> -Original Message-
> From: Saptarshi Guha [mailto:saptarshi.g...@gmail.com]
> Sent: Saturday, March 19, 2011 12:24 AM
> To: XiaoboGu
> Cc: r-help@r-project.org
> Subject: Re: Does RHIPE support R on Windows as the user desktop environment?
>
> Hello,
>
> I should see how to mirror a g
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yi has chosen to send you a free movie ticket, up to $10 value.
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Erin -
If you seriously want to learn about regular expressions, I would
recommend "Mastering Regular Expressions" by Jeffrey Freidl.
(http://oreilly.com/catalog/9781565922570). I don't think there's
anything that's more thorough or authoritive.
- Phi
Finally, I solved my problem using the following procedure and a
database called ej
ej$a <- 1
head(ej)
ano nunico a
1 2008 1 1
2 2009 1 1
3 2008 2 1
4 2009 2 1
5 2008 3 1
6 2009 3 1
library(reshape)
dej <- cast(ej, nunico ~ ano, sum, margins = FALSE)
head(dej
Dear R People:
Could someone recommend a good reference on regular expressions, please?
Thanks in advance,
Sincerely,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com
On 18/03/2011 8:43 PM, Shige Song wrote:
One thing that Linux makes trivially easy is to interpolate R with C++
through the Rcpp package. The GCC compiler collection is part of all
mainstream Linux distro. This is, however, not the case with Windows:
you may be able to do it eventually (not sure
Hi Rita,
This is far from the most efficient or elegant way, but:
## two column data frame, one all NAs
d <- data.frame(1:10, NA)
## use apply to create logical vector and subset d
d[, apply(d, 2, function(x) !all(is.na(x)))]
I am just apply()ing to each column (the 2) of d, the function
!all(is
Thank you very much!
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Dear List Members,I have 55 data frames, each of which with 272 variables and
267 observations. Some of these variables are blanks but the blanks are not the
same for every data frame. I would like to write a procedure in which I import
a data frame, see which variables are blank, and delete th
For some reason when I apply a function to a single row in R it works, but
when that same row is in a data.frame it does not, see below:
apply(dx[954,], 1, query.db)
2571
1092 0.06044147
3890 0.05948577
3273 0.05911261
794 0.05855583
4241 0.05854064
3560 0.05602690
apply(dx[954
Hi,
>From the following code (tweaked from another user):
variable<-sample(rep(1:2,100))
individual<-rep(1:3, length(variable))
group<-rep(LETTERS[1:2],length(variable)/2)
mydata<-data.frame(variable,individual,group)
individual<-as.factor(individual)
group<-as.factor(group)
histogram(~
Hello Thierry,
Based on the code below, it looks like you do not need to worry about
likelihoods from lm() and gls(). They are defined the same way. I agree
with you that caution needs to be exercised. Simply because
mathematically the same likelihood may be defined using different constant.
Hi John,
This is not advanced enough for serious applications, but if you are
just looking for a simple way to make example data, you could do
something like:
gencor <- function(order, mu, sd, r) {
X <- matrix(rnorm(prod(order), mean = mu, sd = sd), nrow = order[1],
ncol = order[2])
R <- matr
One thing that Linux makes trivially easy is to interpolate R with C++
through the Rcpp package. The GCC compiler collection is part of all
mainstream Linux distro. This is, however, not the case with Windows:
you may be able to do it eventually (not sure about this point), but
it takes quite some
How would one generate data to be used in a simulation of a repeated measures
ANOVA given a known (1) within-person correlation with known (2) mean and SD of
data obtained at each of three times of observation?
Thanks,
John
John Sorkin
Chief Biostatistics and Informatics
Univ. of Maryland School
Others can discuss the comparative speed. I recently installed
Linux on a new quad core computer I purchased. I found it surprisingly
easy to learn how to do the things I needed. Now the hard drive is
dying on my Vista notebook computer. In shopping for a replacement, I
likely will in
On 2011-03-18 13:28, Adam Hyland wrote:
As a followup to pi-day, I attempted to make a .gif of a simulation
based estimation of pi by plotting points inside a single quadrant of
a circle (a la http://www.drewconway.com/zia/?p=2667 ). When
rendering the individual x,y pairs with points() I interm
Hi Brigid,
I haven't used it yet, but I would suggest profiling your code with 'Rprof'
on both a Windows and Linux machine.
