R version/sessionInfo()?
/H
On Fri, May 21, 2010 at 10:08 PM, David Reiss wrote:
> Hi,
> I am trying to check a package via R CMD CHECK and it is failing with
>
> Error: '\s' is an unrecognized escape in character string starting "\s"
>
> The culprit looks something like this:
>
> gsub('\\s\\(.*
Alex Ruiz E. wrote:
Dear R helpers,
I created a somewhat big database (+206,700 rows) in MySQL and have
exported into a csv file, but I can't open the whole thing in R. I am
using:
base<-read.csv("/path/to/file.csv", header=F, sep="," nrows=206720)
R doesn't complain but it only opens 128,32
Dear R helpers,
I created a somewhat big database (+206,700 rows) in MySQL and have
exported into a csv file, but I can't open the whole thing in R. I am
using:
> base<-read.csv("/path/to/file.csv", header=F, sep="," nrows=206720)
R doesn't complain but it only opens 128,328 observations (the nu
On 2010-05-21 11:36, apjawor...@mmm.com wrote:
I am not sure if I am correct but I think the labels argument pertains
only to the counterplot function.
It's true that the 'labels' argument is only used in contourplot,
not levelplot. Moreover, there is an example of the use of
levelplot on the h
Many thanks - the problem is solved.
Didn't realize that it was indeed simple. Thanks again.
Cheers,
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the post
Hello,
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
We don't know the function you're using or the data you applied the function to,
which makes it almost impossible to help. Can you use ?dput to
I am receiving this error running a command on a multi-row data-frame. The
data is strings of text (each with new line separator, no spaces, no
numerical characters).
Error in FUN(X[[1L]], ...) :
STRING_ELT() can only be applied to a 'character vector', not a 'integer'
I can run single text str
Hi, Thanks for you insight.
the problem that I have is that the program reports back an error:
Error in FUN(X[[1L]], ...) :
STRING_ELT() can only be applied to a 'character vector', not a 'integer
When I try to read in a multi-row frame, and so I can only feed in one row
at a time. The data
By any chance has anyone put together the R code for reproducing the Firefox
Downloads Analytics Map in R?
Here is a link to an example:
http://www.knitwareblog.com/wp-content/uploads/2008/06/firefox-3-download-map.jpg
I'll try to track down the inputs, but I am interested in putting the code
I probably shouldn't be replying, as I believe this may require system
specific tricks that I don't know. To that end, you should probably provide
the minimum required information (e.g. from sessionInfo() + any other
OS/shell info) requested by the posting guide.
AFAIK, this can't be done in R. Th
Try this:
mydata = as.data.frame(cbind(PrinComp[,1:5],x1=X[,1]))
net5 = lm(x1~.,data=mydata)
y=predict(net5, newdata=data.frame(PrinCompPredict)[,1:5])
(I can't test it, since you didn't provide a reproducible
example.)
Basically your problem is that when you pass a *matrix*
to the modeling fun
On May 21, 2010, at 5:34 PM, Greg Snow wrote:
Look at txtStart and friends in the TeachingDemos package as an
alternative to sink that includes commands as well as output.
Know the quality of Greg Snows work it will probably be better than
mine, but perhaps:
> capfn <- function(inp) {
+
Hello I am creating a linear model with the command
net5 = lm( X[,1] ~ PrinComp[,1:5]) where my vector PrinComp looks like this
> head(PrinComp[,1:5])
PC1 PC2PC3PC4PC5
[1,] 1.8626055 -3.34190998 -0.5448889 2.8296751 0.3994096
[2,] 3.1124144 -1.68113572
Look at txtStart and friends in the TeachingDemos package as an alternative to
sink that includes commands as well as output.
--
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-
On May 21, 2010, at 1:40 PM, Mohamed Abdelaziem Ibrahim wrote:
Pro,
I am PhD student , i am usuing Rational Expectations Model, I want
to model different timeseries with ARMAX models in R because I think
that ARMAX models will map best to these data. I want know the steps
to use R for A
Hi David,
I want to get both the 4 and the "1+3" that created it.
