[R] Extracting approximate Wald test (Chisq) from coxph(..frailty)

[Charles C. Berry]

On Tue, 17 Apr 2007, Mohammad Ehsanul Karim wrote:

> Dear List,
>
> How do I extract the approximate Wald test for the
> frailty (in the following example 17.89 value)?


The example you give silently invokes print.coxph() to produce that 
output.

You _can_ use

 	tmp <- capture.output( print( <your example> ) )

and then further process tmp.

A _better_ solution for most purposes is to look at the object produced by 
coxph() and figure out how to calculate the Wald statistic from that 
object. See

 	?coxph.object

and
 	?str

Another tactic is to look at how print.coxph() does its work and use the 
code in it to produce just the output you desire. Look at

 	page( survival:::print.coxph, "print" )



>
> What about the P-values, other Chisq, DF, se(coef) and
> se2? How can they be extracted?
>
> ######################################################>
> kfitm1
> Call:
> coxph(formula = Surv(time, status) ~ age + sex +
> disease + frailty(id,
>    dist = "gauss"), data = kidney)
>
>                          coef     se(coef)
> age                        0.00489 0.0150
> sex                       -1.69703 0.4609
> diseaseGN                  0.17980 0.5447
> diseaseAN                  0.39283 0.5447
> diseasePKD                -1.13630 0.8250
> frailty(id, dist = "gauss
>                          se2    Chisq DF
> age                       0.0106  0.11  1.0
> sex                       0.3617 13.56  1.0
> diseaseGN                 0.3927  0.11  1.0
> diseaseAN                 0.3982  0.52  1.0
> diseasePKD                0.6173  1.90  1.0
> frailty(id, dist = "gauss        17.89 12.1
>                          p
> age                       0.74000
> sex                       0.00023
> diseaseGN                 0.74000
> diseaseAN                 0.47000
> diseasePKD                0.17000
> frailty(id, dist = "gauss 0.12000
>
> Iterations: 6 outer, 30 Newton-Raphson
>     Variance of random effect= 0.493
> Degrees of freedom for terms=  0.5  0.6  1.7 12.1
> Likelihood ratio test=47.5  on 14.9 df, p=2.82e-05  n=
> 76
>
> ######################################################
>
> Thank you for your time.
> Thanks in advance.
>
> Mohammad Ehsanul Karim
> wildscop at yahoo dot com
> Institute of Statistical Research and Training
> University of Dhaka
>
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> and provide commented, minimal, self-contained, reproducible code.
>

Charles C. Berry                        (858) 534-2098
                                          Dept of Family/Preventive Medicine
E mailto:cberry at tajo.ucsd.edu	         UC San Diego
http://biostat.ucsd.edu/~cberry/         La Jolla, San Diego 92093-0901