igraph R

Question

鉴于是树：

library(igraph)

# setup graph
g= graph.formula(A -+ B,
                 A -+ C,
                 B -+ C,
                 B -+ D,
                 B -+ E
)
plot(g, layout = layout.reingold.tilford(g, root="A"))

顶点"A"是树的根，而顶点"C", "D", "E"被视为终端叶。

问题：

任务是找到根和叶之间的所有路径。我失败了以下代码，因为它只提供最短的路径：

# find root and leaves
leaves= which(degree(g, v = V(g), mode = "out")==0, useNames = T)
root= which(degree(g, v = V(g), mode = "in")==0, useNames = T)

# find all paths
paths= lapply(root, function(x) get.all.shortest.paths(g, from = x, to = leaves, mode = "out")$res)
named_paths= lapply(unlist(paths, recursive=FALSE), function(x) V(g)[x])
named_paths

输出：

$A1
Vertex sequence:
[1] "A" "C"

$A2
Vertex sequence:
[1] "A" "B" "D"

$A3
Vertex sequence:
[1] "A" "B" "E"

问题：

如何找到所有路径，包括顶点序列："A" "B" "C"？

我的理解是，"A" "B" "C"未提供缺失的序列get.all.shortest.paths()作为从"A"到"C"通过顶点序列的路径："A" "C" （在列表元素$A1中找到）更短。所以igraph正常工作。 然而，我正在寻找一种代码解决方案，以R list的形式获取从根到所有叶子的所有路径。

注释：

我知道对于大树，覆盖所有组合的算法可能会变得昂贵，但我的实际应用相对较小。

Answer 1

根据Gabor的评论：

all_simple_paths(g, from = root, to = leaves)

的产率：

[[1]]
+ 3/5 vertices, named:
[1] A B C

[[2]]
+ 3/5 vertices, named:
[1] A B D

[[3]]
+ 3/5 vertices, named:
[1] A B E

[[4]]
+ 2/5 vertices, named:
[1] A C