Dear All,
Are there R packages that can estimate survival model with long-term
survivors? This is sometimes known as "cure" model or "split-population"
model. Thanks.
Shige
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing l
Dear John & Brian,
Ok. Now I start to understand. So basically I cannot use the given SEs
for my purposes. They only make sense on the scale of log-counts.
However in a paper, the log-count scale is not very informative (the
reader want to see the effect on the scale you measure). If I underst
The figures don't obviously scream out `overfitting' to me, and the standard
errors don't look excessively wide, given the data. Unless there is a strong
reason for using `lo', you could also try the `gam' function in package
`mgcv': it attempts to estimate the appropriate degree of smoothing
a
Hi,
Thank you very much for this very rapid and helpful reply.
I'm giving the code a spin around the block. It looks to be working fine.
There is actually also something else on my lineplot.CI wish list. If I might
be so bold to ask:
When plotting multiple data traces with relative high standa
Hallo,
I am running R-2.6. on Windows. I have a code which uses
library(convert). Can anyone tell me which package I need to install to
run this code. Everytime I receive the error message library (convert)
not found.
Thanks, Corinna
__
R-help@r
On Mon, 18 Feb 2008, Gustaf Granath wrote:
> Dear John & Brian,
>
> Ok. Now I start to understand. So basically I cannot use the given SEs for
> my purposes. They only make sense on the scale of log-counts. However in a
> paper, the log-count scale is not very informative (the reader want to se
On Mon, 18 Feb 2008, Gabor Csardi wrote:
> .. (Counting to ten.)
>
> The package is called 'convert'. It seems that this package is not
> on CRAN, however. I think you should ask the person whose code you're
> running.
It's on Bioconductor. So select the BioC software repository on the
Pack
.. (Counting to ten.)
The package is called 'convert'. It seems that this package is not
on CRAN, however. I think you should ask the person whose code you're
running.
Gabor
On Mon, Feb 18, 2008 at 10:17:41AM +0100, Schmitt, Corinna wrote:
>
> Hallo,
>
> I am running R-2.6. on Windows. I
Update: it is a Bioconductor package, so you need to do:
source("http://bioconductor.org/biocLite.R";)
biocLite("convert")
Gabor
On Mon, Feb 18, 2008 at 10:17:41AM +0100, Schmitt, Corinna wrote:
>
> Hallo,
>
> I am running R-2.6. on Windows. I have a code which uses
> library(convert). Can any
Hi,
I am trying to adjust the position of the boxplots when using bwplot(). In
boxplot() there is the alternative to modify the position of the boxes by using
boxplot(y~x, at=...).
However, I don't seem to find a comparable option in bwplot(). I would like in
some instances to add more space
Dear R gurus,
I am trying to plot a legend in the margins of a figure. The basic idea is to
have two (or more) plots on the same figure, but then to have a common legend
at the bottom of the plot. The approach that I have taken is to setup the
figures, then do a "dummy" plot of the legend to de
Hi,
I have two small issues with my R code, no big deal but curiosity
really. Here is a sample code,
>
> x <- rnorm(1:10)
>
> foo <- function(a = 1, b = list(x = c(1:10), y = c(1:10))){
>
> for (ii in seq(along=b$y)){
>
> print(x[ii] + b$x[ii])
>
Hi
I have som problems with a barplot. It can be created easily in Excel but I
cant manage to get it right in R.
For example:
MyData <-
matrix(c(1,2,-1,-1,0,-2,-2,1,1,1,-2,3),ncol=3,dimnames=list(LETTERS[1:4],seq(3)))
I want the barplot to stack the positive and negative values separatly so
tha
[EMAIL PROTECTED] wrote:
> Dear R gurus,
>
> I am trying to plot a legend in the margins of a figure. The basic idea is to
> have two (or more) plots on the same figure, but then to have a common legend
> at the bottom of the plot. The approach that I have taken is to setup the
> figures, then d
i am trying to fit a survival regression model (cox model or parametric
model) in R by including the covariate effects as a function m(x) instead of
just beta*x. is it possible to fit such a model? can someone recommend some
reference? I searched but only found a package called addreg where
the haz
Hi,
When using the "Packages --> Install packages from local zip files" menu
item in the windows-gui:
1) is that supposed to automatically pull in dependencies (in that case I
have to fix something in my package).
