I'll reply to my own post - to make sure no one wastes his/her time on that.
I was able to solve the problem only after I modified the original
function interaction.plot (see below). All I did I added one line
before the final } - asking it to return the means on the numeric
(dependent) variable.
A
Thank you, David, however, I am not sure this approach works.
Let's try it again - I slightly modifed d to make it more clear:
d=data.frame(xx=c(1,1,1,1,2,2,2,2,3,3,3,3),yy=c(3,3,4,4,3,3,4,4,3,3,4,4),zz=c(-1.1,-1.3,0,0.6,-0.5,1,3.3,-1.3,4.4,3.5,5.1,3.5))
d[[1]]<-as.factor(d[[1]])
d[[2]]<-as.factor
See if this helps. After your code, submit this to R:
with(d, text(xx[xx==3],zz[xx==3],paste("3, ",zz[xx==3])))
After that has convinced you that xx and zz are being used properly,
you can try the more general approach:
with(d, text(xx,zz,paste(xx, " , ", zz)))
I would have used ZZ rather
On Wed, Mar 4, 2009 at 10:52 AM, Dimitri Liakhovitski wrote:
> Hello - and sorry for what might look like a simple graphics question.
>
> I am building an interaction plot for d:
>
> d=data.frame(xx=c(3,3,2,2,1,1),yy=c(4,3,4,3,4,3),zz=c(5.1,4.4,3.5,3.3,-1.1,-1.3))
> d[[1]]<-as.factor(d[[1]])
> d[[
Hello - and sorry for what might look like a simple graphics question.
I am building an interaction plot for d:
d=data.frame(xx=c(3,3,2,2,1,1),yy=c(4,3,4,3,4,3),zz=c(5.1,4.4,3.5,3.3,-1.1,-1.3))
d[[1]]<-as.factor(d[[1]])
d[[2]]<-as.factor(d[[2]])
print(d)
interaction.plot(d$xx, d$yy, d$zz,
type
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