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print and coef Methods for survreg Differ

4 messages · biii m@iii@g oii de@@ey@ws, Jeff Newmiller

Original
biii m@iii@g oii de@@ey@ws
# 
Hello,

 

I'm working on a survreg model where the full data are subset for modeling
individual parts of the data separately.  When subsetting, the fit variable
("treatment" in the example below) has levels that are not in the data.  A
work-around for this is to drop the levels, but it seems inaccurate to have
the `coef()` method provide zero as the coefficient for the level without
data.

 

Why does coef(model) provide zero as the coefficient for treatment instead
of NA?  Is this a bug?

 

Thanks,

 

Bill

 

``` r

library(survival)

library(emmeans)

 

my_data <-

  data.frame(

    value=c(rep(1, 5), 6:10),

    treatment=factor(rep(c("A", "B"), each=5), levels=c("A", "B", "C"))

  )

my_data$cens <- c(0, 1)[(my_data$value == 1) + 1]

 

model <- survreg(Surv(time=value, event=cens)~treatment, data=my_data)

#> Warning in survreg.fit(X, Y, weights, offset, init = init, controlvals =

#> control, : Ran out of iterations and did not converge

coef(model)

#> (Intercept)  treatmentB  treatmentC 

#>  0.08588218  2.40341893  0.00000000

model$coef

#> (Intercept)  treatmentB  treatmentC 

#>  0.08588218  2.40341893          NA

model$coefficients

#> (Intercept)  treatmentB  treatmentC 

#>  0.08588218  2.40341893  0.00000000

print(model)

#> Call:

#> survreg(formula = Surv(time = value, event = cens) ~ treatment, 

#>     data = my_data)

#> 

#> Coefficients: (1 not defined because of singularities)

#> (Intercept)  treatmentB  treatmentC 

#>  0.08588218  2.40341893          NA 

#> 

#> Scale= 0.09832254 

#> 

#> Loglik(model)= 4.9   Loglik(intercept only)= -15

#>  Chisq= 39.92 on 2 degrees of freedom, p= 2.15e-09 

#> n= 10

summary(model)

#> 

#> Call:

#> survreg(formula = Surv(time = value, event = cens) ~ treatment, 

#>     data = my_data)

#>               Value Std. Error     z      p

#> (Intercept)  0.0859     0.0681  1.26   0.21

#> treatmentB   2.4034     0.2198 10.93 <2e-16

#> treatmentC   0.0000     0.0000    NA     NA

#> Log(scale)  -2.3195     0.0000  -Inf <2e-16

#> 

#> Scale= 0.0983 

#> 

#> Weibull distribution

#> Loglik(model)= 4.9   Loglik(intercept only)= -15

#>  Chisq= 39.92 on 2 degrees of freedom, p= 2.1e-09 

#> Number of Newton-Raphson Iterations: 30 

#> n= 10

ref_grid(model)

#> Error in ref_grid(model): Something went wrong:

#>  Non-conformable elements in reference grid.

 

my_data_correct_levels <- my_data

my_data_correct_levels$treatment <-
droplevels(my_data_correct_levels$treatment)

 

model_correct <- survreg(Surv(time=value, event=cens)~treatment,
data=my_data_correct_levels)

#> Warning in survreg.fit(X, Y, weights, offset, init = init, controlvals =

#> control, : Ran out of iterations and did not converge

coef(model_correct)

#> (Intercept)  treatmentB 

#>  0.08588218  2.40341893

print(model_correct)

#> Call:

#> survreg(formula = Surv(time = value, event = cens) ~ treatment, 

#>     data = my_data_correct_levels)

#> 

#> Coefficients:

#> (Intercept)  treatmentB 

#>  0.08588218  2.40341893 

#> 

#> Scale= 0.09832254 

#> 

#> Loglik(model)= 4.9   Loglik(intercept only)= -15

#>  Chisq= 39.92 on 1 degrees of freedom, p= 2.65e-10 

#> n= 10

summary(model_correct)

#> 

#> Call:

#> survreg(formula = Surv(time = value, event = cens) ~ treatment, 

#>     data = my_data_correct_levels)

#>               Value Std. Error     z      p

#> (Intercept)  0.0859     0.0681  1.26   0.21

#> treatmentB   2.4034     0.2198 10.93 <2e-16

#> Log(scale)  -2.3195     0.0000  -Inf <2e-16

#> 

#> Scale= 0.0983 

#> 

#> Weibull distribution

#> Loglik(model)= 4.9   Loglik(intercept only)= -15

#>  Chisq= 39.92 on 1 degrees of freedom, p= 2.6e-10 

#> Number of Newton-Raphson Iterations: 30 

#> n= 10

ref_grid(model_correct)

#> 'emmGrid' object with variables:

#>     treatment = A, B

#> Transformation: "log"

```

 

<sup>Created on 2021-02-23 by the [reprex
package](https://reprex.tidyverse.org) (v1.0.0)</sup>
Jeff Newmiller
# 
Model equations do not normally have conditional forms dependent on whether specific coefficients are NA or not. If you assign NA to a coefficient then you will not be able to predict outputs for input cases that you should be able to. Zero allows these expected cases to work... NA would prevent any useful prediction output.
On February 23, 2021 6:45:53 AM PST, bill at denney.ws wrote:

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      biii m@iii@g oii de@@ey@ws
    

    

      
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How should you be able to make a prediction (using this type of model) from a state where there is no data such as treatment="C" in my example?

-----Original Message-----
From: Jeff Newmiller <jdnewmil at dcn.davis.ca.us> 
Sent: Tuesday, February 23, 2021 11:10 AM
To: r-help at r-project.org; bill at denney.ws
Subject: Re: [R] print and coef Methods for survreg Differ

Model equations do not normally have conditional forms dependent on whether specific coefficients are NA or not. If you assign NA to a coefficient then you will not be able to predict outputs for input cases that you should be able to. Zero allows these expected cases to work... NA would prevent any useful prediction output.

        
On February 23, 2021 6:45:53 AM PST, bill at denney.ws wrote:

            

      
      
      
    

        

      
      
    

        
--
Sent from my phone. Please excuse my brevity.

  
  


  


      

    

    

      
      

        


  

        

      Jeff Newmiller
    

    

      
      #
    

  


  

      
You can't. But in order to be able to predict for states that _were_ in the training data, the coefficient cannot be NA.

        
On February 23, 2021 8:40:43 AM PST, bill at denney.ws wrote: