igraph permute（）方法bug

Question

我想节点置换图表。请参阅下面我创建的测试图。当我使用igraph R库中的permute（）方法时，permute（）生成的新图形不会发生任何变化。发生了什么事？

testG <- vector(mode="list", length=6); #assign initial probabilities 
testG = list("gene1"=0, "gene2"=0, "gene3"=0, "gene4"=0, "gene5"=0, "gene6"=0);
adjacency_test <- matrix(c(0,1,1,0,0,0,1,0,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0), nrow=6, ncol=6);
rownames(adjacency_test) <- c("gene1", "gene2", "gene3","gene4","gene5","gene6");
colnames(adjacency_test) <- c("gene1", "gene2", "gene3","gene4","gene5","gene6");
require(igraph)
p <- graph.adjacency(adjacency_test, mode="undirected", weighted=TRUE);
vids.permuted <- c(6,5,4,3,2,1)
p2 <- permute(p, permutation=vids.permuted)
plot(p)
plot(p2)

号码：

P2：

我希望置换图（p2）中的集团再次成为gene6，gene5，gene4，而不是gene1，gene2和gene3，就像原来一样。

发生了什么事？

编辑：

根据下面的回复，这是正确的，我有另一个担心。当我手动重新排列节点名称时，为什么当我检查所有边缘是否相同，并且度数与原始图形与置换图形的度数相同时，igraph是否为真？

p <- graph.adjacency(adjacency_test, mode="undirected", weighted=TRUE);
p2 <- p
V(p2)$name <- V(p)$name[sample(length(V(p)$name), replace=FALSE)]
# p is for sure different from p2 now
plot(p, layout=layout.reingold.tilford)
plot(p2, layout=layout.reingold.tilford)
# But why are the degrees still the same, and edges still the same?
all(E(p)==E(p2))
degs1 <- degree(p)
degs2 <- degree(p2)
all(degs1==degs2)

Answer 1

置换函数交换顶点ID并创建一个同构图。这不是你想要的。要交换顶点标签，请使用

p2=p
V(p2)$name=paste("gene ",6:1)