2) If that's not the default: is there a way to make it so?
Thanks, Joh
__
Hello dear R users!
I did not find a function which gives information about the number of
digits of a value shown by R.
Do you know one?
I need it to solve the problem (see RFAQ 7.31)that 0.2==0.2+0.1-0.1 FALSE
The solution suggested in RFAQ is to use isTRUE(all.equal(0.2,0.2+0.1-0.1))
But if I
Hi Again,
I think I've solved my problem, but please tell me if you think I'm wrong,
or you can see a better way!
A plot of the integrand showed a very sharp peak, so I was running into the
integrand "feature" mentioned in the note. I resolved it by limiting the
range of integration as shown here
On Mon, 18 Feb 2008, Johannes Graumann wrote:
> Hi,
>
> When using the "Packages --> Install packages from local zip files" menu
> item in the windows-gui:
> 1) is that supposed to automatically pull in dependencies (in that case I
> have to fix something in my package).
No (and ?install.packages
Duncan Murdoch wrote:
>> 2) If that's not the default: is there a way to make it so?
>
> Not simply, but of course it's possible with some work. The problem is
> that with repos = NULL, R doesn't know where to look for dependencies.
> So you need to make two passes: First, install the package,
On 18/02/2008 7:14 AM, Johannes Graumann wrote:
> Hi,
>
> When using the "Packages --> Install packages from local zip files" menu
> item in the windows-gui:
> 1) is that supposed to automatically pull in dependencies (in that case I
> have to fix something in my package).
No, it doesn't do that.
help
- Original Message -
From: Vito Ricci
To: Alexandre Lerch Franco
Sent: Monday, February 18, 2008 8:59 AM
Subject: Re: r package
Please send a message to R-help mailing list.
Regards.
Vito Ricci
Se non ora, quando?
Se non qui, dove?
Se non tu, chi?
Personal Web Space: http://v
Dear all, is it possible create, automatically, a variable with all the
Monday and Thursday?
18/02/2008
21/02/2008
25/02/2008
28/02/2008
03/03/2008
.
.
for all months
Best regards
JL
_
Confira vídeos com notícias do NY Times, gols
> I need it to solve the problem (see RFAQ 7.31)that 0.2==0.2+0.1-0.1
FALSE
> The solution suggested in RFAQ is to use
isTRUE(all.equal(0.2,0.2+0.1-0.1))
>
> But if I want to compare inequality:
> 0.2<=0.2 +0.1-0.1 TRUE
> but 0.2<=0.2 +0.1-0.1 FALSE
> bad!
> but in this case all.equal does not w
> > What I mean is if R shows 2.3456 I want to obtain the info that
digits=4
>
> > even if in facts the value has more (internal) digits.
>
> Try:
> x = 1.23456789
> format(x, nsmall=20)
> # [1] "1.2345678899989009"
I've just re-read the question. I suspect what you really wanted was
som
The CRAN Taskviews (http://cran.r-project.org/web/packages/Views/) is
currently unavailable and returns 404 Object not found!
This is possibly due to them now being linked from the side frame and the
URL now being http://cran.r-project.org/web/views/
Unfortunately the webmaster link on the 404 p
On 18/02/2008 7:51 AM, Johannes Graumann wrote:
> Duncan Murdoch wrote:
>
>>> 2) If that's not the default: is there a way to make it so?
>> Not simply, but of course it's possible with some work. The problem is
>> that with repos = NULL, R doesn't know where to look for dependencies.
>> So you n
On Mon, 18 Feb 2008, Sarah J Thomas wrote:
> Hello all:
>
> I have a question regarding the fitted.values returned from the
> zeroinfl() function. The values seem to be nearly identical to those
> fitted.values returned by the ordinary glm(). Why is this, shouldn't
> they be more "zero-inflated"?
Tom.O wrote:
> Hi
>
> I have som problems with a barplot. It can be created easily in Excel but I
> cant manage to get it right in R.
