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

2011-06-14 Thread Mike Marchywka











> Date: Tue, 14 Jun 2011 09:11:25 -0700
> From: dr.jz...@gmail.com
> To: r-help@r-project.org
> Subject: Re: [R] About 'hazard ratio', urgent~~~
>
> Thanks a lot for the great help~


well, if you are referring to my post, as I indicated I made a lot of blunders 
to get that out the
door and posted the first thing that made sense. Note also that by
making the one hazard rate "1" it wasn't clear that the column
I cited was divided by "1" etc. Caveat emptor. I usually
just answer questions like this when I "meant to look at that
anyway" and I have a chance of putting forth an answer which
I and others can verify. In any case, if you are using
some new code it always helps to run known test cases through
it, even if it is just new to you LOL.





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

2011-06-14 Thread karena
Thanks a lot for the great help~

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

2011-06-14 Thread Mike Marchywka













> Date: Mon, 13 Jun 2011 19:44:15 -0700
> From: dr.jz...@gmail.com
> To: r-help@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=10
> 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= 20, 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  1.930.148 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)  zp
gr -0.0266 0.9740.142 -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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>
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> and provide commented, minimal, self-contained, reproducible code.
>   
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[R] About 'hazard ratio', urgent~~~

2011-06-13 Thread karena
Hi, 

I am new to R.

My question is: how to get the 'hazard ratio' using the 'coxph' function in
'survival' package?

thanks,

karena

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