I asked about putting some kind of coloured rug under a dendrogram. Thomas Petzoldt <[EMAIL PROTECTED]> replied: One possibility is to extract the coordinates used by the dendrogram using par("usr") ...
Er, the documentation for par("usr") says 'usr' A vector of the form 'c(x1, x2, y1, y2)' giving the extremes of the user coordinates of the plotting region. When a logarithmic scale is in use (i.e., 'par("xlog")' is true, see below), then the x-limits will be '10 ^ par("usr")[1:2]'. Similarly for the y-axis. But I _know_ the (logical) coordinates of the plotting region; what I need is the coordinates of the leaves of the dendrogram. but as a global alternative in cases like this (many cases and known number of classes), I would suggest a different cluster alorithm, e.g. ?kmeans. That doesn't really help, amongst other things because kmeans is not a hierarchical algorithm. I *DON'T* know the true number of classes. I know how many classes the person who collected the data thinks there are, and I don't need to do any clustering to find them, he gave me a simple rule. What I want to know is how many clusters there OUGHT to be and how similar these clusters are to the ones he thought there were. >From poking around, the "right" number of clusters is somewhere between 2 and 6. (For the record, I _have_ tried kmeans and I've tabulated the kmeans groups against the prespecified groups.) If you want to get a visual idea you may try to apply an ordination method (e.g. princomp or isoMDS the latter from package MASS) and color the objects according to their class found by kmeans. I had already done that (using the prespecified classes, not classes found by kmeans). But it didn't solve my present problem, which was overlaying the *prespecified* classes onto a dendrogram. Two other people gave me answers that are spot on. Unfortunately, I've now lost their messages, so I can't name them. Suggestion 1: use the RowSideColors (or ColSideColors) argument of heatmap(). This gives me two dendrograms (and I can suppress one if I want) and a heat image of the data, and all things considered, it's *better* than what I wanted. (I was aware of heatmap, but I'd failed to notice the relevance, or even the existence, of the ???SideColors arguments.) In this particular case, the graph _beautifully_ displays what I want it to display. Suggestion 2: use the draw.clust function from the maptree packages. I have now installed this package (which R makes *so* easy) and it does exactly what I asked for. Both of these approaches work with any dendrogram. I'm beginning to suspect that if something isn't already available in R, I'll never be able to imagine a need for it. But then I'm a bear of very little brain... ______________________________________________ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html