>
> For example:
> MyData<-
> matrix(c(1,2,-1,-1,0,-2,-2,1,1,1,-2,3),ncol=3,dimnames=list(LETTERS[1:4],seq(3)))
>
> I want the barplot to stack the positive and nega
Ta. I will give that code a bash.
Could you explain why my code didn't work?
> Date: Mon, 18 Feb 2008 00:25:44 -0500
> From: [EMAIL PROTECTED]
> To: [EMAIL PROTECTED]
> Subject: Re: [R] How to make a vector/list/array of POSIXlt object?
> CC: r-help@r-project.org
>
> If the problem is that
On Mon, 2008-02-18 at 12:56 +, lamack lamack wrote:
>
> Dear all, is it possible create, automatically, a variable with all the
> Monday and Thursday?
>
> 18/02/2008
> 21/02/2008
> 25/02/2008
> 28/02/2008
> 03/03/2008
> .
> .
> for all months
Here is one way (for all months in 2008), if you
Neil Shephard wrote:
> The CRAN Taskviews (http://cran.r-project.org/web/packages/Views/) is
> currently unavailable and returns 404 Object not found!
>
> This is possibly due to them now being linked from the side frame and the
> URL now being http://cran.r-project.org/web/views/
>
> Unfortunately
Peter Dalgaard wrote:
> Neil Shephard wrote:
>
>> The CRAN Taskviews (http://cran.r-project.org/web/packages/Views/) is
>> currently unavailable and returns 404 Object not found!
>>
>> This is possibly due to them now being linked from the side frame and the
>> URL now being http://cran.r-projec
[EMAIL PROTECTED] a e'crit :
>>> What I mean is if R shows 2.3456 I want to obtain the info that
>>>
> digits=4
>
>>> even if in facts the value has more (internal) digits.
>>>
>> Try:
>> x = 1.23456789
>> format(x, nsmall=20)
>> # [1] "1.2345678899989009"
>>
>
> I've ju
Try this:
values <- seq(as.Date("2008/01/01"), as.Date("2008/12/31"), by="days")
values[format(values, "%w") %in% c(1,4)]
On 18/02/2008, lamack lamack <[EMAIL PROTECTED]> wrote:
>
>
> Dear all, is it possible create, automatically, a variable with all the
> Monday and Thursday?
>
> 18/02/2008
> 2
two options are the Delta method (see, e.g., function deltamethod() in
package 'msm'), and the Bootstrap method (check package 'boot').
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Ka
Your mapply creates a list so to do it that way convert result to a
POSIXct vector first. Using your strptime2 and dd, tt & dat from mine:
dt <- do.call(c, mapply(strptime2, dd, tt, SIMPLIFY = FALSE, USE.NAMES = FALSE))
data.frame(dt, dat)
# or just
data.frame(dt = strptime2(dd, tt), dat)
On
On Feb 18, 2008 2:02 PM, Peter Dalgaard <[EMAIL PROTECTED]> wrote:
>
> But what gave you the idea that
> http://cran.r-project.org/web/packages/Views/ should work? Google seems
> not to know it.
>
Its the target for the link to the TaskViews from
http://cran.r-project.org/web/packages/index.html (
Hi all!
I try to estimate a statistic of the form: (x1-x2)/(y1-y2), where
x1,x2,y1,y2 represent variable means, so each has an estimate and
standard error associated with it.
How is it possible to estimate the mean and the variance of this ratio?
Thank you!
[[alternative HTML v
Hello!
I would like to visualize the hospitalization within one year of several
patients using a bar chart. For each patient the stay in a hospital
should be illustrated with a dark colour a if there is a stay at home
between 2 hopital stays, it should be illustrated with a bright colour.
e.g.
P
If I undestand your question you can try something about like this:
x <- matrix(rbinom(75, 1,.6), nc=3)
image(x, col=c("gray", "white"), axes=F)
axis(1, at=seq(0,1, l=nrow(x)), labels=paste("Day", 1:nrow(x)))
axis(2, at=seq(0,1, l=ncol(x)), labels=paste("P", 1:3, sep=""))
On 18/02/2008, Stephan
I have a data frame with data similar to this:
NameA GrpA NameB GrpB Dist
A Alpha B Alpha 0.2
A Alpha C Beta0.2
A Alpha D Beta0.4
B Alpha C Beta0.2
B Alpha D Beta0.1
C Beta D Beta0.3
Dis
Matthieu Stigler a e'crit :
> [EMAIL PROTECTED] a e'crit :
What I mean is if R shows 2.3456 I want to obtain the info that
>> digits=4
even if in facts the value has more (internal) digits.
