horizontal dendrogram in R with labels

Question

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:

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"

Maus

I wonder what is hc.poi in your code examples? – Worley

Answer 1

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)

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())

Emulsify

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... – Maus

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! – Maus

@Maus updated my solution. This solution works only if labels are unique. – Emulsify

To change labels, hc$labels <- labs is enough. No need to copy the whole data frame. – Strainer

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 documentation). Also, if you are using stringdistmatrix instead of dist, then remember the argument useNames which labels each string with the string itself. – Worley

@DidzisElferts, this is amazing!!! You should write up your ggplot solution as a small package (or ask to incorporate in, let's say, ggfortify). – Fondafondant

Answer 2

Using dendrapply you can customize your dendro as you like.

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)

Zachery

yes, I appreciate your excellent answer and I've upvoted your post. Sorry I have to choose only one final answer... – Maus