.vector(my_matrix)
)
-Original Message-
From: r-help-boun...@r-projec
Greg,
the animals are a sample of a larger population, as you guessed.
I used lmer to estimate the effects:
> lmer(Count~Animal | Day, dat)
Linear mixed model fit by REML
Formula: Count ~ Animal | Day
Data: dat
AIC BIC logLik deviance REMLdev
1554 1574 -772 15421544
Random
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