Wouldn't the "response" of a survival model be survival times? You
may want to look at survfit and survest.
--
DW
On Jul 16, 2009, at 9:19 PM, Sean Brummel wrote:
> With type="response" in the predict funtion, I was expecting an
> expected survival time given covariates( in my dataset I have a few
> covariates but not in the example)... a natural predictor. The
> prediction function is clearly not returning a probability of
> survival at a given time since my example it is greater than one.
>
>
> Sean
>
>
> On 7/16/09, David Winsemius <dwinsem...@comcast.net> wrote:
>
> On Jul 16, 2009, at 8:19 PM, Sean Brummel wrote:
>
>> 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.
>
>
>
> ?predict.survreg
>
>
> You might get better answers if you specified even more concretely
> what you expected and more expansively why you think so. It sounds
> as though you expect predict to give you a median or mean, but this
> is not what R predict functions generally return. The predict
> function returns the estimated survival at the requested times If
> the newdata parameter is not supplied, then those times are taken to
> be those in the original dataset.
>
>
>
>>
>> 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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