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I am trying to draw a dendrogram from the hclust function output. I hope the dendrogram is horizontally arranged instead of the default, which can be obtain by (for example)

require(graphics)
hc <- hclust(dist(USArrests), "ave")
plot(hc)

I tried to use as.dendrogram() function like plot(as.dendrogram(hc.poi),horiz=TRUE) but the result is without meaningful labels:

enter image description here

If I use plot(hc.poi,labels=c(...)) which is without the as.dendrogram(), I can pass the labels= argument, but now the dendrogram is vertical instead of horizontal. Is there a way to simultaneously arrange the dendrogram horizontally and assign user-specified labels? Thanks!

Update: as an example from the USArrests dataset, suppose I wanna use the abbreviations of the first two letters of the state names as labels, so that I wanna somehow pass labs into the plotting function:

labs = substr(rownames(USArrests),1,2)

which gives 

 [1] "Al" "Al" "Ar" "Ar" "Ca" "Co" "Co" "De" "Fl" "Ge" "Ha"
[12] "Id" "Il" "In" "Io" "Ka" "Ke" "Lo" "Ma" "Ma" "Ma" "Mi"
[23] "Mi" "Mi" "Mi" "Mo" "Ne" "Ne" "Ne" "Ne" "Ne" "Ne" "No"
[34] "No" "Oh" "Ok" "Or" "Pe" "Rh" "So" "So" "Te" "Te" "Ut"
[45] "Ve" "Vi" "Wa" "We" "Wi" "Wy"
alittleboy
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2 Answers2

27

To show your defined labels in horizontal dendrogram, one solution is to set row names of data frame to new labels (all labels should be unique).

require(graphics)
labs = paste("sta_",1:50,sep="") #new labels
USArrests2<-USArrests #new data frame (just to keep original unchanged)
rownames(USArrests2)<-labs #set new row names
hc <- hclust(dist(USArrests2), "ave")
par(mar=c(3,1,1,5)) 
plot(as.dendrogram(hc),horiz=T)

enter image description here

EDIT - solution using ggplot2

labs = paste("sta_",1:50,sep="") #new labels
rownames(USArrests)<-labs #set new row names
hc <- hclust(dist(USArrests), "ave")

library(ggplot2)
library(ggdendro)

#convert cluster object to use with ggplot
dendr <- dendro_data(hc, type="rectangle") 

#your own labels (now rownames) are supplied in geom_text() and label=label
ggplot() + 
  geom_segment(data=segment(dendr), aes(x=x, y=y, xend=xend, yend=yend)) + 
  geom_text(data=label(dendr), aes(x=x, y=y, label=label, hjust=0), size=3) +
  coord_flip() + scale_y_reverse(expand=c(0.2, 0)) + 
  theme(axis.line.y=element_blank(),
        axis.ticks.y=element_blank(),
        axis.text.y=element_blank(),
        axis.title.y=element_blank(),
        panel.background=element_rect(fill="white"),
        panel.grid=element_blank())

enter image description here

Didzis Elferts
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  • thanks, but I still don't get how can we assign user-specified labels to the horizontal dendrogram? The example you gave has build-in labels, but I really wanna pass my own labels... – alittleboy Jan 02 '13 at 07:16
  • Please see the update above. I am sorry that my own data example is hard to post online, so I just made up a label vector that I wanna show on the horizontal dendrogram. Thanks again! – alittleboy Jan 02 '13 at 07:27
  • @alittleboy updated my solution. This solution works only if labels are unique. – Didzis Elferts Jan 02 '13 at 07:44
  • To change labels, ``hc$labels <- labs`` is enough. No need to copy the whole data frame. – h2kyeong Oct 08 '13 at 05:34
  • I think when the OP says "the example you gave has build-in labels", he means that the hclust object stored into `hc` already has 'labels" for the leaves of its tree (as described at the [hclust](https://stat.ethz.ch/R-manual/R-devel/library/stats/html/hclust.html) documentation). Also, if you are using `stringdistmatrix` instead of `dist`, [then remember the argument `useNames`](https://mran.microsoft.com/web/packages/stringdist/stringdist.pdf) which labels each string with the string itself. – Nate Anderson Aug 07 '16 at 11:46
  • @DidzisElferts, this is amazing!!! You should write up your ggplot solution as a small package (or ask to incorporate in, let's say, `ggfortify`). – JelenaČuklina May 10 '18 at 07:38
27

Using dendrapply you can customize your dendro as you like.

enter image description here

colLab <- function(n) {
  if(is.leaf(n)) {
    a <- attributes(n)
    attr(n, "label") <- substr(a$label,1,2)             #  change the node label 
    attr(n, "nodePar") <- c(a$nodePar, lab.col = 'red') #   change the node color
  }
  n
}

require(graphics)
hc <- hclust(dist(USArrests), "ave")
clusDendro <- as.dendrogram(hc)
clusDendro <- dendrapply(clusDendro, colLab)
op <- par(mar = par("mar") + c(0,0,0,2))
plot(clusDendro,horiz=T)
agstudy
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  • yes, I appreciate your excellent answer and I've upvoted your post. Sorry I have to choose only one final answer... – alittleboy Jan 02 '13 at 19:02