Try:
strip.left = strip.custom(factor.levels = colnames(z))
On Mon, Jun 9, 2008 at 12:18 PM, Klaus Nordhausen <[EMAIL PROTECTED]> wrote:
> Dear R masters,
>
> I have large multivariate time series as zoo objects and want to plot them
> using lattice.
>
> Since I have many variates in one obje
Or perhaps this way is more logical since it takes the names from the screens:
test2<-xyplot(z, as.table=TRUE , screens = colnames(z), strip=FALSE ,
strip.left = TRUE, layout=c(1,3,2))
On Mon, Jun 9, 2008 at 2:10 PM, Gabor Grothendieck
<[EMAIL PROTECTED]> wrote:
> Try:
>
> strip.left = strip.cu
Dear R masters,
I have large multivariate time series as zoo objects and want to plot them
using lattice.
Since I have many variates in one object I would like to have the strips on the
left, using strip.left = TRUE. However when I use this the variable names are
converted into numbers. How ca
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