survival-analysis

问题 I'm rerunning code from 2 years ago and am encountering a new error with the survSplit function. The error says that my object cannot be found, even though it is a defined column in my dataframe. Here is an example of my dataframe: f12 <- data.frame(id = 1:6, next.ivl= c(22.348, 1.837, 2.051,1.782,1.692, 1.730), event = c(0,1,1,0,1,1), enter= rep(0,6), end=c(22.348, 1.837,2.051,1.782,1.629,1.730)) My previous code was the following: cutpoints.l <- c(10/12, 1.25, 1.75, 2.25, seq(3,11)) f12

问题 The integrated Brier score (IBS) has been suggested in a paper by Graf et al (1999) as a good measure for prediction accuracy in survival models (see e.g. overview paper by Wiering et al., page 23). It was implemented in the package ipred as function sbrier . However, whereas the brier score definition obviously applies to Cox proportional hazard models, I cannot get sbrier to return the Brier score for a coxph model. Here is the problem set up. library(survival) library(ipred) data("DLBCL",

问题 I'm experiencing the above error message when trying to run a survival analysis with Weibull distribution, in R. My data is a bit tricky in that it contains both left and right censored observations. I followed instructions found on https://stat.ethz.ch/R-manual/R-devel/library/survival/html/Surv.html: "Interval censored data can be represented in two ways. For the first use type = "interval" and the codes shown above. In that usage the value of the time2 argument is ignored unless event=3.

问题 I have a survfit object. A summary survfit for my t=0:50 years of interest is easy enough. summary(survfit, t=0:50) It gives the survival at each t. Is there a way to get the hazard for each t (in this case, the hazard from t-1 to t in each t=0:50)? I want to get the mean and confidence interval (or standard error) for the hazards relating to the Kaplan Meier curve. This seems easy to do when a distribution is fit (eg. type="hazard" in flexsurvreg ) but I can't figure out how to do this for a

问题 I have a Cox proportional hazards model set up using the following code in R that predicts mortality. Covariates A, B and C are added simply to avoid confounding (i.e. age, sex, race) but we are really interested in the predictor X. X is a continuous variable. cox.model <- coxph(Surv(time, dead) ~ A + B + C + X, data = df) Now, I'm having troubles plotting a Kaplan-Meier curve for this. I've been searching on how to create this figure but I haven't had much luck. I'm not sure if plotting a

问题 I am working with a spell dataset that has the following form: clear all input persid start end t_start t_end spell_type year spell_number event 1 8 9 44 45 1 1999 1 0 1 12 12 60 60 1 2000 1 0 1 1 1 61 61 1 2001 1 0 1 7 11 67 71 1 2001 2 0 1 1 4 85 88 2 2003 1 0 1 5 7 89 91 1 2003 2 1 1 8 11 92 95 2 2003 3 0 1 1 1 97 97 2 2004 1 0 1 1 3 121 123 1 2006 1 1 1 4 5 124 125 2 2006 2 0 1 6 9 126 129 1 2006 3 1 1 10 11 130 131 2 2006 4 0 1 12 12 132 132 1 2006 5 1 1 1 12 157 168 1 2009 1 0 1 1 12

问题 I'm relatively new to survival analysis and have been used some standard telco churn data example with a sample below called 'telco': telco <- read.csv(text = "State,Account_Length,Area_Code,Intl_Plan,Day_Mins,Day_Calls,Day_Charge,Eve_Mins,Eve_Calls,Eve_Charge,Night_Mins,Night_Calls,Night_Charge,Intl_Mins,Intl_Calls,Intl_Charge,CustServ_Calls,Churn IN,65,415,no,129.1,137,21.95,228.5,83,19.42,208.8,111,9.4,12.7,6,3.43,4,TRUE RI,74,415,no,187.7,127,31.91,163.4,148,13.89,196,94,8.82,9.1,5,2.46,0

问题 I am trying to understand the function indeptCoxph in the spBayesSurv package. This function fits a Bayesian proportional hazards model. I am getting a little stuck with understanding parts of the R code as well as the Cox model theory. I am working on the authors' example (below). They have first simulated survival time data and I am having trouble following their code for doing this. It seems to me that first they are simulating survival times from an exponential distribution with CDF F(t)

问题 I consider using the lifelines package to fit a Cox-Proportional-Hazards-Model. I read that lifelines uses a nonparametric approach to fit the baseline hazard, which results in different baseline_hazards for some time points (see code example below). For my application, I need an exponential distribution leading to a baseline hazard h0(t) = lambda which is constant across time. So my question is: is it (in the meantime) possible to run a Cox-Proportional-Hazards-Model with an exponential

问题 I'm performing survival analysis in R using the 'survival' package and coxph . My goal is to compare survival between individuals with different chronic diseases. My data are structured like this: id, time, event, disease, age.at.dx 1, 342, 0, A, 8247 2, 2684, 1, B, 3879 3, 7634, 1, A, 3847 where 'time' is the number of days from diagnosis to event, 'event' is 1 if the subject died, 0 if censored, 'disease' is a factor with 8 levels, and 'age.at.dx' is the age in days when the subject was