[R] Extracting approximate Wald test (Chisq) from coxph(..frailty)
Paul Artes
paul_h_artes at yahoo.co.uk
Tue Apr 17 18:02:29 CEST 2007
Assign the output of coxph to some object, and use the $ extractor function
to obtain what you need.
ie:
rtfm <- coxph(formula = Surv(time, status) ~ age + sex + disease +
frailty(id, dist = "gauss"), data = kidney)
Age <- coef(rtfm)["age"]
OR
Sex <- rtfm$coef["sex"]
Hope this helps.
Paul
Mohammad Ehsanul Karim wrote:
>
> Dear List,
>
> How do I extract the approximate Wald test for the
> frailty (in the following example 17.89 value)?
>
> 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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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
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