I am trying to help someone else on the mailing list that is looking for a
way to "sink" the console into word, so he could have word read it to him
(he is blind).
I know how to do the second part, but the first part (using sink wi
On May 21, 2010, at 5:02 PM, Tal Galili wrote:
Hi all,
I am trying to use type message with sink, like this:
sink("all.Rout", type="message")
1+3
sink()
readLines(con = "all.Rout")
So to get the following output:
1+3
[1] 4
Obviously this doesn't work.
What are you trying to do? The
Hi all,
I am trying to use type message with sink, like this:
sink("all.Rout", type="message")
1+3
sink()
readLines(con = "all.Rout")
So to get the following output:
> 1+3
[1] 4
Obviously this doesn't work.
I tried some variations (based on the explanations in the help) but am
missing somet
On 5/21/2010 4:00 AM, Julian Burgos wrote:
Hi Brian,
This is really odd. I keep getting the NA secs answer, only by running
these three lines of code in a new session.
time3=strptime("2009 06 01 00 47 00",format="%Y %m %d %H %M")
time4=strptime("2009 06 01 00 57 00",format="%Y %m %d %H %M")
Eugeniusz Kałuża [Fri, May 21, 2010 at 04:47:21PM
CEST]:
> Dear R users,
> #
> #There is table containing 1000 (lat, lon, time) GPS positions, wchich should
> be recognized
Hi,
I am trying to check a package via R CMD CHECK and it is failing with
Error: '\s' is an unrecognized escape in character string starting "\s"
The culprit looks something like this:
gsub('\\s\\(.*\\)',"","this is a (test, man) dude")
which is correctly escaped with the "\\"'s -- but it see
Hi all,
As a molecular biologist by training, I'm fairly new to R (and statistics!),
and was hoping for some advice. First of all, I'd like to apologise if my
question is more methodological rather than relating to a specific R
function. I've done my best to search both in the forum and elsewhere
Hi!
Thanks for your reply! After running the command below I am certain I am
using a 64-bit R. I am running R through a linux cluster system where R is
globally available for all users. I have asked the system administrators if
they would update their version R but they are not receptive of maki
On May 21, 2010, at 12:21 AM, santana sarma wrote:
Hi David,
SORRY - I am trying to be more clearer this time.
Let's say the dataframe has some rows and columns, with unique
rownames and column names. The rest of the data in the dataframe are
just numbers.
So have you tried any of the s
Pro,
I am PhD student , i am usuing Rational Expectations Model, I want to model
different timeseries with ARMAX models in R because I think that ARMAX models
will map best to these data. I want know the steps to use R for ARMAX models
I coudn't find any solutions in the R help and therefo
Hi, I am currently trying to do dose-response curves
using weighted 4-parameter model (4PL). The weighting was based on
1/(expected variance) derived from historical data. I tried both drm() from drc
package, and nls(), found very different
results derived from drm() vs. nls() using "weights=" arg
The R2wd package can be used to send things to word documents, that may work
for you. If you want this done automatically you could look at the txtStart
and related functions in the TeachingDemos package and just change the parts
that send plain text to the file (sink) with commands from R2wd.
Jan van der Laan [Fri, May 21, 2010 at 11:22:35AM
CEST]:
> Perhaps you mean something like sapply or apply?
>
Neither am I certain what he means exactly, but I would use apply and cousins
only if I don't care in which sequence the elements are processed. His problem
(reading line by line) sound
Awesome. Thats exactly what I want.
On Sat, May 22, 2010 at 12:45 AM, Erik Iverson wrote:
>
> rajesh j wrote:
>>
>>> Hi,
>>>
>>> I wish to plot multiple histograms(representing different data so
>>> different
>>> range along xaxis but y axis is the same) horizontally in ggplot2. I'd
>>> like
>>
rajesh j wrote:
Hi,
I wish to plot multiple histograms(representing different data so
different
range along xaxis but y axis is the same) horizontally in ggplot2. I'd
like
it to look like facets. Is this possible?