>>> Try:
>>> x = 1.23456789
>>> format(x, nsmall=20)
>>> # [1] "1.2345678899989009"
>>
>> I've
On Mon, 18 Feb 2008 14:15:54 + Neil Shephard wrote:
> On Feb 18, 2008 2:02 PM, Peter Dalgaard <[EMAIL PROTECTED]>
> wrote:
> >
> > But what gave you the idea that
> > http://cran.r-project.org/web/packages/Views/ should work? Google
> > seems not to know it.
> >
>
> Its the target for the link
Dear Irene,
if you have a vector of estimates of x1, x2, y1, y2 and the
corresponding estimated variance-covariance matrix (accessible through
the method "vcov"), then one possibility is to use the function
delta.method() in the package alr3 (on CRAN).
This function returns:
1) an estimate of
Dear all,
I am make plots using the function windrose() from the package "circular".
I would like to have the north on the top instead of having 0 degree on
the right side.
Does anybody know how to change the labels in a windrose plot?
thanks,
Huilin
___
Dear all,
how can I perform a repeated measures ANOVA using a covariance matrix as input?
E.g., I have four repeated measures (N = 200) with mean vector tau (no
BS factor):
tau <- rbind(1.10, 2.51, 2.76, 3.52)
and covariance matrix (sigma):
sigma <- matrix(c(0.523, 0.268, 0.267, 0.211,
Try this:
# read test data
Lines <- "NameA GrpA NameB GrpB Dist
A Alpha B Alpha 0.2
A Alpha C Beta0.2
A Alpha D Beta0.4
B Alpha C Beta0.2
B Alpha D Beta0.1
C Beta D Beta0.3
"
DF <- read.ta
Hello,
I'm trying to plot dayly evolution of the temperature over France from
Global Forecast System files ("I'm trying" is the right expression...).
akilonlat03 is the temperature for different latitudes and longitudes à 3
o'clock.
akilonlat06 is the temperature for different latitudes and long
Try this:
with(x,
{tmp <- table(x[Dist==0.2,c('GrpA', 'GrpB')])
tmp[lower.tri(tmp)] <- tmp[upper.tri(tmp)]
tmp})
On 18/02/2008, Karin Lagesen <[EMAIL PROTECTED]> wrote:
>
> I have a data frame with data similar to this:
>
> NameA GrpA NameB GrpB Dist
> A Alpha B Alpha
> I would like to plot akilonlat03 and then akilonlat06 and keep the map of
> France in background.
> My script (see below) doesn't work as image "paints" the background as I
> read somewhere in this forum.
I don't see how you can plot both temperatures on the same plot -
won't they be on top of o
Recently, I have been working with some data that look like two overlapping
gaussian distributions. I would like to either
1) determine the mean and SD for each of the two distributions
OR
2) get some (bayesian ?) statistic that estimates how likely an observation is
to belong to the left-ha
Marlin Keith Cox wrote:
> Thank you in advance for helping me with this.
> I included a part of the actual data. The result of pred.est and
> pred.est1are different, but they should be identical. For
> pred.est, I entered the slope and y intercept and received a value for each
> individual numb
Huilin Chen <[EMAIL PROTECTED]> wrote in
news:[EMAIL PROTECTED]:
> I am make plots using the function windrose() from the package
> "circular". I would like to have the north on the top instead of
> having 0 degree on the right side.
> Does anybody know how to change the labels in a windrose plo
Thank you in advance for helping me with this.
I included a part of the actual data. The result of pred.est and
pred.est1are different, but they should be identical. For
pred.est, I entered the slope and y intercept and received a value for each
individual number in the matrix (Z). For pred.est1
Hi all,
I am quite new to R, and I guess i made a mistake with
it...
