[R] About 'hazard ratio', urgent~~~

[Mike Marchywka]

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> Date: Mon, 13 Jun 2011 19:44:15 -0700
> From: dr.jzhou at gmail.com
> To: r-help at r-project.org
> Subject: [R] About 'hazard ratio', urgent~~~
>
> Hi,
>
> I am new to R.
>
> My question is: how to get the 'hazard ratio' using the 'coxph' function in
> 'survival' package?

You can probably search the docs for hazard terms, for example, 

http://cran.r-project.org/doc/contrib/Fox-Companion/appendix-cox-regression.pdf


and try running known test data through to verify. For example,  
it does seem that the "exp" column contains a decent estimate of hazard
ratio in simple cases( not quite right 3 sig figs but haven't given this
a lot of thought and it is still early here, hoping from input from someone
who can explain better) ,

> ?rexp

> ns=100000
> df<-data.frame(gr<-c(rep(0,ns),rep(1,ns)),t=c(rexp(ns,1),rexp(ns,3)))
> coxph(Surv(t)~gr,df)
Call:
coxph(formula = Surv(t) ~ gr, data = df)

   coef exp(coef) se(coef)   z p
gr 1.09      2.98  0.00503 217 0
Likelihood ratio test=47382  on 1 df, p=0  n= 200000, number of events= 2e+05


> df<-data.frame(gr<-c(rep(0,100),rep(1,100)),t=c(rexp(100,1),rexp(100,2)))
> coxph(Surv(t)~gr,df)
Call:
coxph(formula = Surv(t) ~ gr, data = df)

    coef exp(coef) se(coef)    z       p
gr 0.658      1.93    0.148 4.44 8.8e-06
Likelihood ratio test=19.6  on 1 df, p=9.5e-06  n= 200, number of events= 200
> df<-data.frame(gr<-c(rep(0,100),rep(1,100)),t=c(rexp(100,1),rexp(100,1)))
> coxph(Surv(t)~gr,df)
Call:
coxph(formula = Surv(t) ~ gr, data = df)

      coef exp(coef) se(coef)      z    p
gr -0.0266     0.974    0.142 -0.187 0.85
Likelihood ratio test=0.03  on 1 df, p=0.852  n= 200, number of events= 200
>

>
> thanks,
>
> karena
>
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