### [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)
age0.00489 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
se2Chisq 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
diseasePKD0.6173  1.90  1.0
frailty(id, dist = gauss17.89 12.1
p
age   0.74000
sex   0.00023
diseaseGN 0.74000
diseaseAN 0.47000
diseasePKD0.17000
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

##

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.

```

### Re: [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

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)
age0.00489 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
se2Chisq 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
diseasePKD0.6173  1.90  1.0
frailty(id, dist = gauss17.89 12.1
p
age   0.74000
sex   0.00023
diseaseGN 0.74000
diseaseAN 0.47000
diseasePKD0.17000
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

##

wildscop at yahoo dot com
Institute of Statistical Research and Training
University of Dhaka

__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
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and provide commented, minimal, self-contained, reproducible code.

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```

### Re: [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)
age0.00489 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
se2Chisq 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
diseasePKD0.6173  1.90  1.0
frailty(id, dist = gauss17.89 12.1
p
age   0.74000
sex   0.00023
diseaseGN 0.74000
diseaseAN 0.47000
diseasePKD0.17000
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

##

wildscop at yahoo dot com
Institute of Statistical Research and Training
University of Dhaka

__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
and provide commented, minimal, self-contained, reproducible code.

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

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R-help@stat.math.ethz.ch mailing list
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and provide commented, minimal, self-contained, reproducible code.

```

### Re: [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]

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)
age0.00489 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
se2Chisq 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
diseasePKD0.6173  1.90  1.0
frailty(id, dist = gauss17.89 12.1
p
age   0.74000
sex   0.00023
diseaseGN 0.74000
diseaseAN 0.47000
diseasePKD0.17000
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

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