?Rprof for details.
Hope this helps,
~Caitlin
On Fri, Mar 18, 2011 at 3:00 PM, Brigid Mooney wrote:
> I'm not trying to start a Windows vs. Linux debate, but I've been
On Fri, Mar 18, 2011 at 11:00 PM, Brigid Mooney wrote:
> I'm not trying to start a Windows vs. Linux debate, but I've been
> using R on a Windows machine for a while, and was recently wondering
> if R's performance would be faster on a Linux machine. And similarly,
> if any incremental increase i
It will distort statistical inference to drop any terms on the basis of
P-values, AIC, etc..
If you drop terms, use the hierarchy principle.
High correlations between variables don't necessarily invalidate a test.
Frank
Sacha Viquerat-2 wrote:
>
> hello dear list!
> as I am currently helping s
I'm not trying to start a Windows vs. Linux debate, but I've been
using R on a Windows machine for a while, and was recently wondering
if R's performance would be faster on a Linux machine. And similarly,
if any incremental increase in processing speed would be worth the
time it would take me to m
It is important to pay attention to the convergence message. For example, if
the convergence is due to reaching maximum iteration limit (this is often
the case with Nelder-Mead), then you should increase the maxit from 500 to
1500 or something like that.
Users generally have trouble interpreting t
On Fri, Mar 18, 2011 at 2:19 PM, Sam Steingold wrote:
> Hi,
> I have data with multiple sub-second entries:
>
> 2011/03/15 09:32:15.035619,-0.403103,1.09664,48.6,126.92,117.32
> 2011/03/15 09:32:15.069331,-0.39851,1.09874,48.6,126.92,117.32
> 2011/03/15 09:32:15.289135,-0.402463,1.10084,48.59,126.
As a simplification of a much larger problem, I'm using the following simple
example to explore ways of getting the finite difference results to plot
over the same extents as the finite element solution (see figure). I
haven't discovered (if it exist) a way of getting the finite difference data
to
?do.call ## read it CAREFULLY, please.
do.call(runif,as.list(c(1,range)))
-- Bert
On Fri, Mar 18, 2011 at 9:10 AM, Lisa wrote:
> Hi, everybody,
>
> I just want to pass arguments to a function as below:
>
> range <- c(0.1, 0.5)
>
> runif(1, range)
>
> But it doesn’t work.
>
> Does anyone have a
On Fri, Mar 18, 2011 at 11:27:58AM -0700, LC-Bea wrote:
> Hello!
>
> I´m doing this:
>
> A<-optim(vv,pen.wlsv,method="BFGS")
> B<-optim(vv,pen.wlsv,method="Nelder-Mead")
>
> the function is the same
> and the initial values too
>
> the estimate values are different, very different, at the first
On Mar 18, 2011, at 4:13 PM, Guy Jett wrote:
# I need to create an xyplot() where I control the specific order of
# both my conditioning variables. The default code below plots the
# data correctly (dispersed across all 14 columns), but fails in two
# ways. Both the primary conditioning va
On Mar 18, 2011, at 4:00 PM, armstrwa wrote:
That would certainly help, haha. Thank you for catching that
error.
Do you happen to know what exactly '~' means in R?
?formula
It separates the left hand side of a formul
On Mar 18, 2011, at 12:05 PM, AlexSmith wrote:
Hi,
How do I read just the first line of a data file in R? (it does not
contain
labels)
read.table has an `nrow` argument and readLines has an `n` argument.
Thanks in advance,
Alex.
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As a followup to pi-day, I attempted to make a .gif of a simulation
based estimation of pi by plotting points inside a single quadrant of
a circle (a la http://www.drewconway.com/zia/?p=2667 ). When
rendering the individual x,y pairs with points() I intermittently see
points crop up around (2,0.5)
It seems that you are not alone with that error (
http://stackoverflow.com/questions/3311940/r-rjava-package-install-failing).
