r-使用igraph有效提取重复/多个路径

Question

如何从igraph中提取重复路径？有没有比下面例子中的双循环更有效的方法？我所说的重复/多路径是指结果应该同时包含B1 -> B2a -> B3和B1 -> B2b -> B3。但我的代码并没有产生这种效果。数据：

library(dplyr)
library(igraph)
d <- tribble(~input, ~output,
"A1", "A2",
"A2", "A3a",
"A2", "A3b",
"B1", "B2a",
"B1", "B2b",
"B2a", "B3",
"B2b", "B3")

我的分析：

g <- graph_from_data_frame(d, directed = TRUE)
(finals <- V(g)[degree(g, mode = "out") == 0])
(starts <- V(g)[degree(g, mode = "in") == 0])
res_collect <- vector("list", length(starts) * length(finals))
tmp <- 1
for (i_s in seq_along(starts)) {
for (i_f in seq_along(finals)) {
res <- tryCatch(
{
all_simple_paths(g, from = starts[[i_s]], to = finals[[i_f]])[[1]]
},
error = function(cond) {
message("failed at starts-nr=", i_s, ", finals-nr=", i_f); return(as.integer(NA))
}
)
res_collect[[tmp]] <- res
tmp <- tmp + 1
}
}

问题：

res_collect[!is.na(res_collect)]
# [[1]]
# + 3/8 vertices, named, from 60df173:
#   [1] A1  A2  A3a
# 
# [[2]]
# + 3/8 vertices, named, from 60df173:
#   [1] A1  A2  A3b
# 
# [[3]]
# + 3/8 vertices, named, from 60df173:
#   [1] B1  B2a B3 

缺少1个路径：B1->B2b->B3.

Answer 1

您可以尝试下面的代码

do.call(
c,
lapply(
starts,
all_simple_paths,
graph = g,
to = finals
)
)

它给出

[[1]]
+ 3/4 vertices, named, from fb6335c:
[1] A1  A2  A3a
[[2]]
+ 3/4 vertices, named, from fb6335c:
[1] A1  A2  A3b
[[3]]
+ 3/4 vertices, named, from fb6335c:
[1] B1  B2a B3
[[4]]
+ 3/4 vertices, named, from fb6335c:
[1] B1  B2b B3