Can you give a small example, say, using the diamonds dataset, or one of
y
On May 21, 2010, at 2:18 PM, Robert U wrote:
Dear R-users,
I'm trying to make the following loop: for (x in 0 : 10) and i would
like x to be decimals rather than integers, giving a x range of e.g.
0.0 0.1 0.2 0.3 ... 9.9, 10. Would anyone know how to do that ?
with regards,
Had you not
Robert U wrote:
Dear R-users,
I'm trying to make the following loop: for (x in 0 : 10) and i would
like x to be decimals rather than integers, giving a x range of e.g.
0.0 0.1 0.2 0.3 ... 9.9, 10. Would anyone know how to do that ?
Please start new threads when you have a new question, do no
Try this:
for(x in seq(0, 10, by = 0.1))
On Fri, May 21, 2010 at 3:18 PM, Robert U wrote:
> Dear R-users,
>
> I'm trying to make the following loop: for (x in 0 : 10) and i would like x
> to be decimals rather than integers, giving a x range of e.g. 0.0 0.1 0.2
> 0.3 ... 9.9, 10. Would anyone
Dear R-users,
I'm trying to make the following loop: for (x in 0 : 10) and i would like x to
be decimals rather than integers, giving a x range of e.g. 0.0 0.1 0.2 0.3 ...
9.9, 10. Would anyone know how to do that ?
with regards,
[[alternative HTML version deleted]]
__
rajesh j wrote:
Hi,
I wish to plot multiple histograms(representing different data so different
range along xaxis but y axis is the same) horizontally in ggplot2. I'd like
it to look like facets. Is this possible?
Can you give a small example, say, using the diamonds dataset, or one of
your
Thanks all for your suggestions.
plot(1:8, f, axes=FALSE)
axis(1, at=1:8, labels=time)
axis(2)
That gave me exactly what I wanted.
Anjan
On Fri, May 21, 2010 at 12:27 PM, jim holtman wrote:
> Not exactly sure what you mean by "equally spaced"; here is one way:
>
> > x <- read.table(textConnecti
Hi,
I wish to plot multiple histograms(representing different data so different
range along xaxis but y axis is the same) horizontally in ggplot2. I'd like
it to look like facets. Is this possible?
--
Rajesh.J
[[alternative HTML version deleted]]
Yes, that works beautifully on both the test dataset and my real dataset. This
was exactly what I was looking for. Thank you!
/ Mia
On May 21, 2010, at 6:10 PM, William Dunlap wrote:
>
>> -Original Message-
>> From: r-help-boun...@r-project.org
>> [mailto:r-help-boun...@r-project.org
I am not sure if I am correct but I think the labels argument pertains
only to the counterplot function.
