
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 
