Re: [R] Predict follow up time using parametric model in r
You are right. Specifically, I need to predict the mean and median time to failure from a coxph model and several parametric models using new data. Thanks. El lun., 5 nov. 2018 a las 7:11, Therneau, Terry M., Ph.D. (< thern...@mayo.edu>) escribió: > First, type='expected' gives the expected cumulative hazard for each > subject, given their follow-up time and covariates. It is not the expected > follow-up time, the expected time to death, or the probability of death. > I suspect you are not getting what you think you are. > > A survival model predicts a survival curve for each subject. For Cox > models you get the entire curve with the survfit() method, for survreg > models you get the curve with predict().To get a better answer about > how to "predict follow-up time" you will need to be more clear about what > you actually want, statistically. Mean time to failure? Median? RMST? > > > Terry T. > > > On 11/4/18 5:00 AM, r-help-requ...@r-project.org wrote: > > I am trying to predict follow-up time using several survival models, both > parametric and semi-parametric. I achieve it for semi parametric models > using predict.coxph function in R from survival package using type = > "expected" as indicated in help. However, for parametric models, this > option doesn't exist for the predict.survreg function. Is there any other > option? Maybe using rms package? > > > > [[alternative HTML version deleted]] __ R-help@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.
[R] Predict follow up time using parametric model in r
I am trying to predict follow-up time using several survival models, both parametric and semi-parametric. I achieve it for semi parametric models using predict.coxph function in R from survival package using type = "expected" as indicated in help. However, for parametric models, this option doesn't exist for the predict.survreg function. Is there any other option? Maybe using rms package? [[alternative HTML version deleted]] __ R-help@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.
Re: [R] Error survreg: Density function returned an an invalid matrix
Thanks Terry, I use the following formula for density:
[image: f_X(x)= \begin{cases} \frac{\alpha
x_\mathrm{m}^\alpha}{x^{\alpha+1}} & x \ge x_\mathrm{m}, \\ 0 & x <
x_\mathrm{m}. \end{cases}]
Where *x*m is the minimum value for x. I get this fórmula in
https://en.wikipedia.org/wiki/Pareto_distribution but there are a lot of
books and sites that use the same fórmula. This part of the code use that
formula:
distribution <- function(x, alpha) ifelse(x > min(x) ,
alpha*min(x)**alpha/(x**(alpha+1)), 0)
Also, I support my sintax in the following post:
http://stats.stackexchange.com/questions/78168/how-to-know-if-my-data-fits-pareto-distribution
Another option is transform my variable for time from pareto to exponential
(but this solution it's not very elegant):
If X is pareto distributed then
[image: Y = \log\left(\frac{X}{x_\mathrm{m}}\right)]
it's exponential distributed.
The syntax:
library(foreign)
library(survival)
library(VGAM)
set.seed(3)
X <- rpareto(n=100, scale = 5,shape = 1)
Y <- log(X/min(X))
hist(X,breaks=100)
hist(Y,breaks=100)
b <- rnorm(100,5,1)
c <- rep(1,100)
base <- cbind.data.frame(X,Y,b,c)
mod1<-survreg(Surv(Y+1, c) ~ b, base, dist = "exponential")# +1 it's
because time should be > 1
summary(mod1)
This solution works but I don´t like it.
Thanks.
2015-11-16 7:40 GMT-06:00 Therneau, Terry M., Ph.D. :
> The error message states that there is an invalid value for the density.
> A long stretch of code is not very helpful in understanding this. What we
> need are the definition of your density -- as it would be written in a
> textbook. This formula needs to give a valid response for the range
> -infinity to +infinity. Or more precisely, for any value that the
> maximizer might guess at some point during the iteration.
>
> Terry T.
>
>
> On 11/14/2015 05:00 AM, r-help-requ...@r-project.org wrote:
>
>> Thanks Terry but the error persists. See:
>>
>> >library(foreign)> library(survival)> library(VGAM) > mypareto <-
>>> list(name='Pareto',+ init=
>>>
>>
> remainder of message trucated
>
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[R] Fwd: Error survreg: Density function returned an an invalid matrix
I don´t know how to write a pareto distribution in that form, I want a
pareto function for time because I have a time variable that fits that
distribution. For a weibull and lognormal it is very easy because they are
particular cases from a extreme value and gaussian distributions. I think
it is possible to write a pareto using an exponential distribution,but I´m
not sure, I tried this using :
# OPTION 1.
# Define one distribution in terms of another
library(foreign)
library(survival)
library(VGAM)
my.pareto <- survreg.distributions$exponential
my.pareto$name <- "Pareto"
my.pareto$scale <- NULL
#Using the following transformation:
my.pareto$dtrans <- function(y) min(y)*exp(y)
survregDtest(my.pareto, TRUE)
set.seed(1)
a <- rpareto(100, 1, 6)
b <- rnorm(100,5,1)
c <- rep(1,100)
base <- cbind.data.frame(a,b,c)
mod1 <- survreg(Surv(a, c) ~ b, base, dist = my.pareto)
summary(mod1)
