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
-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 Mi
k]] <- update(if (k==1) Cox0 else Cox[[k-1]], form)
}
From: Frank S.
Sent: Friday, August 30, 2019 6:36:39 PM
To: Andrews, Chris
Cc: r-help@r-project.org
Subject: RE: [R] Efficient way to update a survival model
External Email - Use Caution
Chris, thank y
Enviado: viernes, 30 de agosto de 2019 15:08
> Para: Frank S. ; Vito Michele Rosario Muggeo
>
> Cc: r-help@r-project.org
> Asunto: RE: [R] Efficient way to update a survival model
>
> The updated formula needs to have a different term rather than cos(k * v)
> every time. Here
le Rosario Muggeo
Cc: r-help@r-project.org
Asunto: RE: [R] Efficient way to update a survival model
The updated formula needs to have a different term rather than cos(k * v) every
time. Here is one way to explicitly change the formula.
library("survival")
set.seed(1)
v <- runif(nrow
Sent: Friday, August 30, 2019 5:54 AM
To: Vito Michele Rosario Muggeo
Cc: r-help@r-project.org
Subject: Re: [R] Efficient way to update a survival model
Hi everyone,
Vito, perhaps my previous mail was not clear. It is true that I used a loop,
but the key point is that such a loop
cannot co
De: Vito Michele Rosario Muggeo
Enviado: jueves, 29 de agosto de 2019 8:54
Para: Frank S.
Cc: r-help@r-project.org
Asunto: Re: [R] Efficient way to update a survival model
dear Frank,
update() does not update actually.. It just builds a new call which is
evaluated. To speed up the procedure you co
Cc: r-help@r-project.org
Asunto: Re: [R] Efficient way to update a survival model
dear Frank,
update() does not update actually.. It just builds a new call which is
evaluated. To speed up the procedure you could try to supply starting
values via argument 'init'. The first values come from the pr
dear Frank,
update() does not update actually.. It just builds a new call which is
evaluated. To speed up the procedure you could try to supply starting
values via argument 'init'. The first values come from the previous
fit, and the last one referring to new coefficients is set to zero
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)
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