标签和颜色叶子树状图

Question

I am trying to create a dendrogram, were my samples have 5 group codes (act as sample name/species/etc but its repetitive).

Therefore, I have two issues that a help will be great:

• How can I show the group codes in leaf label (instead of the sample number)?

• I wish to assign a color to each code group and colored the leaf label according to it (it might happen that they will not be in the same clade and by that I can find more information)?

Is it possible to do so with my script to do so (ape or ggdendro):

sample<-read.table("C:/.../DOutput.txt", header=F, sep="")
groupCodes <- sample[,1]
sample2<-sample[,2:100] 
d <- dist(sample2, method = "euclidean")  
fit <- hclust(d, method="ward")
plot(as.phylo(fit), type="fan") 
ggdendrogram(fit, theme_dendro=FALSE)  

A random dataframe to replace my read.table:

sample = data.frame(matrix(floor(abs(rnorm(20000)*100)),ncol=200))
groupCodes <- c(rep("A",25), rep("B",25), rep("C",25), rep("D",25)) # fixed error
sample2 <- data.frame(cbind(groupCodes), sample) 

Answer 1

Here is a solution for this question using a new package called " dendextend ", built exactly for this sort of thing.

You can see many examples in the presentations and vignettes of the package, in the "usage" section in the following URL: https://github.com/talgalili/dendextend

Here is the solution for this question: (notice the importance of how to re-order the colors to first fit the data, and then to fit the new order of the dendrogram)

####################
## Getting the data:

sample = data.frame(matrix(floor(abs(rnorm(20000)*100)),ncol=200))
groupCodes <- c(rep("Cont",25), rep("Tre1",25), rep("Tre2",25), rep("Tre3",25))
rownames(sample) <- make.unique(groupCodes)

colorCodes <- c(Cont="red", Tre1="green", Tre2="blue", Tre3="yellow")

distSamples <- dist(sample)
hc <- hclust(distSamples)
dend <- as.dendrogram(hc)

####################
## installing dendextend for the first time:

install.packages('dendextend')

####################
## Solving the question:

# loading the package
library(dendextend)
# Assigning the labels of dendrogram object with new colors:
labels_colors(dend) <- colorCodes[groupCodes][order.dendrogram(dend)]
# Plotting the new dendrogram
plot(dend)


####################
## A sub tree - so we can see better what we got:
par(cex = 1)
plot(dend[[1]], horiz = TRUE)

Answer 2

You could convert you hclust object into a dendrogram and use ?dendrapply to modify the properties (attributes like color, label, ...) of each node, eg:

## stupid toy example
samples <- matrix(c(1, 1, 1,
                    2, 2, 2,
                    5, 5, 5,
                    6, 6, 6), byrow=TRUE, nrow=4)

## set sample IDs to A-D
rownames(samples) <- LETTERS[1:4]

## perform clustering
distSamples <- dist(samples)
hc <- hclust(distSamples)

## function to set label color
labelCol <- function(x) {
  if (is.leaf(x)) {
    ## fetch label
    label <- attr(x, "label") 
    ## set label color to red for A and B, to blue otherwise
    attr(x, "nodePar") <- list(lab.col=ifelse(label %in% c("A", "B"), "red", "blue"))
  }
  return(x)
}

## apply labelCol on all nodes of the dendrogram
d <- dendrapply(as.dendrogram(hc), labelCol)

plot(d)

EDIT: Add code for your minimal example:

    sample = data.frame(matrix(floor(abs(rnorm(20000)*100)),ncol=200))
groupCodes <- c(rep("A",25), rep("B",25), rep("C",25), rep("D",25))

## make unique rownames (equal rownames are not allowed)
rownames(sample) <- make.unique(groupCodes)

colorCodes <- c(A="red", B="green", C="blue", D="yellow")


## perform clustering
distSamples <- dist(sample)
hc <- hclust(distSamples)

## function to set label color
labelCol <- function(x) {
  if (is.leaf(x)) {
    ## fetch label
    label <- attr(x, "label")
    code <- substr(label, 1, 1)
    ## use the following line to reset the label to one letter code
    # attr(x, "label") <- code
    attr(x, "nodePar") <- list(lab.col=colorCodes[code])
  }
  return(x)
}

## apply labelCol on all nodes of the dendrogram
d <- dendrapply(as.dendrogram(hc), labelCol)

plot(d)