Thanks for the help but...
I did the required transformations at the end of the code. The thing that I
dont understand is: Why is the predicted value (from the predict function)
not either the mean or median. Sorry I was not clear in my explanation.
Thanks,
Sean
On 7/16/09, David Winsemius <dwinsem...@comcast.net> wrote:
>
>
> On Jul 16, 2009, at 7:41 PM, Sean Brummel wrote:
>
> I am trying to generate predictions from a weibull survival curve but it
>> seems that the predictions assume that the shape(scale for
>> survfit) parameter is one(Exponential but with a strange rate estimate?).
>> Here is an examle of the problem, the smaller the shape is the worse the
>> discrepancy.
>>
>> ### Set Parameters
>> scale<-10
>> shape<-.85
>> ### Find Mean
>> scale*gamma(1 + 1/shape)
>>
>> ### Simulate Data and Fit Model
>> y<-rweibull(10000,scale=scale,shape=shape)
>> model<-survreg(Surv(y)~1,dist="weibull")
>>
>> ### Exp of coef and predict are the same
>> exp(model$coef)
>> predict(model,type=c("response"))[1]
>>
>> ### Here is the mean and median of the data
>> mean(y)
>> median(y)
>>
>> ### Fitted Mean and Median from survreg
>> fitScale<-exp(model$coef)
>> fitShape<-1/model$scale
>> fitScale*gamma(1 + 1/fitShape)
>> fitScale*(log(2))^(1/fitShape)
>>
>> Is this done on purpose? If so does anyone know why?
>>
>
> http://finzi.psych.upenn.edu/Rhelp08/2008-October/178487.html
>
> --
>
> David Winsemius, MD
> Heritage Laboratories
> West Hartford, CT
>
>
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