Hi there,
I would like to ask something about how to avoid a possible error message
within a for loop. I am running a simulation and in some repetitions there may
be an error that will cause a crash and stop the whole procedure, what I want
is to simply move on to the next iteration
Hi there,
I perform a stratified cox and then I need the strata as a numeric array
straft.ln
ft.ln -
coxph(Surv(times,deaths)~ages+chemos+chemos:f1+chemos:f2+horms+horms:f1+horms:f2+grades+grades:f1+grades:f2+positives+positives:f1+positives:f2+sizes+sizes:f1+sizes:f2+strata(stra),data=ddd)
Hi there,
I have a rather large data set and perform the following cox model:
test1 - list(tstart,tstop,death1,chemo1,radio1,horm1)
out1-coxph( Surv(tstart,tstop, death1) ~
If I use the following
newdata=data.frame(chemo1=0,
horm1=0,
age1=mean(age1),
grade1=0,
positive1=1,
size1=mean(size1) )
then I get
Error in model.frame.default(Terms, newdata, ...) :
variable lengths differ (found for 'log(tstop + 1)')
In addition: Warning message:
'newdata' had 1 rows
Hi there,
Can anyone please help me because I am going to get crazy with the pbc data
set. I just want to apply simple cox regression in the data set. I am a
beginner in R but I don't think I am doing anything wrong.
I have the book of Fleming and Harrington 1990. I perform cox regression
Hi there,
I am beginner in R and I have some basic question. Suppose I run a common
procedure such as a t test or cox model like below:
out-coxph( Surv(tstart,tstop, death1) ~ x1+x1:log(tstop+1) ,
test1,method=c(breslow))
Which yields the following result:
Call:
coxph(formula =
Hi there,
I am pretty new to R. Actually I started using it yesterday. I have two
questions:
1. Suppose I have a-c(1:10) (or a-array(c(1:10),dim=c(1,10)))
and I want to end up with vector b=[0 0 0 0 0 1 1 1 1 1]. i.e. I want to
substitute alla elements that are 5 with 0 and 5 with 1.
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