从可爱树组 R 的祖先节点获取树

如何解决从可爱树组 R 的祖先节点获取树

我有一棵树，我想得到这棵树的一部分，它是可爱树组的祖先。

library(stats)
library(ape)
tree <- ape::read.tree(text = "((run2:1.2,run7:1.2)master3:1.5,((run8:1.1,run14:1.1)master5:0.2,(run9:1.0,run6:1.0)master6:0.3)master4:1.4)master2;")
plot(tree,show.node.label = TRUE)

我正在使用 cutree 来获取一定数量的组：

cutree(as.hclust(tree),k = 3)
 run2  run7  run8 run14  run9  run6 
    1     1     2     2     3     3 

如何获得树的左侧部分？基本上这个tree2：

tree2 <- ape::read.tree(text = "(master3:1.5,(master5:0.2,master6:0.3)master4:1.4)master2;")
plot(tree2,show.node.label = TRUE)

解决方法

您可以使用 ape::drop.tip 选项使用函数 trim.internal = FALSE：

## Dropping all the tips and their edges leading to the nodes (internals)
tree2 <- drop.tip(tree,tip = tree$tip.label,trim.internal = FALSE)

当然，如果您需要，您也可以指定只删除 hclust::cutree 函数结果后的某个数字：

## Dropping the tips and their edges that have a cut value equal to the scalar k
k = 3
tips_to_drop <- names(which(cutree(as.hclust(tree),k = k) == k))
tree3 <- drop.tip(tree,tip = tips_to_drop,trim.internal = FALSE)

如果您还想删除节点和提示（比如删除所有提示和节点 "master6" 和 "master5"，您可以再次将该树传递给同一个 drop.tip 函数：>

## Removing master6 and master5 from the tree2
tree3 <- drop.tip(tree2,tip = c("master5","master6"),trim.internal = FALSE)

## Or the same but with nested functions
tips_to_drop <- tree$tip.label
nodes_to_drop <- c("master5","master6")
tree3 <- drop.tip(drop.tip(tree,trim.internal = FALSE),tip = nodes_to_drop,trim.internal = FALSE)

,

phangorn 包有一个 Ancestors 函数，但只会返回列表结构中的索引，而不是树结构中的标签。所以我为你标记了它们（检查图表以确保要标记的索引是正确的），但我会让你把它变成一棵树。

library(phangorn)
tree.labels <- c(tree$tip.label,tree$node.label)
tree.ancestors <- Ancestors(tree)
names(tree.ancestors) <- tree.labels
tree.ancestors.labelled <- lapply(tree.ancestors,function(ind.vec) tree.labels[ind.vec])

tree.ancestors.labelled

$run2
[1] "master3" "master2"

$run7
[1] "master3" "master2"

$run8
[1] "master5" "master4" "master2"

$run14
[1] "master5" "master4" "master2"

$run9
[1] "master6" "master4" "master2"

$run6
[1] "master6" "master4" "master2"

$master2
character(0)

$master3
[1] "master2"

$master4
[1] "master2"

$master5
[1] "master4" "master2"

$master6
[1] "master4" "master2"