Reputation: 2056
My problem is to strip my panels with lattice framework.
testData<-data.frame(star=rnorm(1200),frame=factor(rep(1:12,each=100))
,n=factor(rep(rep(c(4,10,50),each=100),4))
,var=factor(rep(c("h","i","h","i"),each=300))
,stat=factor(rep(c("c","r"),each=600))
)
levels(testData$frame)<-c(1,7,4,10,2,8,5,11,3,9,6,12)# order of my frames
histogram(~star|factor(frame), data=testData
,as.table=T
,layout=c(4,3),type="density",breaks=20
,panel=function(x,params,...){
panel.grid()
panel.histogram(x,...,col=1)
panel.curve(dnorm(x,0,1), type="l",col=2)
}
)
What I'm looking for, is:
Upvotes: 4
Views: 1152
Reputation: 263362
You should not need to add the factor call around items in the conditioning section of the formula when they are already factors. If you want to make a cross between two factors the interaction
function is the best approach. It even has a 'sep' argument which will accept a new line character. This is the closest I can produce:
h<-histogram(~star|interaction(stat, var, sep="\n") + n, data=testData ,
as.table=T ,layout=c(4,3), type="density", breaks=20 ,
panel=function(x,params,...){ panel.grid()
panel.histogram(x,...,col=1)
panel.curve(dnorm(x,0,1), type="l",col=2) } )
plot(h)
useOuterStrips(h,strip.left = strip.custom(horizontal = FALSE),
strip.lines=2, strip.left.lines=1)
I get an error when I try to put in three factors separately and then try to use useOuterStrips
. It won't accept three separate conditioning factors. I've searched for postings in Rhelp, but the only perfectly on-point question got an untested suggestion and when I tried it failed miserably.
Upvotes: 1