Defining only JAVA_HOME he would find everything else.
Backup your /etc/R before do this.
Running the R CMD javareconf is to place variables like this in
/etc/R/Makecon
That would certainly help, haha. Thank you for catching that error.
Do you happen to know what exactly '~' means in R?
Thanks again.
Billy
nblarson [via R] wrote:
Flip date and q in your formula, you've got them
backwar
Flip date and q in your formula, you've got them backwards from what you've
said you're trying to model.
armstrwa wrote:
>
>
> Yeah, I did look at the help(loess) page, but I wasn't really sure what to
> do with that. I was inputting it as:
>
>> test<-loess(date ~ q,data.frame(date,q),span=0.
Hi,
I have data with multiple sub-second entries:
2011/03/15 09:32:15.035619,-0.403103,1.09664,48.6,126.92,117.32
2011/03/15 09:32:15.069331,-0.39851,1.09874,48.6,126.92,117.32
2011/03/15 09:32:15.289135,-0.402463,1.10084,48.59,126.92,117.32
2011/03/15 09:32:15.296110,-0.450244,1.10063,48.59,126.9
Hello!
I´m doing this:
A<-optim(vv,pen.wlsv,method="BFGS")
B<-optim(vv,pen.wlsv,method="Nelder-Mead")
the function is the same
and the initial values too
the estimate values are different, very different, at the first decimal
is there an error?
or maybe its because of my data, or funtion to mi
Hi All.
I have created two, 3-D, color graphs using persp(). Z values range from 1
to 100 on the first plot, and 0.5 to 50 on the second plot.
I would like to keep the color range constant between the two graphs, rather
than each graph showing the full range of colors.
So the first graph should go
Dear Orvalho Augusto,
Thank so much for you response!
We had run R CMD javareconf -e, and even R CMD javareconf in administrator.
Actually, we had install the JDK. And in the output, only the item of JNI
cpp flages could not be detected(I listed output below). What does the item
mean? What need we
Yeah, I did look at the help(loess) page, but I wasn't really sure what to do
with that. I was inputting it as:
> test<-loess(date ~ q,data.frame(date,q),span=0.5)
> test
Call:
loess(formula = date ~ q, data = data.frame(date, q), span = 0.5)
Number of Observations: 96
Equivalent Number of Para
Hi R users,
I am getting the following error when using the splsda function in R
v2.12.1:
"Error in switch(classifier, logistic = { : EXPR must be a length 1
vector"
What does this mean and how do I fix this?
Thank you in advance!
Best,
Savi
__
R-he
I wish has simpler solution, apprently simple problem ! thanks for help.
On Fri, Mar 18, 2011 at 10:04 AM, jim holtman wrote:
> I think it was suggested that you save your output to a 'list' and
> then you will have it in a format that can accept variable numbers of
> items in each element and
Hi, everybody,
I just want to pass arguments to a function as below:
range <- c(0.1, 0.5)
runif(1, range)
But it doesn’t work.
Does anyone have any suggestions to offer?
Thanks.
Lisa
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# I need to create an xyplot() where I control the specific order of
# both my conditioning variables. The default code below plots the
# data correctly (dispersed across all 14 columns), but fails in two
# ways. Both the primary conditioning variable (Transect), and the
# secondary conditi
Hi,
How do I read just the first line of a data file in R? (it does not contain
labels)
Thanks in advance,
Alex.
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Good evening! Thanks for your answer, but I don't really understand how to use
vcov matrix for comarision of multiple regression line slopes...
I've read this theme about using t-test in slope comarision:
http://www.mail-archive.com/r-help@r-project.org/msg01713.html
but how to use it in case o
You need to change your second line:
range <- c(0.1, 0.5)
runif(1, range[1], range[2])
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perfect, thanks Henrique!
