> Does > > xyplot(y ~ seq_along(y), xlab = "Index") > > do what you want?
Not exactly, because it does not work once multipanel conditioning comes into play: xyplot(y~seq_along(y)|factor(rep(1:2, each=5)), xlab = "Index") The points in the right panel are plotted from 6:10 while the points in the left panel are plotted from 1:5. Of course I could do something like xyplot(y~rep(1:5, 2) |factor(rep(1:2, each=5)), xlab = "Index") in this toy example, but as pointed out this becomes very cumbersome if the grouping variable does not follow a pattern. BTW: my toy example did not work with multipanel conditioning either, but one can work around that too using the subscripts argument in the panel function (I skipped that exercise for the sake of brevity, but I must admit that it obscured somehow my real intention, sorry for that). However, the more I think of it the more I believe that I have to provide the x's explicitly nevertheless and my solution would be: set.seed(123) y <- rnorm(20) grp <- index <- sample(3, 20, TRUE) index[unlist(lapply(levels(as.factor(grp)), function(n) which(as.factor(grp)==n)))] <- unlist(tapply(grp, grp, seq_along)) xyplot(y ~ index | factor(grp), xlab = "Index") This should work, but it seems to be a rather elaborate solution, especially since an index plot is nothing too fancy. So maybe I'm not seeing the wood for trees, but does anybody know an easier way? Thanks. KR, -Thorn ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
