On Feb 11, 2010, at 5:58 AM, Philipp Rappold wrote:
Göran, thanks!
One more thing that I found: As soon as you have at least one NA in
the independent vars, the trick that you mentioned does not work
anymore. Example:
> testdata
start stop censor groupvar var1
1 0 1 0 1 0.1284928
2 1 2 0 1 0.4896125
3 2 3 0 1 0.7012899
4 3 4 0 1 NA
5 0 1 0 2 0.7964361
6 1 2 0 2 0.8466039
7 2 3 1 2 0.2234271
> aftreg(Surv(start, stop, censor)~var1, data=testdata, id=testdata
$groupvar)
Error in order(id, Y[, 1]) : Different length of arguments (* I
translated this from the German Output *)
Do you think there is a simple hack which excludes all subjects that
have at least on NA in their independent vars? If it was only one
dependent var it would probably be easy by just using subset, but I
have lots of different combinations of vars that I'd like to test ;)
I don't know if it's a "hack", but there are a set of functions that
perform such subsetting:
?na.omit
There is a parameter that would accomplish that goal inside aftreg.
You may want to check what your defaults are for na.action.
--
David.
Best
Philipp
PS: Conerning the benmark: For a large dataset (~ 1600 observations
on ~300 subjects) processing takes about 40 seconds (core 2 duo @
2.46 GHz, T9300). Interestingly, processing the testdata-set above
with only 7 observations on 2 subjects takes 2 minutes...
Göran Broström wrote:
Philipp Rappold wrote:
Dear all,
I have some trouble using the "id"-argument with aftreg
(accelerated failure time regression analysis from the eha library).
As far as I understand it, the id argument is used to group
individuals together if there are time-varying covariates and the
data is arranged in counting process style.
Unfortunately, i cannot figure out how to use the "id"-argument.
The most straight-forward way would be to simply state the
grouping variable, but it throws an error. I've included an
example below: the dataframe for regression is called "test", with
the grouping variable "person".
> test
start end censor person var1
1 0 1 0 1 0.5
2 1 2 0 1 0.4
3 2 3 0 1 0.6
4 3 4 1 1 -0.3
5 0 1 0 2 0.6
6 1 2 0 2 0.7
7 2 3 0 2 0.6
> fit <- aftreg(Surv(start, end, censor)~var1, data=test, id=person)
Error in order(id, Y[, 1]) : argument 1 is not a vector
You have caught the _function_ 'person' (package: utils) instead of
the variable 'person' in the data frame. That explains the odd
error message. If you change the variable name to, e.g., "ID",
you'll get the error message
Error in order(id, Y[, 1]) : object 'id' not found
which would hint you in the right direction. You need to specify
'id' by a full name, in your case 'test$person'. This is of course
a deficiency in the interface of aftreg. I will fix it asap.
So the temporary fix is 'id = test$person'.
Thanks for the report,
Göran
> fit <- aftreg(Surv(start, end, censor)~var1, data=test,
id=test["person"])
Error in `[.data.frame`(id, ord) : undefined columns selected
What would be the correct way to fit this example model?
Thanks + all the best
Philipp
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