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Andrews, Chris

2019-Aug-30 13:08 UTC

[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 at hotmail.com] 
Sent: Friday, August 30, 2019 5:54 AM
To: Vito Michele Rosario Muggeo
Cc: r-help at 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 at hotmail.com>
Enviado: jueves, 29 de agosto de 2019 12:38
Para: Vito Michele Rosario Muggeo <vito.muggeo at unipa.it>
Cc: r-help at r-project.org <r-help at 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 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]]


**********************************************************
Electronic Mail is not secure, may not be read every day, and should not be used
for urgent or sensitive issues

Frank S.

2019-Aug-30 22:36 UTC

head link

[R] Efficient way to update a survival model

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 at med.umich.edu>
Enviado: viernes, 30 de agosto de 2019 15:08
Para: Frank S. <f_j_rod at hotmail.com>; Vito Michele Rosario Muggeo
<vito.muggeo at unipa.it>
Cc: r-help at r-project.org <r-help at 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 at hotmail.com]
Sent: Friday, August 30, 2019 5:54 AM
To: Vito Michele Rosario Muggeo
Cc: r-help at 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 at hotmail.com>
Enviado: jueves, 29 de agosto de 2019 12:38
Para: Vito Michele Rosario Muggeo <vito.muggeo at unipa.it>
Cc: r-help at r-project.org <r-help at 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 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]]


**********************************************************
Electronic Mail is not secure, may not be read every day, and should not be used
for urgent or sensitive issues

	[[alternative HTML version deleted]]

Berry, Charles

2019-Aug-31 17:21 UTC

head link

[R] Efficient way to update a survival model

The i^th model is included in the Cox[[ i ]] object.

You can extract the formula objects with:

frms <- lapply(Cox, formula)

then if you want the existing and incremental terms:

indeps <- lapply(frms, function(x) as.list( x[[ 3 ]] ))

oldTerms <- lapply(indeps, "[[", 2)

newTerms <- lapply(indeps, "[[", 3)
> oldTerms[3:4][[1]]
v + cos(1 * v) + cos(2 * v)

[[2]]
v + cos(1 * v) + cos(2 * v) + cos(3 * v)
> newTerms[ 3:4 ][[1]]
cos(3 * v)

[[2]]
cos(4 * v)
> 
HTH,

Chuck
> On Aug 30, 2019, at 3:36 PM, Frank S. <f_j_rod at hotmail.com> wrote:
> 
> 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 at med.umich.edu>
> Enviado: viernes, 30 de agosto de 2019 15:08
> Para: Frank S. <f_j_rod at hotmail.com>; Vito Michele Rosario Muggeo
<vito.muggeo at unipa.it>
> Cc: r-help at r-project.org <r-help at 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 at hotmail.com]
> Sent: Friday, August 30, 2019 5:54 AM
> To: Vito Michele Rosario Muggeo
> Cc: r-help at 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
> 
> _____

Andrews, Chris

2019-Sep-03 12:14 UTC

head link

[R] Efficient way to update a survival model

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)
for (k in 1:10) {
        form <- as.formula(sprintf(". ~ . + cos(%d * v)", k))
        Cox[[k]] <- update(if (k==1) Cox0 else Cox[[k-1]], form)
}
________________________________________
From: Frank S. <f_j_rod at hotmail.com>
Sent: Friday, August 30, 2019 6:36:39 PM
To: Andrews, Chris
Cc: r-help at r-project.org
Subject: RE: [R] Efficient way to update a survival model

External Email - Use Caution
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 at med.umich.edu>
Enviado: viernes, 30 de agosto de 2019 15:08
Para: Frank S. <f_j_rod at hotmail.com>; Vito Michele Rosario Muggeo
<vito.muggeo at unipa.it>
Cc: r-help at r-project.org <r-help at 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 at hotmail.com]
Sent: Friday, August 30, 2019 5:54 AM
To: Vito Michele Rosario Muggeo
Cc: r-help at 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 at hotmail.com>
Enviado: jueves, 29 de agosto de 2019 12:38
Para: Vito Michele Rosario Muggeo <vito.muggeo at unipa.it>
Cc: r-help at r-project.org <r-help at 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 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)).
>
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> Frank
>
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>
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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

[R] Efficient way to update a survival model