R version 4.4.0 RC (2024-04-16 r86458 ucrt) -- "Puppy Cup"
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> library(survival)
> aeq <- function(x,y) all.equal(as.vector(x), as.vector(y))
> # One more test on coxph survival curves, to test out the individual
> #  option.  First fit a model with a time dependent covariate
> #
> test2 <- data.frame(start=c(1, 2, 5, 2, 1, 7, 3, 4, 8, 8),
+                     stop =c(2, 3, 6, 7, 8, 9, 9, 9,14,17),
+                     event=c(1, 1, 1, 1, 1, 1, 1, 0, 0, 0),
+                     x    =c(1, 0, 0, 1, 0, 1, 1, 1, 0, 0) )
> 
> # True hazard function, from the validation document
> lambda <- function(beta, x=0, method='efron') {
+     r <- exp(beta)
+     lambda <- c(1/(r+1), 1/(r+2), 1/(3*r +2), 1/(3*r+1),
+                 1/(3*r+1), 1/(3*r+2) + 1/(2*r +2))
+     if (method == 'breslow') lambda[9] <- 2/(3*r +2)
+     list(time=c(2,3,6,7,8,9), lambda=lambda)
+     }
> 
> fit <- coxph(Surv(start, stop, event) ~x, test2)
> # A curve for someone who never changes
> surv1 <-survfit(fit, newdata=list(x=0), censor=FALSE)
> 
> true <- lambda(fit$coef, 0)
> 
> aeq(true$time, surv1$time)
[1] TRUE
> aeq(-log(surv1$surv), cumsum(true$lambda))
[1] TRUE
> 
> # Reprise it with a time dependent subject who doesn't change
> data2 <- data.frame(start=c(0, 4, 9, 11), stop=c(4, 9, 11, 17),
+                       event=c(0,0,0,0), x=c(0,0,0,0), patn=c(1,1,1,1))
> surv2 <- survfit(fit, newdata=data2, id=patn, censor=FALSE)
> aeq(surv2$surv, surv1$surv)
[1] TRUE
> 
> 
> #
> # Now a more complex data set with multiple strata
> #
> test3 <- data.frame(start=c(1, 2, 5, 2, 1, 7, 3, 4, 8, 8,
+                             0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0),
+                     stop =c(2, 3, 6, 7, 8, 9, 9, 9,14,17,
+                             1:11),
+                     event=c(1, 1, 1, 1, 1, 1, 1, 0, 0, 0,
+                             0, 1, 1, 0, 0, 1, 1, 0, 1, 0,1),
+                     x    =c(1, 0, 0, 1, 0, 1, 1, 1, 0, 0,
+                             1, 2, 3, 2, 1, 1, 1, 0, 2, 1,0),
+                     grp = c(rep('a', 10), rep('b', 11)))
> 
> fit2 <- coxph(Surv(start, stop, event) ~ x + strata(grp), test3)
> 
> # The above tests show the program works for a simple case, use it to
> #  get a true baseline for strata 2
> fit2b <- coxph(Surv(start, stop, event) ~x, test3,
+                subset=(grp=='b'), init=fit2$coef, iter=0)
> temp <- survfit(fit2b,  newdata=list(x=0), censor=F)
> true2 <- list(time=temp$time, lambda=diff(c(0, -log(temp$surv))))
> true1 <- lambda(fit2$coef, x=0)
> 
> # Separate strata, one value
> surv3 <- survfit(fit2, list(x=0), censor=FALSE)
> aeq(true1$time, (surv3[1])$time)
[1] TRUE
> aeq(-log(surv3[1]$surv), cumsum(true1$lambda))
[1] TRUE
> 
> data4 <- data.frame(start=c(0, 4, 9, 11), stop=c(4, 9, 11, 17),
+                       event=c(0,0,0,0), x=c(0,0,0,0), grp=rep('a', 4),
+                     patid= rep("Jones", 4))
> surv4a <- survfit(fit2, newdata=data4, id=patid, censor=FALSE)
> aeq(-log(surv4a$surv), cumsum(true1$lambda))
[1] TRUE
> 
> data4$grp <- rep('b',4)
> surv4b <- survfit(fit2, newdata=data4, id=patid, censor=FALSE)
> aeq(-log(surv4b$surv), cumsum(true2$lambda))
[1] TRUE
> 
> 
> # Now for something more complex
> # Subject 1 skips day 4.  Since there were no events that day the survival
> #  will be the same, but the times will be different.
> # Subject 2 spends some time in strata 1, some in strata 2, with
> #   moving covariates
> #
> data5 <- data.frame(start=c(0,5,9,11,
+                              0, 4, 3),
+                     stop =c(4,9,11,17, 4,8,7),
+                     event=rep(0,7),
+                     x=c(1,1,1,1, 0,1,2),
+                     grp=c('a', 'a', 'a', 'a', 'a', 'a', 'b'),
+                     subject=c(1,1,1,1, 2,2,2))
> surv5 <- survfit(fit2, newdata=data5, censor=FALSE, id=subject)
> 
> aeq(surv5[1]$time, c(2,3,5,6,7,8))  #surv1 has 2, 3, 6, 7, 8, 9
[1] TRUE
> aeq(surv5[1]$surv, surv3[1]$surv ^ exp(fit2$coef))
[1] TRUE
> 
> tlam <- c(true1$lambda[1:2]* exp(fit2$coef * data5$x[5]),
+           true1$lambda[3:5]* exp(fit2$coef * data5$x[6]),      
+           true2$lambda[3:4]* exp(fit2$coef * data5$x[7]))
> aeq(-log(surv5[2]$surv), cumsum(tlam))
[1] TRUE
> 
>                           
>                           
> 
> proc.time()
   user  system elapsed 
   1.25    0.07    1.32