我想对图进行节点置换。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，新图形 permute() 不会发生任何变化。怎么了？
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)
磷：
p2：
我希望置换图 (p2) 中的集团是gene6、gene5、gene4，而不是gene1、gene2 和gene3，就像原来的那样。
怎么了？
编辑：
根据下面的回答，这是正确的，我还有另一个担心。当我手动重新排列节点名称时，当我检查所有边是否相同，以及原始图与置换图的度数是否相同时，igraph 是否正确？
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)