On Jul 17, 2009, at 12:52 AM, Sean Brummel wrote:
> This conversation is digressing... Here is my question, what does
> the value from the predict function mean? AS STATED IN MY CODE. Try
> running the example that I gave.
>
> predict(model,type=c("response"))[1]=?????????
Look at the help page and work through the examples:
Expected survival time, in the units being measured for time, of an
individual with the same covariates as the first subject. It is not a
probability. If you want a probability rather than a time, you need to
use type ="quantile" and then supply an appropriate set of percentiles.
>
> Sean
>
> On Thu, Jul 16, 2009 at 8:27 PM, David Winsemius <dwinsem...@comcast.net
> > wrote:
> 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
>
>
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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