Question

I am using the survival package for a population with two subgroups. I need the 5 year KM survival percentage for each group. I got this value from:

summary(survfit(Surv(survtime, vital)~subgroup)), time = 60) #in months

However, I also wanted to run a chisq test on the two five year survival rates. Is there an easy way to run this test using the survival package?

Thank you for any advice and help!

Answer 1

Not sure if this is the right way to go about it (buyer beware), but here is an example using the aml dataset included as example data with the survival package.

library(survival)
amlsurv <- with(aml,survfit(Surv(time,status) ~ x))
summary(amlsurv ,time=20)

Which results in, for time=20:

Call: survfit(formula = Surv(time, status) ~ x)

                x=Maintained 
        time       n.risk      n.event
      20.000        7.000        3.000 ...

                x=Nonmaintained 
        time       n.risk      n.event
      20.000        6.000        5.000 ...

Both of these options fail using the survdiff function:

survdiff(summary(aml,time=20))
survdiff(aml,time=20)

Now try limiting to 20 days manually:

aml$status.20 <- aml$status
# set 'deaths' with a time greater than 20 to be 'survivors'
aml$status.20[aml$status == 1 & aml$time > 20] <- 0
amlsurv <- (Surv(aml$time,aml$status.20) ~ aml$x)
summary(survfit(amlsurv))
# results are the same as above, Maintained=3 events,Nonmaintained=5 events

Now you can get your results for the 20 days limit comparing the two groups:

survdiff(amlsurv)
Call:
survdiff(formula = amlsurv)

                     N Observed Expected (O-E)^2/E (O-E)^2/V
aml$x=Maintained    11        3      4.3     0.392     0.874
aml$x=Nonmaintained 12        5      3.7     0.455     0.874

 Chisq= 0.9  on 1 degrees of freedom, p= 0.35