abline() or lines()
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Dear R-help,
I am doing clustering via finite mixture model. Please suggest some packages in
R to find clusters via finite mixture model with continuous variables. And
also I wish to verify the distributional properties of the mixture
distributions
by fitting the model with lognormal, gamma, ex
par(mfrow=c(3,2)) ## will get you 3 rows and 2 columns
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R-h
My code looks like this:
lin = NA
for(i in 1:15){
lin[i] = lm(reservesub[,3]~ reservesub[,i+3])
}
For which I'm given 15 warning messages which say :
"1: In lin[i] = lm(reservesub[, 3] ~ reservesub[, i + 3]) :
number of items to replace is not a multiple of replacement length"
I'm am abl
BSanders wrote:
>
>
> .
> lin[i] = lm(reservesub[,3]~ reservesub[,i+3])
>
> For which I'm given 15 warning messages which say :
>
> "1: In lin[i] = lm(reservesub[, 3] ~ reservesub[, i + 3]) :
> number of items to replace is not a multiple of replacement length"
>
lin = list()
for(i
On Mon, 31 Jan 2011, Nick Matzke wrote:
Hi,
I've written a function which I load with a source command. The function
requires a certain library, phangorn, to work.
Do you mean an R package or an OS library (aka DLL)? I'll assume the
former, but please do use accurate terminology.
I woul
On Mon, 31 Jan 2011, MacQueen, Don wrote:
I'm using list.files() on my home directory, like this:
crnt.files <- list.files(dir.to.check, full.names=TRUE, all.files=TRUE,
recursive=TRUE)
With dir.to.check set to the full path to my home directory.
After a while I get:
Error in list.files(di
Dear Jorge,
Thanks for the reply. But what I mean is to
create plot on "top" of another in to one page.
So in the end there will be only one plot with two curves.
Actually what I am to plot is two ROC curves.
- G.V.
On Tue, Feb 1, 2011 at 3:37 PM, Jorge Ivan Velez
wrote:
> Hi Gundala,
> Yes.
Hi AD,
You might try the following:
# data
a <- c(2,3,5)
b <- c(8,7) # you got this wrong ;)
# option 1
foo <- function(x) as.numeric(paste(x, sep = "", collapse = ""))
# examples
foo(a)
# [1] 235
foo(b)
# [1] 87
foo(a) + foo(b)
# [1] 322
# option 2
foo2 <- function(x, y) foo(x) + foo(y)
#
I am trying to create a function that is able to calculate this sum:
a<-c(2,3,5)
b<-(8,7)
with "a" meaning 235 and "b" 87. So the result of this sum would be 235 + 87
= 322.
a <- c(2,3,5)
b <- c(8,7)
vectorToScalar <- function(x) {
as.numeric(paste(x, collapse = ""))
}
vectorToScalar(a)
I have a R script that contain these lines for plotting:
plot(foo,lwd=2,lty=3,col="red", main="");
plot(bar,lwd=2,lty=3,col="blue");
legend(0.6,0.6,c('Default','Probabilistic'), col=c('red','blue'),lwd=3);
But it generate 1 file (Rplot.pdf) with two pages. Each page for 1 plot.
Is there a way I
baptiste auguie googlemail.com> writes:
>
> Dear list,
>
> I'm trying to visualise some ellipsoidal shapes in 3D. Their position,
> axes, and angular orientation can be arbitrary. I saw an ellipse3d
> function in rgl; however it is heavily oriented towards the
> statistical concept of ellipse o
On Jan 31, 2011, at 11:04 PM, Nick Matzke wrote:
Hi,
I've written a function which I load with a source command. The
function requires a certain library, phangorn, to work.
I would like the function to check if phangorn is loaded as a
library before running. For some reason, just putti
On 01/31/2011 12:15 PM, Elliot Joel Bernstein wrote:
> I'm trying to write a generic function that calls different methods
> depending on the structure of the argument, but not the exact type of
> its contents. For example, the function 'nan2last' below works for a
> numeric vector but not for a ve
Hi
I am trying to create a function that is able to calculate this sum:
a<-c(2,3,5)
b<-(8,7)
with "a" meaning 235 and "b" 87. So the result of this sum would be 235 + 87
= 322.
