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
>>
>>
>
> David Winsemius, MD
> Heritage Laboratories
> West Hartford, CT
>
>
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