[R] problem with interaction in lmer even after creating an "interaction variable"

[Ben Bolker]

a_lampei <anna.lampei-bucharova <at> uni-tuebingen.de> writes:

> 
> Dear all,
> I have a problem with interactions in lmer. I have 2 factors (garden and
> gebiet) which interact, plus one other variable (home), 
> dataframe arr. When
> I put:
> /
> lmer (biomass ~ home + garden:gebiet +  ( 1|Block), data = arr)/
> 
> it writes:
> /Error in lme4::lFormula(formula = biomass ~ home + garden:gebiet 
> + (1 |  : 
>   rank of X = 28 < ncol(X) = 30/
> 
> In the lmer help I found out that if not all combination of
>  the interaction
> are realized, lmer has problems and one should do new variable using
> "interaction", which I did:
> /
> arr$agg <- interaction (arr$gebiet, arr$garden, drop = TRUE)/
> 
> when I fit the interaction term now:
>  /
> lmer (biomass ~ home + agg+  ( 1|Block), data = arr)/
> 
> The error does not change:
> /
> Error in lme4::lFormula(formula = biomass ~ home + agg + (1 | Block),  : 
>   rank of X = 28 < ncol(X) = 29/
> 
> No NAs are in the given variables in the dataset.
> 
> Interestingly it works when I put only given interaction like
> 
> /lmer (biomass ~ agg +  ( 1|Block), data = arr)/
> 
> Even following models work:
> /lmer (biomass ~ gebiet*garden +  ( 1|Block), data = arr)
> lmer (biomass ~ garden + garden:gebiet  +( 1|Block), data = arr)/
> 
> But if I add the interaction term in th enew formate of 
> the new fariable, it
> reports again the same error.
> 
> /lmer (biomass ~ garden + agg  +( 1|Block), data = arr)/
> 
> If I put any other variable from the very same dataframe 
> (not only variable
> "home"), the error is reported again.
> 
> I do not understand it, the new variable is just another 
> factor now, or? And
> it is in the same dataframe, it has the same length.
> 
> Does anyone have any idea?
> 
> Thanks a lot, Anna
> 

  This probably belongs on r-sig-mixed-models.

  Presumably 'home' is still correlated with one of the
columns of 'garden:gebiet'.

  Here's an example of how you can use svd() to find out which
of your columns are collinear:

set.seed(101)
d <- data.frame(x=runif(100),y=1:100,z=2:101)
m <- model.matrix(~x+y+z,data=d)
s <- svd(m)
zapsmall(s$d)
## [1] 828.8452   6.6989   2.6735   0.0000
## this tells us there is one collinear component
zapsmall(s$v)
##            [,1]       [,2]       [,3]       [,4]
## [1,] -0.0105005 -0.7187260  0.3872847  0.5773503
## [2,] -0.0054954 -0.4742947 -0.8803490  0.0000000
## [3,] -0.7017874  0.3692117 -0.1945349  0.5773503
## [4,] -0.7122879 -0.3495142  0.1927498 -0.5773503
## this tells us that the first (intercept), third (y),
## and fourth (z) column of the model matrix are
## involved in the collinear term, i.e.
## 1+y-z is zero