I've searched a function like strsplit but that worked for integers and in
reverse - not spliting but combining.
Can
On Jan 31, 2011, at 8:45 PM, Gundala Viswanath wrote:
Dear sirs,
I have a data that is generated like this:
dat1 <- data.frame(V1 = rep(1, 5), V2 = sample(c(40:45), 5))
dat2 <- data.frame(V1 = sample(c(0,1), 5, replace = TRUE), V2 =
sample(c(40:45), 5, replace = TRUE))
What I want to do
Hi:
Put an equals sign between xlab and '
Dennis
On Mon, Jan 31, 2011 at 6:39 PM, MarquisDM wrote:
>
> Hi everyone,
>
> Sorry for the newbie question but whenever I enter the following code into
> r
> it gives me an unexpected string constant in
> "boxplot(Abs~Conc,ylab='Absorbency',xlab'Etha
On Jan 31, 2011, at 9:39 PM, MarquisDM wrote:
boxplot(Abs~Conc,ylab='Absorbency',xlab'Ethanol(%)')
missing "="..^.
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailma
Hi,
I've written a function which I load with a source command.
The function requires a certain library, phangorn, to work.
I would like the function to check if phangorn is loaded as
a library before running. For some reason, just putting
require(phangorn) into the sourced function doesn'
Hi everyone,
Sorry for the newbie question but whenever I enter the following code into r
it gives me an unexpected string constant in
"boxplot(Abs~Conc,ylab='Absorbency',xlab'Ethanol(%)'" error. I have tried
everything to eliminate it and have searched these forums to no avail, can
anyone tell
See: https://www.msu.edu/~chunghw/downloads.html
Maybe you can find something useful there!
Regards
On 1/31/2011 12:35 PM, Daniel Vecchiato wrote:
Dear R users,
I would like to perform Latent Class Logit Models for the analysis of choice
experiments in environmental valuation.
This kind of a
Dear sirs,
I have a data that is generated like this:
> dat1 <- data.frame(V1 = rep(1, 5), V2 = sample(c(40:45), 5))
> dat2 <- data.frame(V1 = sample(c(0,1), 5, replace = TRUE), V2 =
> sample(c(40:45), 5, replace = TRUE))
What I want to do is to obtain a data frame that contain list of list.
>
Thanks a lot, Kjetil!
On Mon, Jan 31, 2011 at 1:15 PM, Kjetil Halvorsen
wrote:
> The Matrix package (which should already be insatlled on your
> computer, since it is "Recommended") have the function
> nearPD, which should do the job.
>
> Kjetil
>
> On Sat, Jan 29, 2011 at 1:32 AM, Dimitri Liakho
Hi,
I uploaded my package to http://win-builder.r-project.org/ and had it built.
It can work on one machine, but on the other, it reports:
Loading required package: grt
Error: package 'grt' was built for i386-pc-mingw32
In addition: Warning message:
package 'grt' was built under R version 2.12.1
I'm using list.files() on my home directory, like this:
crnt.files <- list.files(dir.to.check, full.names=TRUE, all.files=TRUE,
recursive=TRUE)
With dir.to.check set to the full path to my home directory.
After a while I get:
Error in list.files(dir.to.check, full.names = TRUE, all.files = T
Is there a limit to how many nested function calls R will tolerate? I mean,
if I have a function that calls another function that calls another
function, etc., is there a limit to how deep I can go?
I am getting an error that says a variable is not available. When I do a
traceback(), it shows that
On 2011-01-31 13:49, Dean Castillo wrote:
Hello
I am trying to store output from a loop into an empty matrix. The
current code I am using is:
M<-mat.or.vec(11,89)
for (j in list(3,91))
+ {M[,(j-2)]<-pic(datain[,j], mytree)}
datain is a matrix (11,91). I only want to use the pic() function on
t
On Jan 31, 2011, at 4:49 PM, Dean Castillo wrote:
Hello
I am trying to store output from a loop into an empty matrix. The
current code I am using is:
M<-mat.or.vec(11,89)
R code to create a matix would use the matrix function,
for (j in list(3,91))
+ {M[,(j-2)]<-pic(datain[,j], mytree)}
Hi, I have a vector of data, that I group based on two factors via
tapply. For each such grouping I would like to plot a pie chart. I
can layout these pie charts in a matrix layout, correpsonding to the
levels of the two factors. But I am getting stuck on how to label the
rows and colums. My curre
Indeed, tapply is what I needed. To clarify Phils' question, what I needed was
tapply(x, list(cut.grp1, cut.grp2), function(z) table(z))
On Mon, Jan 31, 2011 at 4:50 PM, Bert Gunter wrote:
> ?tapply is the basic R function for this. There are many other packages
> (e.g. plyr) and functions (e.
