关于IGraph最短路径计算,我缺少一些东西。
假设我生成一个网络(在stackoverflow中的某个地方找到)并执行简单的计算:
library(igraph);
relations <- data.frame(from=c("Bob", "Cecil", "Cecil", "David", "David", "Esmeralda"), to=c("Alice", "Bob", "Alice", "Alice", "Bob", "Alice"));
g = simplify(graph_from_data_frame(d=relations, directed=T), remove.multiple = F, remove.loops = T);
#plotting the network in order to appreciate the directions of the edges
plot(g,edge.arrow.size=0.5);
#V(g)[5] is "Alice" which apparently should not be able to reach any node
print(all_shortest_paths(g,from=V(g)[5],to=V(g),mode="all")$res);
如您所见,找到的最短路径是:
> print(all_shortest_paths(g,from=V(g)[5],to=V(g),mode="all")$res);
[[1]]
+ 2/5 vertices, named, from 823c15d:
[1] Alice Bob
[[2]]
+ 2/5 vertices, named, from 823c15d:
[1] Alice Cecil
[[3]]
+ 2/5 vertices, named, from 823c15d:
[1] Alice David
[[4]]
+ 2/5 vertices, named, from 823c15d:
[1] Alice Esmeralda
[[5]]
+ 1/5 vertex, named, from 823c15d:
[1] Alice
我期望的是,不应返回最短的路径,因为在有向图中Alice没有从其自身伸出的任何边缘。这是由于以下事实:当我计算最短路径时,我正在使用以下选项:
mode="all"
而且这以某种方式甚至适用于有向图?
当然,如果我更改图形构造并设置:
directed=F
即
library(igraph);
relations <- data.frame(from=c("Bob", "Cecil", "Cecil", "David", "David", "Esmeralda"), to=c("Alice", "Bob", "Alice", "Alice", "Bob", "Alice"));
g = simplify(graph_from_data_frame(d=relations, directed=F), remove.multiple = F, remove.loops = T);
#plottin the network in order to appreciate the directions of the edges
plot(g,edge.arrow.size=0.5);
#V(g)[5] is "Alice" which apparently should not be able to reach any node
print(all_shortest_paths(g,from=V(g)[5],to=V(g),mode="all")$res);
返回相同的结果。
到底是怎么回事?我是否太累了,无法直截了当?
这就是什么mode="all"
意思-使用所有边缘而不管方向。
我将使用一个更简单的图形来轻松查看发生了什么。
rel2 <- data.frame(from=c("Bob", "Bob", "David"),
to=c("Alice", "Carol", "Carol"))
g = simplify(graph_from_data_frame(d=rel2, directed=T))
LO = layout_as_bipartite(g, types=c(F,F,T,T))
plot(g, layout=LO)
现在,用最短路径声明
print(all_shortest_paths(g,from=V(g)[3],to=V(g),mode="all")$res)
[[1]]
+ 2/4 vertices, named:
[1] Alice Bob
[[2]]
+ 4/4 vertices, named:
[1] Alice Bob Carol David
[[3]]
+ 1/4 vertex, named:
[1] Alice
[[4]]
+ 3/4 vertices, named:
[1] Alice Bob Carol
即使边沿相反的方向,我们也可以获得将爱丽丝连接到另一个节点的所有路径。
我认为您想要的是:
print(all_shortest_paths(g,from=V(g)[3],to=V(g),mode="out")$res)
[[1]]
+ 1/4 vertex, named:
这仅给出了从Alice到其自身的零长度路径。
为了完整性,
print(all_shortest_paths(g,from=V(g)[3],to=V(g),mode="in")$res)
[[1]]
+ 2/4 vertices, named:
[1] Alice Bob
[[2]]
+ 1/4 vertex, named:
[1] Alice
这遵循仅使用传入边缘的路径,因此我们使用进入Alice的边缘获得了从“ Alice”到“ Bob”的路径,但是由于没有进入Bob的边缘,因此我们一无所获。
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