# It works but I don´t know if it's correct
# OPTION 2.
# Using Density, distribution function, quantile function and random
generation for the Pareto(I) distribution
# from VGAM package.
my.pareto3 <- list(name='Pareto',
init= function(x, weights,alpha,k){
alpha <- length(x)/(sum(log(x))-length(x)*log(min(x)))
k <-min(x)
c(media <-(alpha*k/(alpha-1)),varianza <-
((k/alpha)^2)*(alpha/(alpha-2)))},
density= function (x, alpha,k) {
alpha <- length(x)/(sum(log(x))-length(x)*log(min(x)))
k <-min(x)
pvec <- seq(0.1, 0.9, by = 0.1)
qvec <- qpareto(pvec, alpha, k)
cbind(ppareto(qvec, alpha, k),
1-ppareto(qvec, alpha, k),
dpareto(x, alpha, k),
-(alpha+x)/x,
(alpha+1)*(alpha+2)/x^2)},
deviance=function(x) {stop('deviance residuals not
defined')},
quantile= function(alpha,k) qpareto(seq(0.1, 0.9, by =
0.1), alpha, k))
survregDtest(my.pareto3, TRUE)
mod3 <- survreg(Surv(a, c) ~ b, base, dist = my.pareto3)
# Did not work and I don't want a fixed value for alpha and k paremeters
but the function needs a default.
# I got this error:
Error in logdensity[xok] <- log(shape[xok]) + shape[xok] * log(scale[xok])
- : NAs are not allowed in subscripted assignments
6
dpareto(x, alpha, k)
5
cbind(ppareto(qvec, alpha, k), 1 - ppareto(qvec, alpha, k), dpareto(x,
alpha, k), -(alpha + x)/x, (alpha + 1) * (alpha + 2)/x^2)
4
density(z, parms)
3
derfun(y, yy, exp(vars), sd$density, parms)
2
survreg.fit(X, Y, weights, offset, init = init, controlvals = control, dist
= dlist, scale = scale, nstrat = nstrata, strata, parms = parms)
1
survreg(Surv(a, c) ~ b, base, dist = my.pareto3)
So, I don't know what else can I do.
Thanks.
2015-11-16 11:38 GMT-06:00 Therneau, Terry M., Ph.D. <thern...@mayo.edu>:
> You are still missing the point.
> The survreg routine handles distribution of the form:
>
> (t(y) - m)/s ~ f, where f is a distribution on the real line.
>
> Here t is an optional but fixed transform and m= X\beta. Beta and s=scale
> are the parameters that the routine will fit.
>
> For a log-normal, t=log and f= the density of a Gaussian mean=0, sd=1.
> The distribution function is dnorm(x)
> For a Weibull,t=log and f= the density of the least extreme value
> distribution: exp(-exp(x))
>
> How do you write a Pareto in this form? I assume that you would like
> survreg to solve for some parameters -- how do you map them onto the beta
> and s values that survreg will attempt to optimize? I have not yet grasped
> what it is that you want survreg to DO.
>
> Terry T.
>
>
>
>
>
>
>
>
> On 11/16/2015 08:56 AM, Israel Ortiz wrote:
>
> Thanks Terry, I use the following formula for density:
> [image: f_X(x)= \begin{cases} \frac{\alpha
> x_\mathrm{m}^\alpha}{x^{\alpha+1}} & x \ge x_\mathrm{m}, \\ 0 & x <
> x_\mathrm{m}. \end{cases}]
>
> Where *x*m is the minimum value for x. I get this fórmula in
> https://en.wikipedia.org/wiki/Pareto_distribution but there are a lot of
> books and sites that use the same fórmula. This part of the code use that
> formula:
>
> distribution <- function(x, alpha) ifelse(x > min(x) ,
> alpha*min(x)**alpha/(x**(alpha+1)), 0)
>
> Also, I support my sintax in the following post:
>
>
> http://stats.stackexchange.com/questions/78168/how-to-know-if-my-data-fits-pareto-distribution
>
> Another option is transform my variable for time from pareto to
> exponential (but this solution it's not very elegant):
>
> If X is pareto distributed then
> [image: Y = \log\left(\frac{X}{x_\mathrm{m}}\right)]
>
> it's exponential distributed.
>
> The synta
Re: [R] Error survreg: Density function returned an an invalid matrix
Thanks Terry but the error persists. See:
> library(foreign)> library(survival)> library(VGAM) > mypareto <-
> list(name='Pareto',+ init= function(x, weights,parms){+
> alpha <- length(x)/(sum(log(x)))#this is a MLE for alpha+
> c(media <-(alpha*1/(alpha-1)),varianza <-
> ((1/alpha)^2)*(alpha/(alpha-2)))},+ density=
> function(x,weights) {+alpha <- length(x)/(sum(log(x)))+
> cdf1 <- function(x, alpha) ifelse(x > 1 , 1 - (1/x)**alpha,
> 0 )+cdf2 <- function(x, alpha) ifelse(x > 1, (1/x)**alpha
> ,0)+distribution <- function(x, alpha) ifelse(x > 1 ,
> alpha/(x**(alpha+1)), 0)+firstdev <- function(x, alpha)
> ifelse(x > 1, -(alpha+x)/x, 0)+seconddev <- function(x,
> alpha) ifelse(x > 1, (alpha+1)*(alpha+2)/x^2,0)+
> cbind(cdf1(x,alpha),cdf2(x, alpha),
> distribution(x,alpha),firstdev(x,alpha),seconddev(x,alpha))},+
> devian!