Cheers,
Andy
__
Andy Jaworski
518-1-01
Process Laboratory
3M Corporate Research Laboratory
-
E-mail: apjawor...@mmm.com
Tel: (651) 733-6092
Fax: (651) 7
On May 21, 2010, at 11:48 AM, Thomas Lumley wrote:
On Thu, 20 May 2010, David Winsemius wrote:
Almost. Testing with the examples on svytable help page:
test <- function(Z){
fmla <- as.formula( paste(" ~ ", paste(c(Z, "stype"), collapse=
"+")))
chisq <- svychisq(
Not exactly sure what you mean by "equally spaced"; here is one way:
> x <- read.table(textConnection("timef
+ 0h0.00
+ 0.5h0.54
+ 1h1.15
+ 2h2.33
+ 4h1.57
+ 6h2.19
+ 18h1.45
+ 24h1.79"), header=TRUE, as.is=TRUE)
>
> plot(x$f, xaxt='n')
> axis(1, at=1:8, labels
plot(1:8, f, axes=FALSE)
axis(1, at=1:8, labels=time)
axis(2)
>>> ANJAN PURKAYASTHA 21/05/2010 17:15:48
>>>
Hi,
I need to plot $time on the x-axis and $f on the y-axis for the
following
data:
timef
0h0.00
0.5h0.54
1h1.15
2h2.33
4h1.57
6h2.19
18h1.45
24h1.79
I
This do the trick :
for (i in 1:length(sel) ){
posA<-get(paste(c("Pos",sel[i]),collapse=""))
posB<-get(paste(c("Pos",sel[i]+1),collapse=""))
}
> -Original Message-
> From: David Winsemius [mailto:dwinsem...@comcast.net]
> Sent: Friday, May 21, 2010 4:00 PM
> To: arnaud Ga
Hi, I am currently trying to do dose-response curves
using weighted 4-parameter model (4PL). The weighting was based on
1/(expected variance) derived from historical data. I tried both drm() from drc
package, and nls(), found very different
results derived from drm() vs. nls() using "weights=" arg
Hi,
I need to plot $time on the x-axis and $f on the y-axis for the following
data:
timef
0h0.00
0.5h0.54
1h1.15
2h2.33
4h1.57
6h2.19
18h1.45
24h1.79
I want the order of the data-points to be retained and the x-coordinates of
each of the 8 data-points to be equal
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Mia Bengtsson
> Sent: Friday, May 21, 2010 3:39 AM
> To: Dennis Murphy; Henrique Dallazuanna
> Cc: r-help@r-project.org
> Subject: Re: [R] reshaping data
>
> Thank you Dennis an
On 2010-05-21 8:57, zac...@lmb.uni-muenchen.de wrote:
Dear mailing list,
I am trying to find out, how do a levelplot without labels on the x- and y-axis.
The labels=FALSE does not work...can anyone help me?
m<- matrix(1:25, ncol=5)
levelplot(m, labels=F)
1. This is a function in the lattice
On Thu, 20 May 2010, David Winsemius wrote:
Almost. Testing with the examples on svytable help page:
test <- function(Z){
fmla <- as.formula( paste(" ~ ", paste(c(Z, "stype"), collapse= "+")))
chisq <- svychisq( fmla, design=rclus1,
s
This would do
levelplot(m, xlab="", ylab="")
F
zac...@lmb.uni-muenchen.de wrote:
Dear mailing list,
I am trying to find out, how do a levelplot without labels on the x- and y-axis.
The labels=FALSE does not work...can anyone help me?
m <- matrix(1:25, ncol=5)
levelplot(m, labels=F)
Regards,
AA> In 2008, I have spotted some errors in a package, one which is
AA> likely to have many users (I am not one myself). The more serious
AA> errors are in the documentation, since they lead to a completely
AA> distorted interpretation of the outcome; in addition, there is (at
AA> least) one progr
Dear mailing list,
I am trying to find out, how do a levelplot without labels on the x- and y-axis.
The labels=FALSE does not work...can anyone help me?
m <- matrix(1:25, ncol=5)
levelplot(m, labels=F)
Regards,
Benedikt
__
R-help@r-project.org mailin
Sorry, I made a mistake. Should add "byrow = TRUE".
Using randomly created values can't check the result, a sequence will be
better.
olddata <- data.frame(matrix(1:200, nrow=40, byrow = TRUE))
newdata <- data.frame(matrix(as.vector(t(olddata)),
nrow=nrow(olddata)/10,byrow = TRUE))
dim(olddata)
di
Have a look at ?varClasses. You could you something like varIdent(form =
~ 1|areatrai) or varExp(form = ~areatrai)
R-sig-mixed-models is a better list for this kind of questions.
Thierry
ir. Thierry Onkelinx
Instituut
Which function?
