[R] Evaluation of survival analysis

[Mike Marchywka]

________________________________
> Date: Tue, 30 Nov 2010 12:28:32 +0100
> Subject: [R] Evaluation of survival analysis
> From: hzshasha at googlemail.com
> To: marchywka at hotmail.com; r-help at r-project.org
>
>
>
> ---------- Forwarded message ----------
> From: He Zhang >
> Date: Tue, Nov 30, 2010 at 11:26 AM
> Subject: Re: [R] Evaluation of survival analysis
> To: Mike Marchywka >
> Cc: r-help at r-project.org
>
>
>
>
> On Tue, Nov 30, 2010 at 1:18 AM, Mike Marchywka
> > wrote:
>
>
> Hello Mike,
>
> Thank you very much for your reply and help.
> May i describe the analysis more clearly?
> My data is ecology data and my task is to 1) relate the 8 candidate
> (life history) varaibles with the lifespan of each subject and 2) use
> the known variables to predict lifespan.
> For the 1st task, i used Cox regression "coxph()" to do uni-variate
> analysis first. However, the most variables are correlated with each.
> For involving more variables, principle component analysis is applied.
> After PAC "principal()", I chose three vairalbes according to the
> results (instead of the derived principle components since the
> interpretation of the original variables is easier) .

Thanks, if I get a chance I'll take a look but I'm quite busy
now, just cleaning out my hotmail. I guess I would start
asking questions here however, if you have lots of
putatitive prognostic data and anything resemlbing a 
guess or equation on what response surfaces look like that
may be worth some thought. 
Coxph of course makes sense if you really expect to have
proportional hazards as opposed to some other effect. I think
the example I gave before was a literal survival analysis on
a 2 groups of patients with a well understood disease:
all the control samples event at t1 creating a step survival
curve. What kinds of things could a treatment do to this
that are well modlled with proportional hazards?  etc etc




> For the 2nd task, i wanted to use the chosen variables to predict the
> lifespan. "predict(survreg())" can get the values.
> I attached parts of the results which are the residuals plot and
> predcited values vs. predictors derived from both Cox regression and
> parametric survival.

Did you attach your raw data at any time? 


>
> My problem: 1) not sure if the methods are correct for the tasks since
> the residuals plots are not totally randomly and the predicted hazard
> is less than 0. 2) i dont know how to explain the fitness of the
> model.

In literal survival analysis , bringing people back from the dead is considered
a good thing :) The point of course your second question would often
be answered as, " the computer made a p-value, end of story" is often
the answer. Personally I consider goodness of fit as it relates
to a causal model: the fit is good if it helps you understand causality.



>
> Any suggestion about the methods or results will be really appreciate.
> Thank you again.
>
> Best wishes,
> He
>
>
>
>
>
> ----------------------------------------
> > Date: Mon, 29 Nov 2010 09:26:07 +0100
> > From: hzshasha at googlemail.com
> > To: r-help at r-project.org
> > Subject: [R] Evaluation of survival analysis
> >
> > Dear all,
> >
> > May I ask is there any functions in R to evaluate the fitness of
> "coxph" and
> > "survreg" in survival analysis, please?
> >
> > For example, the results from Cox regression and Parametric survival
> > analysis are shown below. Which method is prefered and how to see
> that / how
> > to compare the methods?
>
> I don't know if anyone answered but personally I like to look
> at pictures and relate to causality. Even the lecture slides I've
> seen ultimately suggest looking at scatter plots of various residuals
> for patterns. If known or suspected dynamics better fit with one
> model or the other that would likely be of interest.
> Generally if you pick enough parameters retrospectively you
> can probably get about what ever answer you want from a quantitative
> comparison.
>
>
> >
> > 1. coxph(formula = y ~ pspline(x1, df = 2))
> >
> > coef se(coef) se2 Chisq DF
> > p
> > pspline(x1, df = 2), line 0.0522 0.00867 0.00866 36.23 1.00 1.8e-
> > pspline(x1, df = 2), nonl 3.27 1.04
> > 7.5e-02
> >
> > Iterations: 4 outer, 13 Newton-Raphson
> > Theta= 0.91
> > Degrees of freedom for terms= 2
> > Likelihood ratio test=34.6 on 2.04 df, p=3.24e-08
> >
> > 2. survreg(formula = y ~ pspline(x1, df = 2))
> >
> > coef se(coef) se2 Chisq DF
> > p
> > (Intercept) 2.8199 0.15980 0.09933 311.37 1.0 0.0e+00
> > pspline(x1, df = 2), line -0.0193 0.00248 0.00248 60.35 1.0 8.0e-
> > pspline(x1, df = 2), nonl 1.43 1.1
> > 2.6e-01
> >
> > Scale= 0.304
> >
> > Iterations: 6 outer, 20 Newton-Raphson
> > Theta= 0.991
> > Degrees of freedom for terms= 0.4 2.1 1.0
> > Likelihood ratio test=48.2 on 1.5 df, p=1.18e-11
> >
> >
> > I really appreciate for your help. Thank you very much in advance.
> >
> > Best wishes,
> > He
>
>
>
>