如何从R中的图像对象中获取树？
如何从R中的图像对象中获取树？
提问于 2019-11-08 15:41:35
我有一个随机有向加权图gg，它有下一个结构：
gg <-
structure(list(10, TRUE, c(0, 0, 1, 2, 2, 5, 5, 6, 6, 6, 6, 9,
9, 9, 9, 9), c(6, 9, 3, 0, 5, 3, 7, 1, 3, 5, 8, 2, 4, 6, 7, 8
), c(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15), c(3,
7, 11, 2, 5, 8, 12, 4, 9, 0, 13, 6, 14, 10, 15, 1), c(0, 2, 3,
5, 5, 5, 7, 11, 11, 11, 16), c(0, 1, 2, 3, 6, 7, 9, 11, 13, 15,
16), list(c(1, 0, 1), structure(list(), .Names = character(0)),
structure(list(name = c("C", "D", "I", "J", "K", "N", "O",
"Q", "S", "T"), color = c("yellow", "red", "red", "red",
"red", "red", "green", "red", "red", "green")), .Names = c("name",
"color")), structure(list(weight = c(0.5, 0.5, 1, 0.333333333333333,
0.333333333333333, 0.333333333333333, 0.333333333333333,
0.25, 0.25, 0.25, 0.25, 0.2, 0.2, 0.2, 0.2, 0.2)), .Names = "weight")),
<environment>), class = "igraph")我需要找到从根(黄色节点)到叶子(红色节点)的所有路径。叶子由(a)边的方向和(b)从根到叶的距离应该是两个边。
在我的例子中，根是C，叶子应该是D, J, N, S, I, K, Q。我试着只定义(a)条件。
root <- "C"
leaves = which(degree(gg, v = V(gg), mode = "out")==0, useNames = T)
leaves
# J K Q S
# 4 5 8 9
plot(gg, layout = layout.reingold.tilford(gg, root=root),
edge.arrow.size=0.2, edge.curved=T,
edge.label = round(E(gg)$weight,2))
问题.如何定义(b)条件并添加到叶集D, N, I, K节点？
回答 1
Stack Overflow用户
回答已采纳
发布于 2019-11-13 15:47:23
有一种方法可以做到这一点:使用shortest_paths获取所有来自根节点的两条边的顶点。
two.edges.from.root = unlist(sapply(shortest_paths(gg,
from = as.numeric(V(gg)["C"]),
mode = "out")$vpath,
function(x) { if(length(x) == 3) { x[3] } }))页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/58769689
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