Now, when i want to open the package manager i got
this :
"Erreur dans package.manager(is.loaded, pkgs,
pkgs.desc, pkgs.url) :
invalid arguments (length mismatch)
De plus : Warning message:
In .find.package(pkgs) : aucun pac
Hi All,
is there a way of predicting memory usage?
I need to build an array of 86000 by 2500 numbers (or I might create
a list of 2 by 2500 arrays 43000 long). How much memory should I
expect to use/need?
Cheers,
Fede
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
I
> Actually (after some trials) there is a little problem when faced with
> zeros...
>
> >getndp(1.0)
> [1] 0
Are you sure this isn't what you want? 1.0 is just 1 in disguise, and
round(1.0, 0) is the same as round(1.0, 1) anyway.
> Note that I thought on a very different way which was starti
This is a simple question. With functions written and compiled in R, the
body of the function can be seen by typing the function name without any
input arguments or by using the body function. How can I look at the body
of code written and compiled externally and called using the .Call function?
If this is numeric, then for just storing one copy, you will require
86000 * 2500 * 8 = 1.7GB of memory. You should have 3-4X that if you
want to analyze it, so you might need about 6GB of physical memory and
a 64-bit version of R. Is there some other alternative? Do you need
all the values at o
On Mon, 18 Feb 2008, gallon li wrote:
> i am trying to fit a survival regression model (cox model or parametric
> model) in R by including the covariate effects as a function m(x) instead of
> just beta*x. is it possible to fit such a model? can someone recommend some
> reference? I searched but o
Hi all,
I have some functions which sometimes take longer to run and it would
be useful to have a progress
bar showing how much time is left for the script to finish (something
like a download progress
bar). I tried searching but could not find it. Is this possible with R?
Note: Using R version
Try this:
plot(rnorm(10), main=expression(paste("Mus., 10 ",mu," g", sep="")))
On 18/02/2008, Samor Gandhi <[EMAIL PROTECTED]> wrote:
> Dear all,
>
> I am very thankful, if you could tell wheather it is possible to write
>
> paste("Mus., 10 ", expression(mu)," g", sep="")
>
> Thank you in a
Howdee,
My question appears at #6 below:
1. I want to model the growth of each of a large number of individuals using
a 4-parameter logistic growth curve.
2. nlme does not converge with the random structure that I want to use.
3. nlsList does not converge for some individuals.
4. I decided to
See RSiteSearch("Progress Bar")
On 18/02/2008, Syed Abid Hussaini <[EMAIL PROTECTED]> wrote:
> I have some functions which sometimes take longer to run and it would be
> useful to have a progress
> bar showing how much time is left for the script to finish (something like a
> download progress
>
I have some functions which sometimes take longer to run and it would be useful
to have a progress
bar showing how much time is left for the script to finish (something like a
download progress
bar). I tried searching but could not find it. Is this possible with R?
Note: Using R version: 2.6.1
"Michael Gormley" <[EMAIL PROTECTED]> wrote in
news:[EMAIL PROTECTED]:
> This is a simple question. With functions written and compiled in
> R, the body of the function can be seen by typing the function name
> without any input arguments or by using the body function. How can
> I look at the b
Dear all,
I am very thankful, if you could tell wheather it is possible to write
paste("Mus., 10 ", expression(mu)," g", sep="")
Thank you in advance,
Samor
-
[[alternative HTML version deleted]]
___
See warnOnly in ?nls.control
On Feb 18, 2008 1:45 PM, Marc Belisle <[EMAIL PROTECTED]> wrote:
> Howdee,
>
> My question appears at #6 below:
>
> 1. I want to model the growth of each of a large number of individuals using
> a 4-parameter logistic growth curve.
>
> 2. nlme does not converge with th
Hi Marc,
use try()
Cheers
Andrew
On Mon, Feb 18, 2008 at 01:45:35PM -0500, Marc Belisle wrote:
> Howdee,
>
> My question appears at #6 below:
>
> 1. I want to model the growth of each of a large number of individuals using
> a 4-parameter logistic growth curve.
>
> 2. nlme does not converge
apparently you want to check the "Introduction to R" document I
found it very useful when I started working with R:
http://cran.r-project.org/doc/manuals/R-intro.pdf
try:
names(df) <- NULL
b
ps: "df" is the name of the function to get the density for an F
distribution...
On Feb 18,
Try this:
names(df) <- NA
or
names(df) <- make.names(seq(ncol(df)))
On 18/02/2008, joseph <[EMAIL PROTECTED]> wrote:
>
>
> I want to remove the column names from a data frame. I do
> it the long way, can any body show me a better way ?
>
>
> df= data.frame(chrN= c("chr1", "chr2", "chr3"), star
I want to remove the column names from a data frame. I do
it the long way, can any body show me a better way ?
df= data.frame(chrN= c(chr1, chr2, chr3), start= c(1,
2, 3), end= c(4, 5, 6), score= c(7, 8, 9))
df
#I write a txt file without row or column names
write.table(df,"df1.txt"
Well, Henrique's solution is nicer than mine. :-)
b
On Feb 18, 2008, at 3:05 PM, joseph wrote:
Hi Benilton
I actually tried "names(df) <- NULL" before I asked for help. Please
see below how it looks like
> names(df) <- NULL
> df
structure(c("chr1", "chr2", "chr3"), class = "AsIs")
struc
Dear List,
Does anybody no how to compare mean survival times for two (more) groups in
R? What test statistics should I use?
Thank you very much!
Joe
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.
Hi Benilton
I actually tried "names(df) <- NULL" before I asked for help. Please see below
how it looks like
> names(df) <- NULL
> df
structure(c("chr1", "chr2", "chr3"), class = "AsIs") structure(c("1", "2",
"3"), class = "AsIs")
1 chr1
Is a finite mixture of 2 gaussians the name you are looking for?
This specific model will not deal with your N component however.
you can find some functions here:
http://cran.r-project.org/web/views/Cluster.html
Thomas
[EMAIL PROTECTED] wrote:
> Recently, I have been working with some data that
Hello,
I use Rbloomberg package to download bloomberg data. Other required packages
are RDCOMClient, zoo and chron. The problem I have is an error file
(RDCOM.err) that is created by starting the download process. The growth of
the file increases with each downloaded ticker and series by 2 Kb.
<[EMAIL PROTECTED]> wrote in
news:[EMAIL PROTECTED]:
>
> Recently, I have been working with some data that look like two
> overlapping gaussian distributions. I would like to either
>
> 1) determine the mean and SD for each of the two distributions
>
> OR
>
> 2) get some (bayesian ?) statis
I can't resist asking, why would you want to remove the variable
names? If it is for printing purposes, then you can probably work
around it on the printing side, depending on what you want to print.
I can't think of another reason for wanting to remove the column
names altogether.
Haris S
Hi Irene,
The result depends on the joint distribution of
x1,x2,y1,y2. The mean (or the variance) not always
exist (if, for example, all of them are independent
and normally distributed).
One possibility to estimate mean and variance is to
use simulation (once you verify that they exist!!!).
Rega
It looks to me like the index range starts at 1 in R.
Is this true?
If so, is there a way to change it to start at 0?
That way, I wouldn't have to make so many
changes when I translate a function from
another language.
--
View this message in context:
http://www.nabble.com/index-range-tp15550797
If you want to compute (157+221)! then sum up the log:
> a <- 1:(157+221)
> sum(log10(a))
[1] 811.8165
This is about 6.55e811 which exceeds the range of floating point
numbers (1.797693e+308). You might check out the Brobdingnag package.
On Feb 18, 2008 6:23 PM, Hyojin Lee <[EMAIL PROTECTED]> w
Hi,
I'm trying to calculate p-value to findout definitely expressed genes
compare A to B situation.
I got this data(this is a part of data) from whole organism , and each
number means each expression values (that means, we could think 'a' gene
is 13 in A situation, and it turns 30 in B situation)
On Feb 18, 2008 6:52 PM, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> You can define origin 0 objects yourself if you like.
> Here is a partial implementation:
>
> "[.orig0" <- function(x, i)
The 0 should be a 1.
>if (is.numeric(i)) .subset(x, i+0) else .subset(x, i)
> orig0 <- function(x)
You can define origin 0 objects yourself if you like.
Here is a partial implementation:
"[.orig0" <- function(x, i)
if (is.numeric(i)) .subset(x, i+0) else .subset(x, i)
orig0 <- function(x)
structure(x, class = c("orig0", setdiff(class(x), "orig0")))
x <- orig0(1:5)
x[0:3] # 1:4
Note tha
Xing Yuan gmail.com> writes:
>
> Dear List,
>
> Does anybody no how to compare mean survival times for two (more) groups in
> R? What test statistics should I use?
>
> Thank you very much!
>
> Joe
The question is probably a little bit too vague. You should
certainly look at the survival p
Or see the "Oarray" package (note that's
a letter O, not a zero!)
Ben Bolker
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provid
it could be a finite mixture.
take a look at flexmix package.
On Feb 18, 2008 11:03 AM, <[EMAIL PROTECTED]> wrote:
>
> Recently, I have been working with some data that look like two overlapping
> gaussian distributions. I would like to either
>
> 1) determine the mean and SD for each of the tw
"=?UTF-8?B?5a6L5pe25q2M?=" <[EMAIL PROTECTED]> wrote in
news:[EMAIL PROTECTED]:
> Dear All,
>
> Are there R packages that can estimate survival model with long-term
> survivors? This is sometimes known as "cure" model or
> "split-population" model. Thanks.
The usual Cox model would certainly al
If your numbers are not integers then they are double
precision which means 8 bytes per number. In such a
case you will need 86000*2500*8 = 1.7Gb of memory
(plus a small amount for book keeping). You will need
more (and sometimes much more) if you intend to do
some operations on your arrays). So yo
Maybe:
http://post.queensu.ca/~pengp/software.html
On Feb 18, 2008 3:08 AM, 宋时歌 <[EMAIL PROTECTED]> wrote:
> Dear All,
>
> Are there R packages that can estimate survival model with long-term
> survivors? This is sometimes known as "cure" model or "split-population"
> model. Thanks.
>
> Shige
>
_
Howdy,
I am trying to invert a matrix for the purposes of least squares. I
have tried a number of things, and the variety of results has me
confused.
1. When I try solve() I get the following:
>Error in solve.default(t(X) %*% X) : system is computationally
singular: reciprocal condition number = 3
Hi R help,
I run my data in nnet with skip layer, factor response (with 0 & 1
values) and explicitly put softmax=T to compare the result of the
default nnet with no softmax specification. I assume this should give
me the same result. I got the result the default one, but not the
softmax versi
Ben Domingue asks:
> I am trying to invert a matrix for the purposes of least squares. I
> have tried a number of things, and the variety of results has me
> confused.
Don't be.
> 1. When I try solve() I get the following:
> >Error in solve.default(t(X) %*% X) : system is computationally
> sing
If d is your dataframe, how about
ind <- d$Dist == 2
aggregate(d,by=list(d$NameA,dNameA),FUN=length)
Regards,
Moshe.
--- Karin Lagesen <[EMAIL PROTECTED]>
wrote:
>
> I have a data frame with data similar to this:
>
> NameA GrpA NameB GrpB Dist
> A Alpha B Alpha 0.2
> A
Hi list,
I'm wanting to use a matrix as input to some java code, but I seem to be
unable to do this (See code below). When searching online for a solution I
found that rJava 0.5-2 (the version under development not yet in CRAN) is
adding "direct support for raw vectors as method parameters". Is th
Thanks for your prompt reply Martin. That should work, but still involves
changing the java code.
I was just hoping there might be an obvious solution I was missing. Sorry
for the mailing list wrongness. I didn't realise there was an rJava mailing
list.
Thanks,
ben
On 19/02/2008, Martin Morgan <
any input?
(and also, why would all my posts get bounced off while I clearly
subscribed to this list?)
Best regards,
On 18 Feb 2008, at 09:53, baptiste Auguié wrote:
> Hi,
>
> I have two small issues with my R code, no big deal but curiosity
> really. Here is a sample code,
>
>
>>
>> x <- rn
Dear all,
I encountered the following problem while using the "R" software: "unrecognised
record type 7, subtype 16 encountered in system file".
Can you please help me with this?
Thanks in advance,
Catrin Theuser
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