Trimms a tree (dendrogram, hclust) from a set of leaves based on their labels.

prune(dend, ...)

# S3 method for default
prune(dend, ...)

# S3 method for dendrogram
prune(dend, leaves, reindex_dend = TRUE, ...)

# S3 method for hclust
prune(dend, leaves, ...)

# S3 method for phylo
prune(dend, ...)

# S3 method for rpart
prune(dend, ...)



tree object (dendrogram/hclust/phylo)


passed on


a character vector of the label(S) of the tip(s) (leaves) we wish to prune off the tree.


logical (default is TRUE). If TRUE, the leaves of the new dendrograms include the rank of the old order.dendrogram. This insures that their values are just like the number of leaves. When FALSE, the values in the leaves is that of the original dendrogram. Thie is useful if prunning a dendrogram but then wanting to use order.dendrogram with the original values. When using prune.hclust, then reindex_dend is used by default since otherwise the as.hclust function would return an error.


A pruned tree


I was not sure if to call this function drop.tip (from ape), snip/prune (from rpart) or just remove.leaves. I ended up deciding on prune.

See also


hc <- hclust(dist(USArrests[1:5, ]), "ave")
dend <- as.dendrogram(hc)

par(mfrow = c(1, 2))
plot(dend, main = "original tree")
plot(prune(dend, c("Alaska", "California")), main = "tree without Alaska and California")

# this works because prune uses reindex_dend = TRUE by default
as.hclust(prune(dend, c("Alaska", "California")))
#> Call:
#> as.hclust.dendrogram(x = prune(dend, c("Alaska", "California")))
#> Cluster method   : NA 
#> Distance         : NA 
#> Number of objects: 3 
prune(hc, c("Alaska", "California"))
#> Call:
#> hclust(d = dist(USArrests[1:5, ]), method = "ave")
#> Cluster method   : average 
#> Distance         : euclidean 
#> Number of objects: 3