Question

我在R中使用iGraph来计算一组10个节点的k核。其中8个节点连接，度数为5或更大，另外两个节点的度数为3。

根据我对k核的理解，不应该连接8个节点，并且至少有5个节点都在k核中，值为5吗？然而，这8个中的两个被赋予k核值4和3。

请参阅下面的可重现图表：

el = matrix(c("Ns-1","Ns-1","Ns-1","Ns-1","Ns-1","Ns-1","Ns-1","Ns-1","Ns-14","Ns-14","Ns-15","Ns-15","Ns-15","Ns-17","Ns-17","Ns-17","Ns-17","Ns-2","Ns-2","Ns-2","Ns-2","Ns-4","Ns-4","Ns-4","Ns-5","Ns-5","Ns-5","TAMU-7","Ns-14","Ns-15","Ns-17","Ns-2","Ns-4","Ns-5","TAMU-7","TAMU-8","Ns-15","Ns-17","Ns-17","Ns-4","Ns-18","Ns-2","Ns-4","Ns-5","Ns-18","Ns-4","Ns-5","TAMU-7","TAMU-8","Ns-5","TAMU-7","TAMU-8","TAMU-7","TAMU-8","Ns-18","TAMU-8"),nrow=28,ncol=2)

graph = graph.edgelist(el[,1:2],directed=F)

l <- layout.kamada.kawai(graph)
V(graph)$x <- l[,1]
V(graph)$y <- l[,2]

V(graph)$kCore = graph.coreness(graph)
V(graph)$degree = degree(graph)

layout(matrix(c(1,2), 1, 2))
plot(graph,
 main="Degree",
 vertex.size = 20,
 vertex.label = V(graph)$degree,
 vertex.label.color = "black",
 vertex.label.font = 1,
 vertex.label.family = "sans",
 vertex.label.cex = 1,
 vertex.color = "white")

plot(graph,
 main="K Cores",
 vertex.size = 20,
 vertex.label = V(graph)$kCore,
 vertex.label.color = "black",
 vertex.label.font = 1,
 vertex.label.family = "sans",
 vertex.label.cex = 1,
 vertex.color = "white")

...生成相同图形的图像（https://www.dropbox.com/s/unkkgodf8ppteyq/igraph.png?dl=0），左边显示度数，右边显示k-core。

基本上，左上角和右上角的节点应该同时具有3度和k核。剩下的应该是5，但是顶部中间的两个不是。

这是iGraph中的错误，还是我对k-core / graph.coreness算法的理解错误？

Answer 1

根据我对k核的理解，不应该连接8个节点，并且至少有5个节点都在k核中，值为5吗？然而，这8个中的两个被赋予k核值4和3。

没有。如果删除程度小于 k ，迭代的所有节点，k-core就是图形的一部分，这意味着如果某些节点的度数低于 k 由于删除了其他节点，您还必须删除这些节点。剩下的k-core是一个子图，其中所有节点的度数至少为 k 。在您的特定情况下，删除图中最左侧的节点并移除最顶层节点（在右上角）会降低5以下其他两个节点的度数 - 因此它们也将被删除。

iGraph-k-core with graph.coreness低估了一些核心

1 个答案: