ents and
p-values for the 5 baseline variables, so I assumed
that it was due to the small number of levels (in fact, too few ). However,
when computing anova(model.rand, model.fix),
the output indicates a p-value < 0.001 in favour of the "model.rand". What's
happ
Chris, thank you so much for your answer!!
Best,
Frank S.
De: Andrews, Chris
Enviado: martes, 3 de septiembre de 2019 14:14
Para: Frank S.
Cc: r-help@r-project.org
Asunto: Re: [R] Efficient way to update a survival model
library("survival")
set
Charles, thank you for your suggestion!
Frank S.
De: Berry, Charles
Enviado: s�bado, 31 de agosto de 2019 19:21
Para: Frank S.
Cc: Andrews, Chris ; r-help@r-project.org
Asunto: Re: Efficient way to update a survival model
The i^th model is included in the Cox
that, when computing Cox[[1]], the term
Cox[[k -1]]
does not exist. Is it possible to perform some "trick" (e.g. re-indexing) in
order to achieve this?
Best,
Frank
De: Andrews, Chris
Enviado: viernes, 30 de agosto de 2019 15:08
Para: Frank S. ; Vito
date(Cox0, substitute(. ~ . + Z[, 1:k]), data = pbc)
attr(Cox[[k]]$coefficients, "names")[2:(k+1)] <- paste0("sin(", 1:k, "* v)")
}
Cox
Best,
Frank
De: Frank S.
Enviado: jueves, 29 de agosto de 2019 12:38
Para: Vito Michele Rosa
hat OK?
Additionally, in my original question I wondered about the possibility of
reducing the
10 lines of code to one general expression or some loop. Is it possible?
Best,
Frank
De: Vito Michele Rosario Muggeo
Enviado: jueves, 29 de agosto de 2019 8:54
Para: Frank S.
Hello everybody, I come with a question which I do not know how to conduct in
an efficient way. In order to
provide a toy example, consider the dataset "pbc" from the package "survival".
First, I fit the Cox model "Cox0":
library("survival")
set.seed(1)
v <- runif(nrow(pbc), min = 0, max = 2)
Co
do.call(
what = rbind,
args = lapply(
X = paste0("dt_sp_", 1:length(sp)),
FUN = get,
envir = environment()
)
)
________
De: Frank S.
Envi
Many thanks Ista and Bert for your nice solutions!
As Ista commented in a previous mail, the 0.87 value in my example is not
fixed, but for each subject
it depends on the difference "2007-01-01 - fini". However, both of your
solutions take into account this
fact.
lt;- 1 # 1st case
exposure[fini >= as.Date("2006-01-01") & fini <= as.Date("2006-06-30")]
<- difftime(as.Date("2007-01-01"), fini, units="days")/365.25 # 2nd case
exposure[fini >= as.Date("2006-07-01") & fini <= as.Dat
Dear list,
I have one folder named "scripts_JMbayes", wich contains 10 R scripts.
I can read them properly by doing:
> pathnames <- list.files(pattern="[.]R", path="Mydir/scripts_JMbayes",
> full.names = TRUE)
> sapply(pathnames, USE.NAMES = FALSE, FUN = source,)
However, R generates the fol
Cheers
> Petr
>
>
> > -----Original Message-
> > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Frank
> > S.
> > Sent: Thursday, January 14, 2016 12:48 PM
> > To: r-help@r-project.org
> > Subject: [R] Overlapping subject-specific
Dear R users,
First of all, excuse me if my doubt is very trivial, but so far I haven't been
able to solve it.
My question is this: I have a data frame which contains repeated measurements
on 4 subjects coded
as "id", and I want to plot, for each subject, not only the corresponding
"counts" va
Many thanks to David L Carlson, Ben Gunter and David Winsemius for your quick
and very elegant solutions!!
With your list answers I am learning sa lot of things that will help me in the
future to program.
Best,
Frank S.
> Subject: Re: [R] Random selection of a fixed number of values
LSE, breaks =
0:ceiling(max(data$value
and the size vector:
size <- c(10, 7, 5, 5, 3)
But I'm not able to get it by using sample function. Does anyone have some idea?
Thank you very much for any suggestions!!
Frank S.
Thank you for all your observations and comments!!
As you suggest, the option x <- x*c(rep(1,19), -1) is a more elegant and a fast
way!
Frank S.
> From: dwinsem...@comcast.net
> Date: Thu, 18 Jun 2015 21:33:57 -0700
> To: marc_schwa...@me.com
> CC: r-help@r-project.org
&g
he desired result. II
wonder wether there is a cool way to do so, that is, for example with apply or
sign function.
Thans in advanced for your help!
Frank S.
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Thank you very much Bill,
Your answer to my question is exactly what I was trying to do in my R code.
Best regards.
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rse. Does anybody can help me? Thanks a bunch!! Frank S.
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Hi,
I think I got it!
The clue: Function split!
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PLEASE do read the posting guide
Hi everybody,
I have (as an example) the following
two data tables:
all <-
data.table(ID = c(rep(c(100:105),c(3,2,2,3,3,3))),
value =
c(100,120,110,90,45,35,270,50,65,40,25,55,75,30,95,70))
DT <-
data.table(ID = 100:105, code=c(2,1,3,2,3,1))
My aim is to construct as many sub
Jim, Thanks for the comment about else!
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Dear Berend and Petr,
I do apologise for the disorderly code I posted. I have tried to solve it in a
new mail.
Frank S.
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Hi to all members of R list,
I�m working with data.table package, and with 6
variables:
ID: Subject identifier
born: Birthdate
start: Starting date
register: date of measurement
value: Value of measurement
end: date of expiration of the measurements.
So, the natural order of d
Hi to all members of R list,
I�m working with data.table package, and with 6
variables: "ID" (Identifier), "born" (Birthdate), "start" (Starting date),
"register" (date of measurement), "value"and "end" (date of expiration). So,
the natural order of dates would be: born
=< start =< register =
Thanks Richard!
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Hi to all members of the list,
I have a data frame with subjects who can get into a certain study from
2010-01-01 onwards. Small example:
DF <- data.frame(id=as.factor(1:3), born=as.Date(c("1939/10/28", "1946/02/23",
"1948/02/29")))
id born
1 1 1939-10-28
2 2 1946-02-23
3 3 1948
Many thanks for you help Uwe!
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Hi everyone, After trying to find the solution during days, I decided to write
in this help list in order to ask if anyone can help me.I would want to
construct an R function, with "initial", "final" and "specific" dates as 3
arguments (for example, becauseI'm not really sure it is the best wa
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