Mohammad Ehsanul Karim
2007-Apr-17 07:18 UTC
[R] Extracting approximate Wald test (Chisq) from coxph(..frailty)
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 n76
######################################################
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
Paul Artes
2007-Apr-17 16:02 UTC
[R] Extracting approximate Wald test (Chisq) from coxph(..frailty)
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 > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > >-- View this message in context: http://www.nabble.com/Extracting-approximate-Wald-test-%28Chisq%29-from-coxph%28..frailty%29-tf3589257.html#a10038426 Sent from the R help mailing list archive at Nabble.com.
Charles C. Berry
2007-Apr-17 16:43 UTC
[R] Extracting approximate Wald test (Chisq) from coxph(..frailty)
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 > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > 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
Mohammad Ehsanul Karim
2007-Apr-17 21:55 UTC
[R] Extracting approximate Wald test (Chisq) from coxph(..frailty)
Dear list, I need to extract the approximate Wald test (Chisq) so that I can put it in a loop. str seemed like a great idea, but I cannot seem to find the approximate Wald test for frailty (in the example data below: 17.89 and its p-value 0.12000) there. I cannot seem to find it in capture.output either as numeric form. Do I need to modify some given values? If yes, please give me a clue for the example: library(survival) kfitm1<-coxph(formula = Surv(time, status) ~ age + sex +disease + frailty(id, dist = "gauss"), data = kidney) str(kfitm1) capture.output( print(kfitm1) ) Mohammad Ehsanul Karim (R - 2.3.1 on windows) wildscop at yahoo dot com Institute of Statistical Research and Training University of Dhaka ________________________________ On Tue, 17 Apr 2007, Mohammad Ehsanul Karim wrote: 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" ) 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"] 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-05n> 76