On 2011-01-31 12:42, moleps wrote:
Dear all,
Given
rr<-data.frame(r1<-rnorm(1000,10,5),r2<-rnorm(1000,220,5))
How can I add a column (rr$p) for the joint probability of each r1& r2 pair?
If you take the values in each pair to be observations
from two independent Normal distributions, it's
Hello
I am trying to store output from a loop into an empty matrix. The
current code I am using is:
> M<-mat.or.vec(11,89)
> for (j in list(3,91))
+ {M[,(j-2)]<-pic(datain[,j], mytree)}
datain is a matrix (11,91). I only want to use the pic() function on
the columns 3:91.
When I use this code outp
After ordering the table of membership degrees , i must get the difference
between the first and second coloumns , between the first and second largest
membership degree of object i. This for K=2,K=3,to K.max=6.
This difference is multiplyed by the Crisp silhouette index vector (si). Too
it de
Hi,
I'm a beginner with R.
I have two different tables with
the variable name dmr1 and tp2 a given following. v1 is the common column
field of
the both tables. In the first table column v3 is always v2+1 while in
the second table v2 and v3 hold the range. I want to know which rows
of tb1 in
On Mon, Jan 31, 2011 at 05:35:35PM +0100, Mauluda Akhtar wrote:
> Hello,
>
> I'm a new R user.
>
> I have two different dummy tables with the variable name tb1 and tb2.
Hello.
First, let me put your data into an R command using dput().
tb1 <-
structure(list(V1 = structure(c(1L, 1L, 1L, 1L,
Hi,
I doing some analysis and truying to use Survival analysis (Survival
Package) to show the relation between some metrics and the presence of an
event A in a classe.
I build a Cox model with Coxph.
My problem is with the function predict.
Because i would like to know the accuracy of the model (by
I'm trying to write a generic function that calls different methods depending
on the structure of the argument, but not the exact type of its contents. For
example, the function 'nan2last' below works for a numeric vector but not for a
vector of Date objects. Is there any way to make it work on
Hi Sarah,
Here is how I would do it. Not elegent, but fairly transparent, and it
seems to give the desired result.
DF <- as.data.frame(DF)
pick.value <- function(x){
if(0 %in% x$PA1) {
x <- x[x$PA1 == 0,]
}
x <- x[x$Area == max(x$Area, na.rm=T),]
S <- x[sample(1:nrow(x),
Rajarshi -
It's not clear to me what you mean by "the distribution of
levels obs.". Does
as.data.frame(table(x$obs,cut.grp1,cut.grp2))
give you something like what you want?
- Phil Spector
Statistical Computi
Hi, I have a data.frame that has a categorical variable, for which I
would like to look at the distribution of levels of this variable,
based on a grouping of two other variables.
As an example:
x <- data.frame(obs=sample(c('low', 'high'),100, replace=TRUE),
grp1=sample(1:10, 100, replace=TRUE),
Daniel,
searching for 'latent class analysis' on
http://www.r-project.org/search.html
gives many results: the CRAN taskviews on psychometrics and the one on
clustering both contain relevant links to packages that may do what you are
looking for.
In particular, for general LCA there is package e107
Hi Silvano:
Could you tell me what "correlation=corSymm(form = ~ 1 |id)" represents? In
our case, team is random effect, trt, pairs, grade, school are fixed effect,
and each team is within school.
I still got the different results from both SAS and R.
> unstruct <- gls(score~trt+pairs+grade+s
Greetings,
Suppose I fit an fGarch model via garchFit function for a time series X.