> getAnywhere(confint)
A single object matching ‘confint’ was found
It was found in the following places
package:stats
namespace:stats
with value
function (object, parm, level = 0.95, ...)
UseMethod("confint")
> methods(confint)
[1] confint.default confint.glm*
Dear All
I would like to know, how to establishing socket in R language.
Distributed and parallel Grid data mining is my research project.
I want to work in nws , for sharing classifiers between different nodes.
ws= newWorkSpace(' R Space').
It shows one error "socket not established".
Can anyon
Dear R users,
#
#There is table containing 1000 (lat, lon, time) GPS positions, wchich should
be recognized and labeled in every row of that #recognized position by label
fr
This maybe a solution. I am currently working with something like :
For (i in length(select)){
posA<-get(paste(c("Pos",select[i]),collapse=""))
posB<-get(paste(c("Pos",select[i+1]),collapse=""))
}
Not yet finalized.
TY for the tip about "select" not to use as an object name
> -
Hello
I am trying to reproduce a mixed regression model published in (J Evol
Biol 12:94-102) with PASW 17.03. The authors used SAS to fit the model
using the following syntaxis:
proc mixed nobound;
class area individ trait;
model a = side trait area side*trait side*area trait*area/ddfm =
satterth
Hi,
I want to view the code for the function confint in the stats package.
Typing just confint doesn't work. How can I view the code?
Thanks,
Walt
Walter R. Paczkowski, Ph.D.
Data Analytics Corp.
44 Hamilton Lane
Plainsboro, NJ 08536
(V) 6
A dummy way is to resequence or rematrix
olddata <- data.frame(matrix(rnorm(200), nrow=40))
newdata <- data.frame(matrix(as.vector(t(olddata)), nrow=nrow(olddata)/10))
dim(olddata)
dim(newdata)
-
A R learner.
--
View this message in context:
http://r.789695.n4.nabble.com/Concatenation-tp
Hi Robert,
Consider the following:
x <- c(1, 3, 5, 4, 6)
ifelse(x < 6 & x > 1, 1, 2)
[1] 2 1 1 1 2
HTH,
Jorge
On Fri, May 21, 2010 at 9:36 AM, Robert U <> wrote:
> Dear
> R-users,
>
>
>
> I've been trying to write a script but i encounter much problems with basic
> functions.. That is, i try
Thank you, David and Gabor, for the creative solutions, and for introducing
me to pmin()
Jonathan
On Thu, May 20, 2010 at 9:55 PM, Gabor Grothendieck wrote:
> Try this one liner. The first argument of rep is the sorted
> intersection and the second argument is the calculated from the
> paral
On May 21, 2010, at 9:19 AM, arnaud Gaboury wrote:
Dear group,
Here is my environment :
ls()
[1] "l" "PLglobal" "Pos100415" "Pos100416" "Pos100419"
"Pos100420" "position" "select" "Trad100415" "Trad100416"
"Trad100419" "Trad100420" "trade" "y"
With objects :
l
[1
Example:
if (abs(-1-1) <= 3 & abs(1+1) <= 3) {print(1)} else {print(2)}
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statis
You need parentheses enclosing the entire if() statement:
if(something & somethingelse) {
do this
}
Sarah
On Fri, May 21, 2010 at 9:36 AM, Robert U wrote:
> Dear
> R-users,
>
>
>
> I've been trying to write a script but i encounter much problems with basic
> functions.. That is, i try to wri
Dear
R-users,
I've been trying to write a script but i encounter much problems with basic
functions.. That is, i try to write a simple IF ELSE statement, with 2
requirements for the IF :
if (abs(x + Dataset$LAT[1]) <= 3) and (abs(y + Dataset$LONG[1]) <= 3) {z
<- 1} else {z <-0}
This does n
Dear group,
Here is my environment :
> ls()
[1] "l" "PLglobal" "Pos100415" "Pos100416" "Pos100419"
"Pos100420" "position" "select" "Trad100415" "Trad100416"
"Trad100419" "Trad100420" "trade" "y"
With objects :
> l
[1] "100415" "100416" "100419" "100420" "100421" "10
> > ... interactions between covariables and time.
