我有以下无向图g
,它是从河流的形状文件生成的。我想将 g
转换为有向图,其中顶点 55 是根,所有边的“方向”都朝向根顶点(想象一下水从网络的所有部分流向根)。
无向图示例:
library(igraph)
g <- structure(list(From = c(48, 37, 39, 32, 38, 36, 49, 46, 31, 26,
33, 35, 18, 23, 45, 41, 42, 47, 52, 51, 50, 54, 16, 14, 8, 10,
5, 6, 17, 11, 20, 24, 2, 3, 1, 0, 44, 4, 7, 29, 30, 34, 40, 53,
43, 15, 9, 28, 27, 12, 13, 19, 21, 22, 25), To = c(32, 32, 31,
31, 33, 33, 45, 45, 23, 23, 26, 26, 16, 16, 35, 35, 41, 41, 50,
50, 47, 47, 6, 6, 5, 5, 2, 2, 10, 10, 11, 11, 1, 1, 0, 55, 30,
3, 3, 28, 28, 29, 29, 40, 40, 7, 7, 22, 22, 9, 9, 13, 13, 19,
19)), class = "data.frame", row.names = c(NA, -55L))
g <- graph.data.frame(g, directed = FALSE)
l <- igraph::layout_as_tree(g, flip.y = FALSE)
plot(g,
vertex.size = 10,
vertex.color = "darkgray",
layout = l)
我可以执行以下操作来创建有向图,但有些边的方向是正确的,而另一些则不是。
g2 <- get.adjacency(g, sparse = F)
g2[upper.tri(g2)] <- 0
g2 <- igraph::graph.adjacency(g2)
plot(g2,
vertex.size = 10,
vertex.color = "darkgray",
layout = l)
我可以从邻接矩阵中的顶点如何标记中看出问题的结果,但无法提出解决方案。
我的问题:是否可以将无向图转换为有向图,其中所有边的方向都朝向选定的顶点(在本例中为顶点 55)?
如果顶点被重命名等也很好。
最佳答案
对于连接顶点a和b的每条边,如果a到55的最短距离小于b到55的最短距离,则b应该朝向a。我不熟悉igraph
,但我根据这个理由想出了一种方法:
d <- distances(g)[, '55']
dd <- outer(d, d, FUN = '>')
g2 <- get.adjacency(g, sparse = F)
g2 <- g2 * dd
g2 <- igraph::graph.adjacency(g2)
plot(g2,
vertex.size = 10,
vertex.color = "darkgray",
layout = l)
注意:所有箭头的方向都可以通过改变 outer(d, d, FUN = '>')
来反转。至outer(d, d, FUN = '<')
.
关于r - 在有向图中设置方向,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/61201346/