I'm wondering is there any easy way to apply the fitted model to a different
time series Y to calculate conditional variances and standardized residuals?
Thanks.
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Bill Pikounis provided a clever and elegant solution: in the program
barplot.default, replace the statement
width <- rep(width, NR)
that occours around line 51 ( NR = nrow(height) ) with the statement width
<- width. I renamed the program
barplotX.fn
and attached it to this ema
Thanks very much Professor Ripley.
I was running R 2.10 under XP. As part of troubleshooting I uninstalled and
installed 2.11. I also updated cygwin. This is a system-specific problem,
editing still works on my home laptop.
I am running CRAN build of R.
Yes I did mean Vim.
I believe it is a s
Dear all,
Given
rr<-data.frame(r1<-rnorm(1000,10,5),r2<-rnorm(1000,220,5))
How can I add a column (rr$p) for the joint probability of each r1 & r2 pair?
I know how to add the column.. I just dont know how to compute the p value for
joint probabilities given the two samples.
//M
__
One option is the nearPD function in the Matrix package.
Other options include robust estimation of the covariance matrix.
You should Google this. It's been discussed before.
Kevin Wright
On Mon, Jan 31, 2011 at 11:30 AM, Mike Miller
> wrote:
> Is there an R function for computing a varianc
Hi all,
I am hoping to get R installed configured at work, on an HP-Unix server. R
was installed but was not configured correctly because we do not have a
Fortran compiler for HP unix. The IT guys sent me an email with some
questions that I hope to get some help in anwsering.
"
We need the inform
Hey there,
I have a matrix which is from NMR data (first column represents the ppm
values and the subsequent their respective intensities for my various
samples) that I would like to bin. Make every 10 points ( on my x axis)
become one by averaging them out. Any suggestions?
Thanks!
--
Marcelo A
On Jan 31, 2011, at 2:26 PM, David Winsemius wrote:
On Jan 31, 2011, at 2:18 PM, Bogaso Christofer wrote:
Sorry if I did not clarify that. Here I have a data frame with many
columns,
which was taken from some outside DB. Now I want to split that data
frame
and create a "list" object (to m
On Jan 31, 2011, at 2:18 PM, Bogaso Christofer wrote:
Sorry if I did not clarify that. Here I have a data frame with many
columns,
which was taken from some outside DB. Now I want to split that data
frame
and create a "list" object (to make my further calculation easier),
on basis
of a typ
Try this:
length(unique(sapply(list(rnorm(4), rnorm(5), rnorm(6)), length))) == 1
On Mon, Jan 31, 2011 at 5:14 PM, Megh Dal wrote:
> I am looking for an elegant way how I can test the equality of lengths of
> multiple vectors.
>
> For example, this is working fine:
>
> > length(rnorm(4)) == len
I am looking for an elegant way how I can test the equality of lengths of
multiple vectors.
For example, this is working fine:
> length(rnorm(4)) == length(rnorm(5))
[1] FALSE
However this is not:
> length(rnorm(4)) == length(rnorm(5)) == length(rnorm(6))
Error: unexpected '==' in "length(rnorm(
Ted et.That solution is for (in "missing data language") MCAR
(Missing Completely At Random), i.e. the probability
of being missing does not depend on any of the variables
in the data.
For MAR (Missing At Random), the probability of being
missing may depend on the values of covariates but must
not
I see that others have already responded, but will add my point of
view. You indicated that you wanted to take the difference between
pairs of columns and did not specify exactly how many there were; in
your example there were 4 columns (2 pairs). If there were only two,
then the solution from De
Sorry if I did not clarify that. Here I have a data frame with many columns,
which was taken from some outside DB. Now I want to split that data frame
and create a "list" object (to make my further calculation easier), on basis
of a typical column of that DB. I cannot post my original DB here (due
Dear R users,
I would like to perform Latent Class Logit Models for the analysis of choice
experiments in environmental valuation.
This kind of analysis is usually performed with NLogit Software
(http://www.limdep.com).