Nico
On 3/18/2011 11:17 AM, Henrique Dallazuanna wrote:
Try this:
subset(pop, (ave(Area, Area, FUN = length) == 1 | ave(Area, Area, FUN
= function(x)cumsum(prop.table(x)))< 0.7& Area %in% 1:3))
On Fri, Mar 18, 2011 at 2:48 PM, Nicolas Gutierrez wrote:
Hello,
O
That works
# hydeUrls = vector of urls to file names
#destList = vector of destination file names
for(files in 1:length(hydeUrls)){
filesFailed[files]<- FALSE
filesFailed[files] <-
tryCatch(download.file(url=hydeUrls[files],destfile=destList[files],quiet=FALSE,mode="wb"),warning
Hi Patrick,
On Fri, Mar 18, 2011 at 12:17 PM, Patrick Connolly
wrote:
> On Wed, 16-Mar-2011 at 07:58PM -0700, Joshua Wiley wrote:
> |> ## A small example is always nice
> |> dat <- ts(1:12, frequency = 12,
> |> start = c(1998, 1), end = c(2010, 12))
> |>
> |> ## Achim and Gabor's wonderful pack
On Wed, 16-Mar-2011 at 07:58PM -0700, Joshua Wiley wrote:
|> Hi Erin,
|>
|> I am not sure what a "seq.Date object" is. My first thought is that
|> you are talking about the date method for seq(), but there are
|> hundreds of packages I do not know. In any case, here is what I think
|> you want.
Try this:
tryCatch(log(rnorm(10)), warning = function(m)deparse(m$call[[2]]))
Where log(rnorm(10)) is your expr to evaluate.
On Fri, Mar 18, 2011 at 4:04 PM, steven mosher wrote:
> I've read the help and the archives on tryCatch but I'm still stuggling
> trying to understand how it
> works exac
On Fri, Mar 18, 2011 at 10:48:44AM -0700, Nicolas Gutierrez wrote:
> Hello,
>
> One more question.. I have the data.frame "pop":
>
> xloc yloc gonad indEneW Area
> 123 20 516.74 1 0.02 20.21 1
> 223 20 1143.20 1 0.02 20.21 1
> 323 20 250.00 1
tryCatch will only execute it's error function on errors. If you are
getting warnings, you may find it helpful to set:
options(warn = 2)
?options
tryCatch(warning("!"), error = function(x) print("Error!"))
--
Jonathan P. Daily
Technician - USGS Leetown Science
I've read the help and the archives on tryCatch but I'm still stuggling
trying to understand how it
works exactly and how I can use it to get the result I need.
I have a data.frame of urls which point to 11 .zip files. Basically I use
RCurl to get the list of
files from a ftp and then reduce that
Mighty thanks Peter. This does precisely what I wanted.
Thanks again.
On Mon, Mar 14, 2011 at 4:52 PM, Peter Langfelder <
peter.langfel...@gmail.com> wrote:
> On Mon, Mar 14, 2011 at 1:06 PM, Pavan G wrote:
> > Hello All,
> > I have a histogram with values above and below 0. I would like to colo
On Fri, Mar 18, 2011 at 10:48:44AM -0700, Nicolas Gutierrez wrote:
> Hello,
>
> One more question.. I have the data.frame "pop":
>
> xloc yloc gonad indEneW Area
> 123 20 516.74 1 0.02 20.21 1
> 223 20 1143.20 1 0.02 20.21 1
> 323 20 250.00 1
Try this:
subset(pop, (ave(Area, Area, FUN = length) == 1 | ave(Area, Area, FUN
= function(x)cumsum(prop.table(x))) < 0.7 & Area %in% 1:3))
On Fri, Mar 18, 2011 at 2:48 PM, Nicolas Gutierrez wrote:
> Hello,
>
> One more question.. I have the data.frame "pop":
>
> xloc yloc gonad ind En
Hello,
One more question.. I have the data.frame "pop":
xloc yloc gonad indEneW Area
123 20 516.74 1 0.02 20.21 1
223 20 1143.20 1 0.02 20.21 1
323 20 250.00 1 0.02 20.21 1
422 15 251.98 1 0.02 18.69 2
522 15 598.08
Run
R CMD javareconf -e
as the being suggest by the output. That will try to guess where is Java. Or
you can install the JDK_HOME and JAVA_HOME variables.
