similar to: parametric proportional hazard regression

I am trying to use R to do loglogistic hazard estimation. My plan is to generate a loglogistic hazard sample data and then use survreg to estimate it. If everything is correct, survreg should return the parameters I have used to generate the sample data. I have written the following code to do a time invariant hazard estimation. The output of summary(modloglog) shows the factor loading of

Hi to all, I am working on design package using survival function. First using PSM and adopting a weibull specification for the baseline hazard , I have got the following results(since weibull has both PH and AFT propreties ,in addition I have used the PPHSm command): Value Std. Error z p (Intercept) 1.768 1.0007 1.77 7.73e-02 SIZE -0.707 0.0895 -7.90 2.80e-15

Dear All, I would like to fit some parametric survival models using left truncated, right censored data in R. However I am having problems finding a function to fit parametric survival models which can handle left truncated data. I have tested both the survreg function in package survival: fit1 <- survreg(Surv(start, stop, status) ~ X + Y + Z, data=data1) and the psm function in package

Greetings -- I would like to estimate a proportional hazard model with a weibull or lognormal baseline. I have looked at both the coxph() and survreg() functions and neither appear (to me ) to do it. Am I missing something in the docs or is there another terrific package out there that will do this. Many Thanks. Carl Mason

Hola a tod@s, Hace poco en la lista se discutieron algunas aproximaciones para determinar la distribución de probabilidad que potencialmente podrían haber generado los datos (ver [1]) y una de ellas fué el AIC. Haciendo uso del programa enviado por Pablo Verde (ver [1]), estos son los resultados para mis datos: weibull , AIC = 69839.44 exponential , AIC = 79488.77 gaussian , AIC = 69413.03

Dear R users, Please can you help me with a relatively straightforward problem that I am struggling with? I am simply trying to plot a baseline survivor and hazard function for a simple data set of lung cancer survival where `futime' is follow up time in months and status is 1=dead and 0=alive. Using the survival package: lung.wbs <- survreg( Surv(futime, status)~ 1, data=lung,

Hi, I'm estimating a loglogistic aft (accelerated failure time) model, just a simple plain vanilla one (without time dependent covariates), I'm comparing the results that I obtain between aftreg (eha package) and survreg(surv package). If I don't use any covariate the results are identical , if I add covariates all the coefficients are the same until a precision of 10^4 or 10^-5 except

Hello I would like to compare the results obtained with a classical non parametric proportionnal hazard model with a parametric proportionnal hazard model using a Weibull. How can we obtain the equivalence of the parameters using coxph(non parametric model) and survreg(parametric model) ? Thanks Virginie

Dear R Help group I need to run a command line script from within R session. I am not clear how i can acheive this. I tried shell and system function, but i am missing something critical.can someone provide help? My intention is to create a pdf file of a plot in R and then attach existing files from my system as attachment into the newly created pdf file. Any help would be greatly appreciated..

Dear all, What is the cumulative distribution (with parameterization) used within survreg with respect to the log-logistic distribution? That is, how are the parameters linked to the survivor function? Best regards, Mario [[alternative HTML version deleted]]

Hi I'm trying to predict the values for a survreg object called loglogistic_na. Here is the definition of loglogistic_na and following that the syntax used for the predict function. But upon execution I don't get any output. Not sure what I'm doing wrong: loglogistic_na <- survreg(Surv(time_na,event_na) ~ t_na, dist="loglogistic") summary(loglogistic_na)

I am trying to use the sas.get function to bring a SAS dataset into R. After calling the sas.get function I got an error message stating that 'sas' is not a recognized internal or external command, etc. I am guessing that I need to specify the internal command to invoke SAS in the sasprog option for the sas.get function, but I don't know how to determine what that command might be.

Hola, ¿qué tal? Tengo una pregunta de esta

I have a dataset at a hospital level (as opposed to the patient level) that contains number of patients experiencing events (call this number y), and the number of patients eligible for such events (call this number n). I am trying to model logit(y/n) = XBeta. In SAS this can be done in PROC LOGISTIC or GENMOD with a model statement such as: model y/n = <predictors>;. Can this be done

Hi all, Is there an equivalent to predict(...,type="linear") of a Proportional hazard model for a Cox model instead? For example, the Figure 13.12 in MASS (p384) is produced by: (aids.ps <- survreg(Surv(survtime + 0.9, status) ~ state + T.categ + pspline(age, df=6), data = Aidsp)) zz <- predict(aids.ps, data.frame(state = factor(rep("NSW", 83), levels =

Dear R help list, I am modeling some survival data with coxph and survreg (dist='weibull') using package survival. I have 2 problems: 1) I do not understand how to interpret the regression coefficients in the survreg output and it is not clear, for me, from ?survreg.objects how to. Here is an example of the codes that points out my problem: - data is stc1 - the factor is dichotomous

Hi experts, http://r.789695.n4.nabble.com/file/n4034318/Parametric_survival_analysis_2nd-order_efffect.JPG Parametric_survival_analysis_2nd-order_efffect.JPG As we know a normal survival regression is the equation (1) Well, I'ld like to modify it to be 2nd-order interaction model as shown in equation(2) Question: Assume a and z is two covariates. x = dummy variable (1 or 0) z = factors

Thanks for sharing the questions and responses! Is it possible to appreciate how much the coefficients matter in one or the other model? Say, using Biau's example, using coxph, as.factor(grade2 == "high")TRUE gives hazard ratio 1.27 (rounded). As clinician I can grasp this HR as 27% relative increase. I can relate with other published results. With survreg the Weibull model gives a

1. The weibull is the only distribution that can be written in both a proportional hazazrds for and an accelerated failure time form. Survreg uses the latter. In an ACF model, we model the time to failure. Positive coefficients are good (longer time to death). In a PH model, we model the death rate. Positive coefficients are bad (higher death rate). You are not the first to be confused

Hello, I'm quite new to R and want to make a Weibull-regression with the survival package. I know how to build my "Surv"-object and how to make a standard-weibull regression with "survreg". However, I want to fit a translated or 3-parametric weibull dist to account for a failure-free time. I think I would need a new object in survreg.distributions, but I don't know how