R help - Apr 2010 - Kaplan-Meier survfit problem

When I try to the code from library(survival) of library(ISwR),

the following code 

survfit(Surv(days,status==1))

that could produce Kaplan-Meier estimates shows the following error

"Error in survfit(Surv(days, status == 1)) : 
  Survfit requires a formula or a coxph fit as the first argument"

How it can be done in R.2.10
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you need:

survfit(Surv(days, status == 1) ~ 1)


I hope it helps.

Best,
Dimitris


On 4/19/2010 4:44 AM, ericyujin99 wrote:
>
> When I try to the code from library(survival) of library(ISwR),
>
> the following code
>
> survfit(Surv(days,status==1))
>
> that could produce Kaplan-Meier estimates shows the following error
>
> "Error in survfit(Surv(days, status == 1)) :
>    Survfit requires a formula or a coxph fit as the first argument"
>
> How it can be done in R.2.10
-- 
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center

Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014

2011/12/11 Esteban Cervetto <estebancster@gmail.com>
> I am working with uncensored data.
>
> I have duration of workers compensation. Then I have for each the number
> of days thet it doesn't work.
> This sample is not censored at right because I query only work accidents
> with date of return of work (saned)
>
> That is because I have only one vector: the number of days that the worker
> doesn't work.
> Reading works that uses this library, noticed that it need's a vector
to
> mark the type of termination. That is because I did a formula  Y(x) = 1
>  as.numeric(T.**201110))~1
>
> Your words were inspired me to do this: I solvet it putting to T.201110
> this proper vector and a vector of ones.
>
    T.201110$censor <- apply(T.201110,1,function(row) 1)    ##is there a
best method to do that? it takes much time

    I believe that this vector is superfluous, because the result of the
formula is ever 1

> km1<-survfit(Surv(T.201110$dias,T.201110$censor)~1)
>
>
> 2011/12/10 David Winsemius <dwinsemius@comcast.net>
>
>>
>> On Dec 10, 2011, at 6:39 PM, capitantyler wrote:
>>
>>  done it, again, i have the next problem
>>>
>>> my traduction:
>>> "The object (list) cannot be corced as "double"
>>>
>>> Original:
>>> *km1 <- survfit(Surv(as.numeric(T.**201110))~1)*
>>> Error en Surv(as.numeric(T.201110)) :
>>>  el objeto (list) no puede ser coercionado a 'double'
>>>
>>
>> I do not read that language, but I am surprised to see a single vector
>> being used as an argument to Surv(). When I use Surv(. , .) it is with
two
>> vectors, an interval and a censor variable.
>>
>>
>>
>>> note that need it convert to numeric class, otherwise:
>>>
>>> *km1 <- survfit(Surv((T.201110))~1)*
>>> Error en Surv((T.201110)) : Time variable is not numeric
>>> --
>>> View this message in context: http://r.789695.n4.nabble.com/**
>>>
Kaplan-Meier-survfit-problem-**tp2015369p4181476.html<http://r.789695.n4.nabble.com/Kaplan-Meier-survfit-problem-tp2015369p4181476.html>
>>>
>>
>> Context? Yes. We do want context, but we don't want to go to no
steenking
>> Nabble.
>>
>>
>>  Sent from the R help mailing list archive at Nabble.com.
>>>
>>>  Arrrgh. The rhelp mailing list is NOT at Nabble. Nabbel si a
commercial
>> mirror of the real thing. And it isn't really an archive. either,
since
>> they start discarding posts after a year or two.
>>
>> --
>> David Winsemius, MD
>> West Hartford, CT
>>
>>
>
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