Good luck.
Caveman
On Fri, Mar 18, 2011 at 4:39 PM, jcheng liu wrote:
> Dear all,
> Installing rJava fails. The message was listed below.
On 11-03-18 08:42 AM, Federico Bonofiglio wrote:
> I Dears,
> if that wouldn't take u too much effort I'd like to ask for a swift opinion:
>
> I have a alinear (mixed effect) model that I wish to run as a piecewise one.
>
> When I predict the values Iget quite some odds and disturbing results:
>
Many thanks!
2011/3/18 Marc Schwartz
> On Mar 18, 2011, at 6:06 AM, Dunia Scheid wrote:
>
> > Hello,
> >
> > I am trying to compute the mdeian of the survival time from the function
> > survfit:
> >
> >> fit <- survfit(Surv(time, status) ~ 1)
> >> fit
> > Call: survfit(formula = Surv(time, statu
Hello,
I should see how to mirror a google group (if indeed possible).
There are few things stopping from having RHIPE work on the remote
while the client sends queries from a Windows client.
But as of now, it does not work.
Regards
Saptarshi
On Fri, Mar 18, 2011 at 7:50 AM, XiaoboGu wrote:
The easiest thing is to use 'save' so that you write the object out as
binary. If you don't need 'text', then save/load is the way to
operate with the data.
On Fri, Mar 18, 2011 at 10:53 AM, Ram H. Sharma wrote:
> Thanks, Jim for the idea.
>
> I tried with save as list. I can not write to a tabl
Well, you might start by reading the Help file for loess and
specifying the model via a formula, as described there.
The docs say your input is coerced to a formula -- but obviously not
the one you want. Specify it explicitly.
-- Bert
On Fri, Mar 18, 2011 at 8:34 AM, armstrwa wrote:
> Hi,
>
> I
brilliant, thanks guys! diverge_hsv is perfect.
Frank
On Fri, 2011-03-18 at 16:47 +0100, Achim Zeileis wrote:
> On Fri, 18 Mar 2011, Frank Schwach wrote:
>
> > Hi,
> >
> > I'm plotting a heatmap with values ranging from -10 to +10 and I would
> > like the negative values to show up in shades of
On Fri, 18 Mar 2011, Frank Schwach wrote:
Hi,
I'm plotting a heatmap with values ranging from -10 to +10 and I would
like the negative values to show up in shades of blue and the positive
ones in shadea of red. Basically, I want exactly what the RColorBrewer
palette RdBu does but with more of a
Some ideas for you...
RedWhiteBlue <- colorRampPalette(c("firebrick","white","#023868"))
RedGrayBlue <- colorRampPalette(c("firebrick", "lightgray", "#023868"))
RedWhiteBlue2 <- colorRampPalette(c("red","white","blue"))
RedGrayBlue2 <- colorRampPalette(c("red", "lightgray", "blue"))
pie(rep(1,21),
Hi,
I am trying to create a loess smooth from hydrologic data. My goal is to
create a smooth line that describes discharge at a certain point in time. I
have done this using the 'lowess' function and had no problem, but I'm
having some difficulty with loess. I am inputting the date ('date') and
Many thanks to all
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On Mar 18, 2011, at 10:53 AM, Ram H. Sharma wrote:
Thanks, Jim for the idea.
I tried with save as list. I can not write to a table with
"write.table", I
could not find a function that is write.list or equivalent. Even if
it is
list I think it would be difficult to post-processing than as
On Fri, Mar 18, 2011 at 3:37 PM, Frank Schwach wrote:
> Hi,
>
> I'm plotting a heatmap with values ranging from -10 to +10 and I would
> like the negative values to show up in shades of blue and the positive
> ones in shadea of red. Basically, I want exactly what the RColorBrewer
> palette RdBu do
Hi,
I'm plotting a heatmap with values ranging from -10 to +10 and I would
like the negative values to show up in shades of blue and the positive
ones in shadea of red. Basically, I want exactly what the RColorBrewer
palette RdBu does but with more of a gradual change (the RdBu can only
give me 11
Thanks, Jim for the idea.
