Hi there,
I have a question on how to extract the linear term in the penalized
spline in coxph. Here is a sample code:
n=100
set.seed(1)
x=runif(100)
f1 = cos(2*pi*x)
hazard = exp(f1)
T = 0
for (i in 1:100) {
T[i] = rexp(1,hazard[i])
}
C = runif(n)*4
cen = T<=C
y = T*(cen) + C*(1-cen)
data.tr=cbind(y,cen,x)
fit=coxph(Surv(data.tr[,1], data.tr[,2])~pspline(data.tr[,3]) )
If I use summary(fit), it will show the following results:
summary(fit)
Call:
coxph(formula = Surv(data.tr[, 1], data.tr[, 2]) ~ pspline(data.tr[,
3]))
n= 100
coef se(coef) se2 Chisq DF p
pspline(data.tr[, 3]), li 0.495 0.437 0.437 1.28 1.00 2.6e-01
pspline(data.tr[, 3]), no 43.79 3.08 1.9e-09
exp(coef) exp(-coef) lower .95 upper .95
ps(data.tr[, 3])2 0.4404 2.270 0.08164 2.376
ps(data.tr[, 3])3 0.2065 4.842 0.01701 2.507
ps(data.tr[, 3])4 0.0951 10.512 0.00695 1.302
ps(data.tr[, 3])5 0.0493 20.274 0.00387 0.628
ps(data.tr[, 3])6 0.0280 35.741 0.00230 0.340
ps(data.tr[, 3])7 0.0192 52.068 0.00156 0.237
ps(data.tr[, 3])8 0.0219 45.605 0.00178 0.271
ps(data.tr[, 3])9 0.0473 21.156 0.00399 0.561
ps(data.tr[, 3])10 0.1432 6.983 0.01250 1.640
ps(data.tr[, 3])11 0.3936 2.541 0.03436 4.509
ps(data.tr[, 3])12 0.9449 1.058 0.06885 12.969
ps(data.tr[, 3])13 2.2406 0.446 0.07643 65.683
Iterations: 3 outer, 9 Newton-Raphson
Theta= 0.697
Degrees of freedom for terms= 4.1
Rsquare= 0.385 (max possible= 0.994 )
Likelihood ratio test= 48.6 on 4.08 df, p=7.74e-10
Wald test = 45.1 on 4.08 df, p=4.29e-09
My question is how to extract the linear coefficient (0.495) in
pspline(data.tr[, 3]). I tried coef(fit) but it fails to display this
term. Your help is greatly appreciated!
Lei Liu
Associate Professor
Division of Biostatistics and Epidemiology
Department of Public Health Sciences
University of Virginia School of Medicine
http://people.virginia.edu/~ll9f/
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