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2025-01-12 00:52:51 +08:00
R version 3.4.1 (2017-06-30) -- "Single Candle"
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> options(na.action=na.exclude) # preserve missings
> options(contrasts=c('contr.treatment', 'contr.poly')) #ensure constrast type
> library(survival)
>
> #
> # Test out the strata capabilities
> #
> tol <- survreg.control()$rel.tolerance
> aeq <- function(x,y,...) all.equal(as.vector(x), as.vector(y), ...)
>
> # intercept only models
> fit1 <- survreg(Surv(time, status) ~ strata(sex), lung)
> fit2 <- survreg(Surv(time, status) ~ strata(sex) + sex, lung)
> fit3a<- survreg(Surv(time,status) ~1, lung, subset=(sex==1))
> fit3b<- survreg(Surv(time,status) ~1, lung, subset=(sex==2))
>
> fit1
Call:
survreg(formula = Surv(time, status) ~ strata(sex), data = lung)
Coefficients:
(Intercept)
6.062171
Scale:
sex=1 sex=2
0.8167551 0.6533036
Loglik(model)= -1152.5 Loglik(intercept only)= -1152.5
n= 228
> fit2
Call:
survreg(formula = Surv(time, status) ~ strata(sex) + sex, data = lung)
Coefficients:
(Intercept) sex
5.494409 0.380171
Scale:
sex=1 sex=2
0.8084294 0.6355816
Loglik(model)= -1147.1 Loglik(intercept only)= -1152.5
Chisq= 10.9 on 1 degrees of freedom, p= 0.000963
n= 228
> aeq(fit2$scale, c(fit3a$scale, fit3b$scale), tolerance=tol)
[1] TRUE
> aeq(fit2$loglik[2], (fit3a$loglik + fit3b$loglik)[2], tolerance=tol)
[1] TRUE
> aeq(fit2$coef[1] + 1:2*fit2$coef[2], c(fit3a$coef, fit3b$coef), tolerance=tol)
[1] TRUE
>
> #penalized models
> fit1 <- survreg(Surv(time, status) ~ pspline(age, theta=.92)+
+ strata(sex), lung)
> fit2 <- survreg(Surv(time, status) ~ pspline(age, theta=.92)+
+ strata(sex) + sex, lung)
> fit1
Call:
survreg(formula = Surv(time, status) ~ pspline(age, theta = 0.92) +
strata(sex), data = lung)
coef se(coef) se2 Chisq DF p
(Intercept) 6.9036 0.8469 0.5688 66.45 1.00 3.6e-16
pspline(age, theta = 0.92 -0.0124 0.0067 0.0067 3.45 1.00 6.3e-02
pspline(age, theta = 0.92 2.53 2.65 4.0e-01
Scale:
sex=1 sex=2
0.807 0.654
Iterations: 1 outer, 4 Newton-Raphson
Theta= 0.92
Degrees of freedom for terms= 0.5 3.6 2.0
Likelihood ratio test=6.54 on 3.1 df, p=0.09 n= 228
> fit2
Call:
survreg(formula = Surv(time, status) ~ pspline(age, theta = 0.92) +
strata(sex) + sex, data = lung)
coef se(coef) se2 Chisq DF p
(Intercept) 6.3729 0.84471 0.59118 56.92 1.00 4.5e-14
pspline(age, theta = 0.92 -0.0111 0.00666 0.00666 2.77 1.00 9.6e-02
pspline(age, theta = 0.92 2.46 2.68 4.2e-01
sex 0.3686 0.11711 0.11685 9.91 1.00 1.6e-03
Scale:
sex=1 sex=2
0.800 0.636
Iterations: 1 outer, 5 Newton-Raphson
Theta= 0.92
Degrees of freedom for terms= 0.5 3.7 1.0 2.0
Likelihood ratio test=16.8 on 4.2 df, p=0.002 n= 228
>
> age1 <- ifelse(lung$sex==1, lung$age, mean(lung$age))
> age2 <- ifelse(lung$sex==2, lung$age, mean(lung$age))
> fit3 <- survreg(Surv(time,status) ~ pspline(age1, theta=.92) +
+ pspline(age2, theta=.95) + sex + strata(sex), lung)
> fit3a<- survreg(Surv(time,status) ~pspline(age, theta=.92), lung,
+ subset=(sex==1))
> fit3b<- survreg(Surv(time,status) ~pspline(age, theta=.95), lung,
+ subset=(sex==2))
> fit3b<- survreg(Surv(time,status) ~pspline(age, theta=.95),
+ lung[lung$sex==2,], x=T)
> #
> # The above line is tricky, and it took me a long time to realize
> # it's necessity. The range of age1 = range(age) = 39-82. That for
> # age2 = range of females = 41-77. The basis functions for pspline are
> # based on age. If I used data=lung, subset=(sex==2) in fit3b (earlier
> # form of the test, the pspline function is called before the subset
> # occurs, and fit3b has a different basis for the second spline than
> # fit3 does; leading to failure of the all.equal tests below. A theta
> # of .95 on one basis is not exactly the same as a theta of .95 on the
> # other. Coefficients were within 1%, but not the same.
>
> aeq(fit3$scale, c(fit3a$scale, fit3b$scale))
[1] TRUE
> aeq(fit3$loglik[2], (fit3a$loglik + fit3b$loglik)[2])
[1] TRUE
> pred <- predict(fit3)
> aeq(pred[lung$sex==1] , predict(fit3a))
[1] TRUE
> aeq(pred[lung$sex==2], predict(fit3b))
[1] TRUE
>
>
>
>
>
> proc.time()
user system elapsed
2.032 0.131 2.443