I attach the results I usually obtain using NLogit and NLogit model
speci
Dear all
How can I obtain the data from the function "rearrange" in package quantreg
More especifically, based on the example below (available in the help of the
rearrange function), how can I access the data generated by
"rearrange(zp)" ?
data(engel)
z <- rq(foodexp ~ income, ta
Hello All,
I am having a problem creating text for my forest plot using the "forestplot"
function in the rmeta package. I think my problem is that I have too many
columns of text, so I was wondering if there is a way to shrink my text or to
change the default setting.
Here is an example:
My dataframe looks like this one:
SightingID<-c(2001,2002,2003,2004,2005,2006,2007,2008,2009,2010,2011,2012,2013)PA1<-c(0,1,0,0,1,1,1,1,0,0,-99,1,1)PA2<-c(1,NA,1,1,NA,-99,-99,NA,1,1,1,NA,NA)PlotID<-c(1,1,2,2,2,3,3,3,4,4,4,4,5)Area<-c(0.2,0.3,0.25,0.2,0.3,0.4,0.3,0.35,0.4,0.4,0.5,0.3,0.2)DF<-cbind(
On Jan 31, 2011, at 1:56 PM, Bogaso Christofer wrote:
Thanks David for this reply. However if my data frame has only 2
columns
then it is working fine. It is not working for a general setting:
dfrm <- data.frame(x=rnorm(18), y=rep(c("a", "b", "c"), each=6),
z=rep(c("x", "y", "z"), each=2))
t
Thanks David for this reply. However if my data frame has only 2 columns
then it is working fine. It is not working for a general setting:
dfrm <- data.frame(x=rnorm(18), y=rep(c("a", "b", "c"), each=6),
z=rep(c("x", "y", "z"), each=2))
tapply(dfrm[,1], dfrm$y, c) # this is working fine
> tapply(
The Matrix package (which should already be insatlled on your
computer, since it is "Recommended") have the function
nearPD, which should do the job.
Kjetil
On Sat, Jan 29, 2011 at 1:32 AM, Dimitri Liakhovitski
wrote:
> Dear all:
>
> In what I am doing I sometimes get a (Hessian) matrix that has
Hi All,
I would like to apply Bayesian network on some data. However, some of the
features are based on time. E.g. Number of time he/she visit library.
As it can be Total number of this person visits, Average weekly number of
visit, Daily number of visits, or even Number of visits in th
On Jan 31, 2011, at 1:03 PM, Bogaso Christofer wrote:
Dear all, let say I have following data frame:
> dfrm <- data.frame(x=rnorm(18), y=rep(c("a", "b", "c"), each=6))
> tapply(dfrm$x, dfrm$y, c)
$a
[1] 0.9711995 1.4018345 -1.4355713 -0.5106138 -0.8470171
[6] 1.1634586
$b
[1] -0.8058164
On Mon, Jan 31, 2011 at 9:30 AM, Mike Miller wrote:
> Is there an R function for computing a variance-covariance matrix that
> guarantees that it will have no negative eigenvalues? In my case, there is
> a *lot* of missing data, especially for a subset of variables. I think my
> tactic will be t
Dear all, let say I have following data frame:
> data.frame(x=rnorm(18), y=rep(c("a", "b", "c"), each=6))
x y
1 -1.072152537 a
2 0.382985265 a
3 0.058877377 a
4 -0.006911939 a
5 -2.355269051 a
6 -0.303095553 a
7 0.484038422 b
8 0.733928931 b
9 -1.136014346 b
Is there an R function for computing a variance-covariance matrix that
guarantees that it will have no negative eigenvalues? In my case, there
is a *lot* of missing data, especially for a subset of variables. I think
my tactic will be to compute cor(x, use="pairwise.complete.obs") and then
pr
Dear list,
I'm trying to visualise some ellipsoidal shapes in 3D. Their position,
axes, and angular orientation can be arbitrary. I saw an ellipse3d
function in rgl; however it is heavily oriented towards the
statistical concept of ellipse of confidence, whilst I am just
concerned with the geometr
> "JH" == Johannes Huesing
> on Fri, 28 Jan 2011 06:37:31 +0100 writes:
JH> David Winsemius [Thu, Jan 27, 2011 at
10:08:00PM CET]:
>> You got a perfectly sensible reply from Thereau, the author of the
>> package, a day after your posting and then failed to respond to h
Eiiti Kasuya kyushu-u.org> writes:
>
> When quasi family (not quasipoisson or quasibinomial) is used in glm,
> what is the appropriate test statistic in anova.glm?