I tried with save as list. I can not write to a table with "write.table", I
could not find a function that is write.list or equivalent. Even if it is
list I think it would be difficult to post-processing than as table.
outx<- as.list(apply(datafr1, 2, fout))
write.table
Dear all,
Installing rJava fails. The message was listed below. I wonder why cpp flag
could not be detected, although I had installed. Thanks!!!
-Jiacheng
R CMD INSTALL rJava_0.9-0.tar.gz
* installing to library
/data1/mri_researchers/wexler_data/jiacheng/R-2.12.0/library
* installing *source* p
Hi,
Since we can’t access Google Groups here in China, so please forgive me if
anyone has asked this question before.
Sincerely
Xiaobo Gu
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PLEASE do read the posting gui
Sorry again. I think if you click on the slow download is free.However I attach
the CSV in txt.
> CC: r-help@r-project.org
> From: dwinsem...@comcast.net
> To: tintin...@hotmail.com
> Subject: Re: [R] Spatial cluster analysis of continous outcome variable
> Date: Fri, 18 Mar 2011 09:48:46 -0400
I think it was suggested that you save your output to a 'list' and
then you will have it in a format that can accept variable numbers of
items in each element and it is also in a form that you can easily
process it to create whatever other output you might need.
On Fri, Mar 18, 2011 at 7:24 AM, Ra
On Mar 17, 2011, at 11:40 PM, Sebastián Daza wrote:
Hi everyone,
Is there any command to identify the pattern of responses of a
database with this format:
year id
20081
20091
20082
20092
20083
20093
20084
20094
20104
Two methods assuming this is in a
It isn't quite clear what you're doing. Do you
already have distances calculated elsewhere
and imported into R?
If so, then if you import the *full symmetric* distance
matrix, you can convert it into a distance object using
as.dist().
If you only have the lower triangle, you can use eg
full() fro
On Mar 17, 2011, at 10:45 PM, Jon Toledo wrote:
I read it but it said PDF file and Ps, didn´t specify which other
files, so I attached a csv file, which I thought would work.
The last time I read it, it said .pdf and .txt. I doubt that it has
changed since then. I admit that I originally t
That's a good solution, but if you're really, really sure that the timestamps
are in the format you gave, it's quite a bit faster to use substr and paste,
because you don't have to do any searching in the string.
HTH
Rex
> x = rep("09:30:00.000.633",100)
> system.time(y<-paste(substr(x,1,12)
Hello, everybody,
I hope somebody could help me with a dist() function.
I have a data frame of size 2*4087 (col*row), where col corresponds to the
treatment and rows are
species, values are Hellinger distances, I should reconstruct a distance
matrix
with a dist() function. I know that "euclidean"
Hi Dennis and R-users
Thank you for more help. I am pretty close, but challenge still remain is
forcing the output with different length to output dataframe.
> x <- data.frame(apply(datafr1, 2, fout))
Error in data.frame(var1 = c(-0.70777998321315, 0.418602152926712,
2.08356737154810, :
argume
@font-face {
font-family: "ＭＳ 明朝";
}@font-face {
font-family: "Verdana";
}@font-face {
font-family: "Verdana";
}@font-face {
font-family: "Monaco";
}p.MsoNormal, li.MsoNormal, div.MsoNormal { margin: 0cm 0cm 0.0001pt;
font-size: 10pt; font-family: Verdana; }.MsoChpDefault { f
I Dears,
if that wouldn't take u too much effort I'd like to ask for a swift opinion:
I have a alinear (mixed effect) model that I wish to run as a piecewise one.
When I predict the values Iget quite some odds and disturbing results:
The predicted stright line after the break point is not straigh
Thank you very much Duncan, I sorted it out. Your advice helped me to be
sure about the column range so I could track down my error.
