Be sure you reply or forward your message to me to the r-help listhost. I
might not have time to review it tonight, while others might.
Jeremy
On Mon, Jul 9, 2012 at 2:38 PM, JPF [via R] <
ml-node+s789695n4635888...@n4.nabble.com> wrote:
> Please, find an example with a data set here.
>
> The data sets are prepared to be read with R (dataR) and STATA (dataSTATA)
> directly. The only difference is that STATA treats blanks as NAs.
>
> *in R, by using aftreg:*
>
> dataR<-read.table("clipboard",h=T)
> > su<-aftreg(Surv(sta,sto,S) ~
> a+b+c+d+g+h+j+k+l+m+factor(pcont),dist="weibull",
> data=data.frame(dataR),id=ID)
> > summary(su)
>
> Call:
> aftreg(formula = Surv(sta, sto, S) ~ a + b + c + d + g + h +
> j + k + l + m + factor(pcont), data = data.frame(thern),
> dist = "weibull", id = ID)
>
> Covariate W.mean Coef Exp(Coef) se(Coef) Wald p
> a 39.416 0.001 1.001 0.002 0.598
> b 55.879 0.002 1.002 0.001 0.002
> c 49.554 -0.001 0.999 0.001 0.251
> d 51.266 0.000 1.000 0.000 0.758
> g 14.701 -0.006 0.994 0.002 0.011
> h 32.358 0.005 1.005 0.001 0.000
> j 51.768 -0.022 0.978 0.001 0.000
> k 53.832 -0.004 0.996 0.001 0.000
> l 18.851 0.016 1.017 0.001 0.000
> m 0.809 0.907 2.478 0.132 0.000
> factor(pcont)
> 1 0.007 0 1 (reference)
> 2 0.272 -0.051 0.950 0.158 0.745
> 3 0.201 0.653 1.922 0.155 0.000
> 4 0.253 0.181 1.198 0.156 0.246
> 5 0.267 0.691 1.996 0.162 0.000
>
> log(scale) 3.369 29.038 0.225 0.000
> log(shape) 1.925 6.858 0.055 0.000
>
> Events 190
> Total time at risk 3129
> Max. log. likelihood -298.47
> LR test statistic 681
> Degrees of freedom 14
> Overall p-value 0
>
>
>
> *in STATA, by using streg:*
>
> insheet using "dataSTATA.txt"
> (18 vars, 4862 obs)
>
> .
> . stset ntime, failure(s) id(id)
> id: id
> failure event: s != 0 & s < .
> obs. time interval: (ntime[_n-1], ntime]
> exit on or before: failure
> ------------------------------------------------------------------------------
>
> 4862 total obs.
> 0 exclusions
> ------------------------------------------------------------------------------
>
> 4862 obs. remaining, representing
> 351 subjects
> 283 failures in single failure-per-subject data
> 4862 total analysis time at risk, at risk from t = 0
> earliest observed entry t = 0
> last observed exit t = 30
>
> .
> . gen f1 = 0
> . replace f1 = 1 if pcont==1
> (520 real changes made)
>
> .
> . gen f2 = 0
> . replace f2= 1 if pcont==2
> (1267 real changes made)
>
>
> . gen f3 = 0
> . replace f3= 1 if pcont==3
> (771 real changes made)
>
>
> . gen f4 = 0
> . replace f4= 1 if pcont==4
> (960 real changes made)
>
>
> . gen f5 = 0
> . replace f5= 1 if pcont==5
> (1344 real changes made)
>
>
> . streg f1 f2 f3 f4 f5 a b c d g h j k l m, dist(weibull) time nolog
>
> failure _d: s
> analysis time _t: ntime
> id: id
> note: f5 dropped due to collinearity
>
> Weibull regression -- accelerated failure-time form
>
> No. of subjects = 228 Number of obs =
> 3129
> No. of failures = 190
> Time at risk = 3129
> LR chi2(14) =
> 226.94
> Log likelihood = -47.541886 Prob > chi2 =
> 0.0000
>
> ------------------------------------------------------------------------------
>
> _t | Coef. Std. Err. z P>|z| [95% Conf.