> Help of anova.glm tells “For models with known dispersion (e.g.,
> binomial and Poisson fits) the chi-squared test is most appropr
Hello,
I'm a new R user.
I have two different dummy tables with the variable name tb1 and tb2.
tb1<
v1v2 v3 v4
"chr1" 2223 3
"chr1" 3637 1
"chr1" 5455 0
"chr1" 7778 1
"chr2" 8081 4
"chr2" 8586 0
"chr2" 99 1
Here are the specifications from Excel 2010. There are others for older
versions of Excel so you may need to do a search specific to your
application.
http://office.microsoft.com/en-us/excel-help/excel-specifications-and-limits-HP010342495.aspx?CTT=5&origin=HP005199291
You should be OK looking
Dear Sarah,
What you will need is a series of logical conditions. ?Logic or ?"|"
should pull up the documentation on the logical operators available to
use. Because this list does not accept HTML emails (see the posting
guide), your data frame did not come through in any coherent form.
Can you
We need to know much more (see the posting guide)
- What version of R? Did you update recently?
- Are you running a Windows build of R, or did you compile your own
from the sources as a Cygwin build?
On Mon, 31 Jan 2011, beehatch wrote:
Hi all,
I've been a happy user of R under cygwin for
Thanks. It helped me a lot.
Ramya
On Mon, Jan 31, 2011 at 6:56 AM, djmuseR [via R] <
ml-node+3248651-1296211736-40...@n4.nabble.com
> wrote:
> Hi:
>
> I won't speak for Jim, as he's more than capable of responding to this
> himself, but I'll give it a shot:
>
> (1) It's not just the 'double per
My data frame looks like:
SightingID PA1 PA2 PlotID InOverlap Area12001 1 -99392
Y0.222002 1 -99388 Y0.2532008 1
NA104 N0.3442010 1 NA 71 N
0.1852012 1 NA 6
Please I would like that my email address be withdrawn from contact
list.Sincerely yours
Fernanda Melo Carneiro contato: (62) 3521-1480 e 8121-7374www.ecoevol.ufg.br
Laboratório de Ecologia Teórica e Síntese (UFG)
[[alternative HTML version deleted]]
_
Petr Pikal wrote:
>
> Hi
>
> r-help-boun...@r-project.org napsal dne 31.01.2011 09:44:00:
>
>>
>>
>> David Winsemius wrote:
>> >
>> >
>> > On Jan 30, 2011, at 5:27 PM, ADias wrote:
>> >
>> >> dados<-
>> >> data
>> >> .frame
>> >> (Store
>> >> =
>> >> c
>> >> ("Setubal
>> >> ","lx
Hi David,
Thanks for reply. It really was the fda library.
Best Wishes,
Jari
Lainaus Jari Soininen :
Hi all,
This is very embarrassment, but I have used some years ago this type
of function call:
" Arima(trend + max(seasonal), order=c(2,0,1), xreg=fourier(1:n,c*2,m))"
Unfortunately,
On Jan 31, 2011, at 10:46 AM, huang min wrote:
It seems Prof. Lumley has moved to Auckland.
But his webpage also says: "He is still an Affiliate Professor at the
University of Washington."
On Mon, Jan 31, 2011 at 11:33 PM, John Fox wrote:
Dear Joe,
Take a look at the mitools package
It seems Prof. Lumley has moved to Auckland.
On Mon, Jan 31, 2011 at 11:33 PM, John Fox wrote:
> Dear Joe,
>
> Take a look at the mitools package, written by Thomas Lumley, who's a
> faculty member at your university.
>
> Best,
> John
>
>
> John Fox
> Senator Wi
Dear Joe,
Take a look at the mitools package, written by Thomas Lumley, who's a
faculty member at your university.
Best,
John
John Fox
Senator William McMaster
Professor of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
Dear R-users!