For anyone who is interested: Inside my plot meta script I reference the arg
x column as M[[x]]. It turned out that the other data set used slightly
different column
Your lm args are wrong:
Try this:
predict(lm(Dens_min ~ Velocity, dfTestes3sitesCriptic))
or
predict(lm(Dens_min ~ Velocity_corrected, dfTestes3sitesCriptic))
On Fri, Mar 18, 2011 at 9:45 AM, barbara costa wrote:
> Hello,
> does anyone knows this predict is not resulting?
>
> # lm predict
>
Hello,
does anyone knows this predict is not resulting?
# lm predict
dfTestes3sitesCriptic$Velocity_corrected <- ifelse
(dfTestes3sitesCriptic$Season == "A" & dfTestes3sitesCriptic$BeforeAfter ==
"Before", (dfTestes3sitesCriptic$Velocity * mVel3ABefAfter), (ifelse
(dfTestes3sitesCriptic$Season ==
hello dear list!
as I am currently helping someone with their statistical analysis of a
count survey, I stumbled upon a very basic question upon model optimization:
when fitting a model like:
model<-lmer(abundance~(x+y+z)^3,family=poisson,...)
in which x,y,z are continuous abiotic parameters
On Mar 18, 2011, at 6:06 AM, Dunia Scheid wrote:
> Hello,
>
> I am trying to compute the mdeian of the survival time from the function
> survfit:
>
>> fit <- survfit(Surv(time, status) ~ 1)
>> fit
> Call: survfit(formula = Surv(time, status) ~ 1)
>
> records n.max n.start events median 0.95
Try this:
sub("\\.(\\d+)$", "\\1", ts)
On Thu, Mar 17, 2011 at 11:01 PM, rivercode wrote:
>
> Have timestamp in format HH:MM:SS.MMM.UUU and need to remove the last "." so
> it is in format HH:MM:SS.MMMUUU.
>
> What is the fastest way to do this, since it has to be repeated on millions
> of row
On 18/03/2011 6:37 AM, Michael Bach wrote:
Dear R users,
I want to do a fitted.contour plot of selected columns of a dataframe M with
M$AM and M$Irradiance as x and y axes respectively. The level of the contour
shall be determined by M$PR.
Some words on my data first. Dataframe M looks like:
h
Hello,
I am trying to compute the mdeian of the survival time from the function
survfit:
> fit <- survfit(Surv(time, status) ~ 1)
> fit
Call: survfit(formula = Surv(time, status) ~ 1)
records n.max n.start events median 0.95LCL 0.95UCL
111 111 111 20 NA NA NA
Hi Антон,
my approach would be to use the estimated variance of the slope
estimator. For your regression models m1 and m2 you can enter
vcov(m1)
vcov(m2)
to get the variance-covariance matrix of m1 and m2. Then, assuming that
the slope estimator is normal, you compute the p-value for slope_theo
Dear R users,
I want to do a fitted.contour plot of selected columns of a dataframe M with
M$AM and M$Irradiance as x and y axes respectively. The level of the contour
shall be determined by M$PR.
Some words on my data first. Dataframe M looks like:
head(M$Irradiance)
[1] 293 350 412 419 477 509
On Mar 18, 2011; 10:55am Thierry Onkelinx wrote:
>> Furthermore, I get an error when doing an anova between a lm() and a
>> lme() model.
Hi Thierry,
You get this error because you have not done the comparison the way I said
you should, by putting the lme$obj model first in the call to anova(). T
Apologies to all for the multiple posting. Don't know what caused it. Maybe
it _is_ time to stop using Nabble after all...
Regards, Mark.
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Sent from the R help mailing list archive
Hi folks,
I am having trouble to plot a mixed model analysis of covariance (ANCOVA).
To do so I use the function predict in nlme but the line that is being drawn
is totally out of control!!!
here is my script (where MASS_S is dry mass and MASS_F is fresh mass):
MEN<-read.table("Mentha_lme2.txt",
Thank you both, it works!
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Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project
> It would be nice to have a standard directory where R can
> write things this way. A semi-standard directory is given
> by Sys.getenv("R_LIBS_USER"), which defaults to ~/R/.../.
> Maybe ~/R/ could serve as that convention? That way we
> (various developers etc) would also n
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