ce=function(x) {stop('deviance residuals not defined')},+
quantile= function(p, alpha) ifelse(p < 0 | p > 1, NaN, 1*(1-p)**(-1/alpha)))>
> survregDtest(mypareto, TRUE)[1] TRUE> set.seed(1)> a <- rpareto(100, 1, 1) >
b <- rnorm(100,5,1)> c <- rep(1,100)> base <- cbind.data.frame(a,b,c)>
mod1<-survreg(Surv(a, c) ~ b, base, dist = mypareto)Error in survreg.fit(X, Y,
weights, offset, init = init, controlvals = control, :
Density function returned an invalid matrix
2015-11-04 7:52 GMT-06:00 Therneau, Terry M., Ph.D. :
> Hi, I want to perform a survival analysis using survreg procedure from
>> survival library in R for a pareto distribution for a time variable, so I
>> set the new distribution using the following sintax:
>>
>> library(foreign)
>> library(survival)
>> library(VGAM)
>>
>> mypareto <- list(name='Pareto',
>> init= function(x, weights,parms){
>>
> etc.
>
> The survreg routine fits location-scale distributions such that (t(y) -
> Xb)/s ~ F, where t is an optional transformation, F is some fixed
> distribution and X is a matrix of covariates. For any distribution the
> questions to ask before trying to add the distribution to survreg are
> - can it be written in a location-scale form?
> - if so, how do the parameters of the distribution map to the location
> (Xb) and scale (s).
>
> In fitting data we normally have per-subject location (X b) but an
> intercept-only model is of course possible.
>
> If y is Weibull then log(y) fits into the framework, which is how survreg
> fits it. The transformation of parameters location and scale parameters
> for log(y) back to the usual Weibull parameterization for y often trips
> people up (see comments in the Examples section of ?survreg).
>
> The log of a Pareto can be written in this form (I think?). The two
> parameters are the scale a and lower limit b, with survival function of
> S(x)= (b/x)^a, for x >= b. If y = log(x) the survival function for y is
> S(y) = (b/exp(y))^a = exp[-(y - log(b))/(1/a)], which has location log(b)
> and scale 1/a. But even if I am correct the discontinuity at b will cause
> the underlying Newton-Raphson method to fail.
>
> Terry Therneau
>
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[R] Error survreg: Density function returned an invalid matrix
Hi, I want to perform a survival analysis using survreg procedure from
survival library in R for a pareto distribution for a time variable, so I
set the new distribution using the following sintax:
library(foreign)
library(survival)
library(VGAM)
mypareto <- list(name='Pareto',
init= function(x, weights,parms){
alpha <-
length(x)/(sum(log(x))-length(x)*log(min(x)))#this is a MLE for alpha
c(media <-(alpha*min(x)/(alpha-1)),varianza <-
((min(x)/alpha)^2)*(alpha/(alpha-2)))},
density= function(x,weights) {
alpha <- length(x)/(sum(log(x))-length(x)*log(min(x)))
cdf1 <- function(x, alpha) ifelse(x > min(x) , 1 -
(min(x)/x)**alpha, 0 )
cdf2 <- function(x, alpha) ifelse(x > min(x),
(min(x)/x)**alpha ,0)
distribution <- function(x, alpha) ifelse(x >
min(x) , alpha*min(x)**alpha/(x**(alpha+1)), 0)
firstdev <- function(x, alpha) ifelse(x > min(x),
-(alpha+x)/x, 0)
seconddev <- function(x, alpha) ifelse(x > min(x),
(alpha+1)*(alpha+2)/x^2,0)
cbind(cdf1(x,alpha),cdf2(x, alpha),
distribution(x,alpha),firstdev(x,alpha),seconddev(x,alpha))},
deviance=function(x) {stop('deviance residuals not
defined')},
quantile= function(p, alpha) ifelse(p < 0 | p > 1, NaN,
min(x)*(1-p)**(-1/alpha)))
I tested new distribution using survregDtest and it was successful:
survregDtest(mypareto, TRUE)
#TRUE
But I get the following error when I use it:
set.seed(1)
a <- rpareto(100, 1, 6)
b <- rnorm(100,5,1)
c <- rep(1,100)
base <- cbind.data.frame(a,b,c)
mod1<-survreg(Surv(a, c) ~ b, base, dist = mypareto)
Error in survreg.fit(X, Y, weights, offset, init = init,
controlvals = control, : Density function returned an invalid matrix
Why this happened even when the test was successful? and how can I solve
that?
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