>
> A model such as "coxph(Surv(ptime, pstat) ~ age + age*ptime, "
> is invalid -- it is not at all what you think.
Actually what i'm trying to fit is
coxph(Surv(start,end,event)~age+age:start)
to model a time-varying effect \beta(t)
Also see mixedsort and mixedorder in the gtools package.
dd <- data.frame(b = c("chr2", "chr1", "chr15", "chr13"),
x = c("A", "D", "A", "C"), y = c(8, 3, 9, 9),
z = c(1, 1, 1, 2))
dd[gtools::mixedorder(dd$b),]
R> b x y z
2 chr1 D 3 1
1 chr2 A 8 1
4 chr13 C 9 2
3 chr15 A 9 1
The caret package can do a lot of that for you:
http://www.jstatsoft.org/v28/i05/paper
http://cran.r-project.org/web/packages/caret/index.html
http://cran.r-project.org/web/packages/caret/vignettes/caretTrain.pdf
Max
On Fri, May 21, 2010 at 8:05 AM, Roel Meeuws wrote:
> Dear R expert
But you should both be warned that would only be correct for a species
where there were 24 chromosomes. This would work for humans and
species with a higher number:
dd[ order(as.numeric(substring(dd$b, 4), substring(dd$b, 4))), ]
--
David Winsemius, MD
West Hartford, CT
On May 20, 2010, a
Dear R expert
I have come across the GBM package for R and it seemed appropriate for my
research. I am trying to predict the number of FPGA resources required by a
Software Function if it were mapped onto hardware. As input I use software
metrics (a lot of them). I already use several regression
> ... interactions between covariables and time.
A model such as "coxph(Surv(ptime, pstat) ~ age + age*ptime, "
is invalid -- it is not at all what you think. If cph flags this as an
error that is a good thing: I should probably add the same message to
coxph.
> Is is somewhat sensible to
Hi Brian,
This is really odd. I keep getting the NA secs answer, only by running
these three lines of code in a new session.
> time3=strptime("2009 06 01 00 47 00",format="%Y %m %d %H %M")
> time4=strptime("2009 06 01 00 57 00",format="%Y %m %d %H %M")
> diff(c(time3,time4))
Time difference of
Try this:
lm(y ~., tmp)
On Fri, May 21, 2010 at 3:54 AM, Yuan Jian wrote:
> Hi,
>
> if I know the colnames x and y in the following example, I can easily to do
> lm.
> tmp <- data.frame(x=c(1,1.2),y=c(1,2))
> lm(y ~ x, data=tmp)
>
> when the colnames are variable, what should I do? for example
Thank you Dennis and Henrique for your help!
Both solutions work! I just need to find a way of removing the empty "cells"
from the final "long" dataframe since they are not NAs.
Maybe there is an easier way of doing this of the data is not treated as a
dataframe? The original data file that is
TY Ivan
select <- l[which(l==x):which(l==y)] returns the correct result.
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Ivan Calandra
> Sent: Friday, May 21, 2010 11:20 AM
> To: r-help@r-project.org
> Subject: Re: [R] se
Perhaps you mean something like sapply or apply?
When d is indeed a data.frame with one column: sapply(d[,1], mash)
Regards,
Jan van der Laan
On Thu, May 20, 2010 at 12:47 AM, sedm1000 wrote:
>
> I hope that somebody can help me with this - I think very simple - issue...?
>
> I am running a
Hi,
This should work:
select <- l[which(l==x):which(l==y)]
The problem is that you tried to select the values, but gave x and y as
indexes. That's why you didn't get the expected result.
See ?which for more info.
