[R-sig-ME] error in lmer: length(f1) == length(f2) is not TRUE

[joana martelo]

I've changed the code of my factor levels to trial: A,B,C...... and fish.id:
A1,A2....B1,B2....., and used data=.... and it worked!

Thank you so much Ben!
Joana

-----Mensagem original-----
De: r-sig-mixed-models-bounces at r-project.org
[mailto:r-sig-mixed-models-bounces at r-project.org] Em nome de Ben Bolker
Enviada: quarta-feira, 17 de Outubro de 2012 19:03
Para: r-sig-mixed-models at r-project.org
Assunto: Re: [R-sig-ME] error in lmer: length(f1) == length(f2) is not TRUE

joana martelo <jmmartelo at ...> writes:

> 
> Thanks for your answers, but it still doesn't work. 
> Trial and fish.id are factors, but I get the same error:
> 
> > Error: length(f1) == length(f2) is not TRUE In addition: Warning
> > messages:
> > 1: In fish.id:trial :
> >   numerical expression has 2518 elements: only the first used
> > 2: In fish.id:trial :
> >  numerical expression has 2518 elements: only the first used
> 
> I've tried to find the answer in other mailing lists, but they all 
> give the
same answer....
> 
> Any ideas of what the mistake might be?
> 

  Not sure, but I strongly recommend that you put your response variable
along with all of your predictor variables into a data frame together and
use the data= argument to glmer().  It's extremely easy to have a modified
version of one of the predictors floating around in the workspace.  It's
also possible (although I think this would technically constitute a bug in
lmer ...) that this is caused by NAs somewhere in the dataset, although I
couldn't provoke it with this
sample:

d <- data.frame(y=rnorm(200),f1=sample(LETTERS[1:5],200,replace=TRUE),
    f2=sample(letters[1:5],200,replace=TRUE))
d$f2[100] <- NA
lmer(y~1+(1|f1/f2),data=d)

  If putting your variables into a data frame and using na.omit() on it
don't make the problem go away, then we'll need a reproducible example to
get any farther ...

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