[R] Efficient way to update a survival model
Frank S.
|_j_rod @end|ng |rom hotm@||@com
Sat Aug 31 00:36:39 CEST 2019
Chris, thank you for your elegant solution!
Just one minor question:
I wonder how to include within the loop of your solution the 10 models, that is, writing
for (k in 1:10) so that you can get {Cox[[1]], ..., Cox[[10]]}. However, I'm aware that some
change has to be done due to the fact 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 <chrisaa using med.umich.edu>
Enviado: viernes, 30 de agosto de 2019 15:08
Para: Frank S. <f_j_rod using hotmail.com>; Vito Michele Rosario Muggeo <vito.muggeo using unipa.it>
Cc: r-help using r-project.org <r-help using 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(pbc), min = 0, max = 2)
Cox0 <- coxph(Surv(pbc$time,pbc$status == 2) ~ v, data = pbc)
Cox <- vector("list", 10)
Cox[[1]] <- update(Cox0, . ~ . + cos(1 * v))
for (k in 2:10) {
form <- as.formula(sprintf(". ~ . + cos(%d * v)", k))
Cox[[k]] <- update(Cox[[k-1]], form)
}
Cox
-----Original Message-----
From: Frank S. [mailto:f_j_rod using hotmail.com]
Sent: Friday, August 30, 2019 5:54 AM
To: Vito Michele Rosario Muggeo
Cc: r-help using 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 compute the desired result. For example, for k = 3 the following loop
Cox <- list()
Cox[[1]] <- coxph(Surv(time,status == 2) ~ v + cos(v), data = pbc)
for (k in 2:10) {
Cox[[k]] <- update(Cox[[k-1]], . ~ . + cos(k * v), data = pbc)
}
leads to a model Cox[[3]] which accounts for terms {v, cos(v), cos(3*v)}, but does not include the term cos(2*v).
I think that this could be one way to solve my question:
library("survival")
set.seed(1)
v <- runif(nrow(pbc), min = 0, max = 2)
Cox0 <- coxph(Surv(pbc$time,pbc$status == 2) ~ v, data = pbc)
k.max <- 9
Z <- outer(v, 1:k.max, function (x, y) {sin(x * y)}) # Matrix with the outer product of the two arrays
Cox <- list()
for (k in 1:k.max){
Cox[[k]] <-
update(Cox0, substitute(. ~ . + Z[, 1:k]), data = pbc)
attr(Cox[[k]]$coefficients, "names")[2:(k+1)] <- paste0("sin(", 1:k, "* v)")
}
Cox
Best,
Frank
________________________________
De: Frank S. <f_j_rod using hotmail.com>
Enviado: jueves, 29 de agosto de 2019 12:38
Para: Vito Michele Rosario Muggeo <vito.muggeo using unipa.it>
Cc: r-help using r-project.org <r-help using r-project.org>
Asunto: RE: [R] Efficient way to update a survival model
Hi Vito,
Thanks for your reply! Following your suggestion, I have tried:
Cox[[3]] <- update(Cox[[2]], . ~ . + cos(3 * v), init=c(coef(Cox[[1]]), 0, 0), data = pbc)
Cox[[3]] <- update(Cox[[2]], . ~ . + cos(3 * v), data = pbc)
and both expressions lead to the same result. Is that OK?
Additionally, in my original question I wondered about the possibility of reducing the
10 lines of code to one general expression or some loop. Is it possible?
Best,
Frank
________________________________
De: Vito Michele Rosario Muggeo <vito.muggeo using unipa.it>
Enviado: jueves, 29 de agosto de 2019 8:54
Para: Frank S. <f_j_rod using hotmail.com>
Cc: r-help using r-project.org <r-help using r-project.org>
Asunto: Re: [R] Efficient way to update a survival model
dear Frank,
update() does not update actually.. It just builds a new call which is
evaluated. To speed up the procedure you could try to supply starting
values via argument 'init'. The first values come from the previous
fit, and the last one referring to new coefficients is set to zero (or
any other appropriate value).
Something like (untested), for instance
update(Cox[[2]], . ~ . + cos(3 * v), init=c(coef(Cox[[1]]),0), data = pbc)
Hope this helps,
best,
vito
"Frank S." <f_j_rod using hotmail.com> ha scritto:
> Hello everybody, I come with a question which I do not know how to
> conduct in an efficient way. In order to
> provide a toy example, consider the dataset "pbc" from the package
> "survival". First, I fit the Cox model "Cox0":
>
> library("survival")
> set.seed(1)
> v <- runif(nrow(pbc), min = 0, max = 2)
> Cox0 <- coxph(Surv(pbc$time,pbc$status == 2) ~ v, data = pbc)
>
> Then, from the above model, I can fit recursively 10 additional models as:
>
> Cox <- list()
>
> Cox[[1]] <- update(Cox0, . ~ . + cos(1 * v), data = pbc)
> Cox[[2]] <- update(Cox[[1]], . ~ . + cos(2 * v), data = pbc)
> Cox[[3]] <- update(Cox[[2]], . ~ . + cos(3 * v), data = pbc)
> Cox[[4]] <- update(Cox[[3]], . ~ . + cos(4 * v), data = pbc)
> ...
> Cox[[10]] <- update(Cox[[9]], . ~ . + cos(10* v), data = pbc)
>
> Since in practice I have to repeat above step until Cox[[100]], say,
> do you know an efficient way to
> wrap this code chunk in a loop or similar?
>
> I had tried:
>
> set.seed(1)
> v <- runif(nrow(pbc), min = 0, max = 2)
> Cox0 <- coxph(Surv(pbc$time,pbc$status == 2) ~ v, data = pbc)
>
> Cox <- list()
> Cox[[1]] <- update(Cox0, . ~ . + cos(1 * v), data = pbc)
> for (k in 1:10) {
> Cox[[k + 1]] <- update(Cox[[k]], . ~ . + cos((k + 1) * v), data = pbc)
> }
>
> However, from Cox[[3]] onwards, the intermediate values of integer k
> are not included here (for
> instance, the model Cox[[10]] would only include the cosinus terms
> for cos(1*v) and cos(10*v)).
>
[[elided Hotmail spam]]
>
> Frank
>
> [[alternative HTML version deleted]]
>
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