HTH,
Ivan
Le 5/21/2010 11:07, arnaud Gaboury a écrit :
Dear group,
Here is a
Dear group,
Here is a list :
l <-
list("100415", "100416", "100419", "100420", "100421", "100422",
"100423", "100426", "100427", "100428", "100429", "100430",
"100503", "100504", "100505", "100506", "100507", "100510",
"100511", "100512", "100513")
I need to access part of it with
Hi!
The problem is that paste() returns a character string.
You can use get() this way:
lm(y ~ get(paste("aa",1,sep="_")), data = tmp)
Or:
lm(tmp[[2]]~tmp[[1]], data=tmp)
HTH,
Ivan
Le 5/21/2010 09:54, Yuan Jian a écrit :
Hi,
if I know the colnames x and y in the following example, I can eas
On 20.05.2010 00:47, sedm1000 wrote:
I hope that somebody can help me with this - I think very simple - issue...?
I am running a package that only accepts one line at a time, but I would
like to run this package on a dataframe of>500 lines.
Dataframe "d" is a single column:
APPLES
PEARS
AUB
On 21.05.2010 09:54, Yuan Jian wrote:
Hi,
if I know the colnames x and y in the following example, I can easily to do lm.
tmp<- data.frame(x=c(1,1.2),y=c(1,2))
lm(y ~ x, data=tmp)
when the colnames are variable, what should I do? for example
colnames(tmp)[1]<- paste("aa",1,sep="_")
lm(y ~ pas
I think kernlab can do. Also the svmlight implementation that has a
rather crude interface in package klaR.
Uwe Ligges
On 20.05.2010 01:48, Xiyan Lon wrote:
I am learning classification using SVM for research (survey).
The data that I have had some class. I know that the library (e1071)
can o
On Fri, 21 May 2010 03:29:49 -0400
Yong Wang wrote:
> Dear list
>
> I need to 1) run several R scripts sequentially due to results waiting
> and 2) run them in unix background since my ssh frequently timeout for
> some reason.
If this is the main obstacle you should learn about screen!
http://
On 20.05.2010 19:16, Biau David wrote:
Hi,
I am trying to describe a data.frame by obtaining multiple crosstable summary
statistics at once. I have tried table, xtab, crosstable, summaryBy and
describe but none of these functions seems to allow muliple conparisons at once.
Here, is what
On 21.05.2010 09:29, Yong Wang wrote:
Dear list
I need to 1) run several R scripts sequentially due to results waiting
and 2) run them in unix background since my ssh frequently timeout for
some reason.
if paste following codes to unix
R --vanilla script1&
R --vanilla script2&
R --vanilla scr
At first, I'd try with an R version from 2010 rather than one from 2007.
Next, I'd try to be sure to really have a 64-bit version of R rather
than a 32 bit one which is what I suspect.
Best,
Uwe Ligges
On 20.05.2010 20:10, Yesha Patel wrote:
I've looked through all of the posts about this is
Hi,
if I know the colnames x and y in the following example, I can easily to do lm.
tmp <- data.frame(x=c(1,1.2),y=c(1,2))
lm(y ~ x, data=tmp)
when the colnames are variable, what should I do? for example
colnames(tmp)[1] <- paste("aa",1,sep="_")
lm(y ~ paste("aa",1,sep="_"), data = tmp)
it gives
Dear list
I need to 1) run several R scripts sequentially due to results waiting
and 2) run them in unix background since my ssh frequently timeout for
some reason.
if paste following codes to unix
R --vanilla script1 &
R --vanilla script2 &
R --vanilla script3 &
will result in simultaneous inst
Well, it seems a simple "c" will do.
e.g.
df <- matrix(rnorm(100), ncol=4)
c(t(df[1:10,])) # concatenate rows
c(df) # concatenate columns
>
>From: santana sarma
>To: "Shi, Tao" ; r-help@r-project.org
>Sent: Thu, May 20, 2010 11:45:25 PM
>Subject: Re: Concatenation
>
>
>Ok -let's forget
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