I've been trying to produce tables from glm-results using mtable from the
memsic package but I'm not quite happy with the default of signif.codes in
R. What I want to do is to change this line:
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1
to that line:
Signif. co
R gurus:
I'm thinking about using R for website traffic analysis but didn't find
anything in my web searches specific to R.
If I have the webpages (simple example would contain something like three (3)
static pages with a couple of links each) and parse the apache access log file
(httpd-acce
> survmod <- survreg(Surv(sapply(bleedtimes, min, 10), bleedtimes < 10)
> ~ device + frailty.gaussian(pat))
> residuals(survmod)
> Error in residuals.survreg.penal(survmod) :
> Residuals not available for sparse models
Thanks for the reproducable error.
a. With respect to your earli
You're right, it's not exactly what you wanted.
But... data.frames are lists, so you would access each element of it as
you intended to:
HIV$hiv.dat1$predictions
My opinion is that it's easier to work with data.frames when possible
(as opposed to lists), and I don't see why you would break do
Hi Ivan,
Thanks for the reply.
> Would that do?
But I won't do.
> names(dat1) <- names(dat2) <- c("labels","predictions")
> HIV <- list(hiv.dat1=dat1, hiv.dat2=dat2)
The above snippet produces this instead:
> names(dat1) <- names(dat2) <- c("labels","predictions")
> HIV <- list(hiv.dat1=dat1
Well, AutoHotKey [1] is a free program/scripting language that runs
scripts to perform basic window handling, file I/O, and program execution.
It also allows you to bind the running of these scripts to keys or
mouse-strokes with the option of making each window-specific. NppToR is
actually a co
When quasi family (not quasipoisson or quasibinomial) is used in glm,
what is the appropriate test statistic in anova.glm?
Help of anova.glm tells “For models with known dispersion (e.g.,
binomial and Poisson fits) the chi-squared test is most appropriate, and
for those with dispersion estimated by
Dear All,
I've always used this code:
year=c(1948:1953,2000,2100,2200,2300)
numdays=ifelse((year%%4==0 & year%%100!=0) | year%%400==0,366,365)
> numdays
[1] 366 365 365 365 366 365 366 365 365 365
Toby
From: r-help-boun...@r-project.org [r-help-boun...
Thank you !
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Hi users,
When I use the scaling = by.level, appear in the R, values of the scales by
level, but I'm viewing the levels 10, 11, 12, 13 and 14. I do like to have a
global scale, using the value of Level 14 ".
I would like to see everyone on the same scale level of 14. If I leave th
I am trying to read some file names from an specific directory and such names
contains metacharacters.
The file names is like V4.35_T01-400720.csv
In total I have 14 files for which the value T01 goes up to T14. I need to read
the files into a string vector that looks like
>names"V4.35_T01-4007
Hi all,
This is very embarrassment, but I have used some years ago this type
of function call:
" Arima(trend + max(seasonal), order=c(2,0,1), xreg=fourier(1:n,c*2,m))"
Unfortunately, now I'm not able to find the suitable library where
this specific fourier-call is implemented. Can you hel
Hello Alaios,
I would like to figure out the command "version" inside of R:
version
_
platform x86_64-pc-mingw32
arch x86_64
os mingw32
system x86_64, mingw32
status
major 2
minor 11.1
year 2010
month 05
d
Hi all,
I've been a happy user of R under cygwin for a number of years now. I
prefer the bash interface and readline history, and I find Vi to be the
most efficient editor.
Recently, the edit command has started giving me errors:
> edit(file='foo.R')
Error in edit(name, file, title, editor) : u
I tried using the "Snowball" package for performing stemming in text mining.
But when I tried to load the package the following error is thrown:
Error : .onLoad failed in loadNamespace() for 'Snowball', details:
call: NULL
error: .onLoad failed in loadNamespace() for 'rJava', details:
call:
This worked fine. Thanks.
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On Jan 31, 2011, at 1:07 AM, Maithula Chandrashekhar wrote:
Hi all, in R I have Sys.timezone() function to get the current working
Time zone.
Only on some systems.
However I want to have a vector to get the list of all
available time zones, like say, LETTERS gives me all letters. Is there
a
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