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Frank S.

2019-Aug-28 19:51 UTC

[R] Efficient way to update a survival model

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)).

Thanks in advance for any help!

Frank

	[[alternative HTML version deleted]]

Vito Michele Rosario Muggeo

2019-Aug-29 06:54 UTC

head link

[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 at 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)).
>
> Thanks in advance for any help!
>
> Frank
>
> 	[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

Frank S.

2019-Aug-29 10:38 UTC

head link

[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 at unipa.it>
Enviado: jueves, 29 de agosto de 2019 8:54
Para: Frank S. <f_j_rod at hotmail.com>
Cc: r-help at r-project.org <r-help at 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 at 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]]
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.


	[[alternative HTML version deleted]]

R help - Aug 2019 - Efficient way to update a survival model

[R] Efficient way to update a survival model

[R] Efficient way to update a survival model

[R] Efficient way to update a survival model