> Interval]
> -------------+----------------------------------------------------------------
>
> f1 | .7194758 .2271213 3.17 0.002 .2743262
> 1.164625
> f2 | .4011278 .0604573 6.63 0.000 .2826336
> .519622
> f3 | .0142088 .0676573 0.21 0.834 -.1183972
> .1468148
> f4 | .2984225 .0814102 3.67 0.000 .1388614
> .4579835
> a | .0030444 .0024399 1.25 0.212 -.0017377
> .0078265
> b | -.0008451 .0009987 -0.85 0.397 -.0028026
> .0011124
> c | .0015207 .0009827 1.55 0.122 -.0004055
> .0034468
> d | .0005143 .0007139 0.72 0.471 -.0008848
> .0019135
> g | .0024349 .0025024 0.97 0.331 -.0024698
> .0073395
> h | -.0024076 .0012125 -1.99 0.047 -.0047842
> -.0000311
> j | .013318 .0012565 10.60 0.000 .0108553
> .0157806
> k | .002104 .0010438 2.02 0.044 .0000581
> .0041498
> l | -.0017612 .0010071 -1.75 0.080 -.003735
> .0002127
> m | -.0694492 .090727 -0.77 0.444 -.2472708
> .1083724
> _cons | 1.812124 .2000892 9.06 0.000 1.419956
> 2.204291
> -------------+----------------------------------------------------------------
>
> /ln_p | 1.536971 .0756138 20.33 0.000 1.38877
> 1.685171
> -------------+----------------------------------------------------------------
>
> p | 4.650481 .3516405 4.009916
> 5.393373
> 1/p | .2150315 .0162593 .1854127
> .2493818
> ------------------------------------------------------------------------------
>
>
> I do an accelerated failure time model with both programs. The problem of
> "survreg" is that it cannot handle time-dependent covariates:
> https://stat.ethz.ch/pipermail/r-help/2010-July/247216.html
>
>
>
> Thanks,
>
>
>
> On Mon, Jul 9, 2012 at 10:19 AM, Terry Therneau <[hidden
> email]<http://user/SendEmail.jtp?type=node&node=4635888&i=0>>
> wrote:
>
> > No input data, no output listed...... my crytal ball is too cloudy to
> > answer this.
> >
> > Terry T
> >
> >
> > On 07/09/2012 09:17 AM, Javier Palacios Fenech wrote:
> >
> > Please.
> >
> > find an example here. With exactly the same data set, I run two hazard
> > models following the instructions for each function.
> >
> > aftreg(formula = Surv(sta, sto, S) ~ a + b + c + d + e + f + g
> > , factor(F), data = data.frame(SURV),
> > dist = "weibull", id = ID)
> >
> > streg f1 f2 f3 f4 f5 a b c d g f g, dist(weibull) time nolog
> > (note: F= f1, f2,f3,f4,f5)
> >
> > Results are different. Really different. With aftreg some estimates are
> > significant, and with STATA they are not. Many estimates do not even
> have
> > the same sign, therefore predicting contrary effects. Which model should
> I
> > trust?
> >
> > Best,
> >
> > J
> >
> >
> >
> >
> >
> > On Mon, Jul 9, 2012 at 9:59 AM, Terry Therneau <[hidden
> > email]<http://user/SendEmail.jtp?type=node&node=4635888&i=1>>
> wrote:
> >
> >> Without more information, we can only guess what you did, or what you
> are
> >> seeing on the page that is "different".
> >>
> >> I'll make a random guess though. There are about 5 ways to
> paramaterize
> >> the Weibull distribution. The standard packages that I know, however,
> tend
> >> to use the one found in the Kalbfleisch and Prentice book The
> Statistical
> >> Analysis of Failure time Data. This includes the survreg funciton in R
> and
> >> lifereg in SAS, and likely stata tthought I don't know that package.
> The
> >> aftreg function in the eha package uses something different.
> >>
> >> About 1/2 the weibull questions I see are due to a change in
> parameters.
> >>
> >> Terry T.
> >>
> >> ---- begin included message -----
> >>
> >>
> >>
> >>
> >> Dear Community,
> >>
> >> I have been using two types of survival programs to analyse a data set.
> >>
> >> The first one is an R function called aftreg. The second one an STATA
> >> function called streg.
> >>
> >> Both of them include the same analyisis with a weibull distribution.
> Yet,
> >> results are very different.
> >>
> >> Shouldn't the results be the same?
> >>
> >> Kind regards,
> >> J
> >>
> >>
> >
> >
> > --
> > *Javier Palacios Fenech*
> > PhD in Management
> > Research Area: Marketing
> > www.javierpalacios.com
> >
> >
> >
> >
>
> --
> *Javier Palacios Fenech*
> PhD in Management
> Research Area: Marketing
> www.javierpalacios.com
>
> ______________________________________________
> [hidden email] <http://user/SendEmail.jtp?type=node&node=4635888&i=2>mailing
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> PLEASE do read the posting guide
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>
> *dataR.txt* (447K) Download
> Attachment<http://r.789695.n4.nabble.com/attachment/4635888/0/dataR.txt>
> *dataSTATA.txt* (434K) Download
> Attachment<http://r.789695.n4.nabble.com/attachment/4635888/1/dataSTATA.txt>
>
>
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-----
Jeremy T. Hetzel
Boston University
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