如何根据一个连通分量内节点的最大度数，在图中添加连接两个连通分量之间的边

Question

我有一个包含四个不同聚类大小组件的图形。

我可以通过以下代码查看细节：

cl <- components(graph1)

详细信息看起来像这样：

$membership
ID_00104 ID_00136 ID_00169 ID_00178 ID_00180 ID_06663 ID_06791 ID_09099 ID_00910 ID_00790 ID_01013 ID_01130 ID_01260 ID_00394 ID_00860 ID_00959 ID_01222 ID_00288 ID_00324 ID_00663 ID_00846 ID_01047 ID_06781 ID_06786 
       1        2        2        3        4        1        1        1        2        3        4        4        4        4        4        4        4        4        4        4        4        4        4        4 

$csize
[1]  4  3  2 15

$no
[1] 4

使用以下代码，我还可以获取节点的degree数量。

degree(graph1)

输出结果为

ID_00104 ID_00136 ID_00169 ID_00178 ID_00180 ID_06663 ID_06791 ID_09099 ID_00910 ID_00790 ID_01013 ID_01130 ID_01260 ID_00394 ID_00860 ID_00959 ID_01222 ID_00288 ID_00324 ID_00663 ID_00846 ID_01047 ID_06781 ID_06786 
       3        2        2        1       14        1        1        1        2        1        1        1        1        1        1        1        1        1        1        1        1        1        1        1 

我可以使用以下代码将所有组件的任意两个节点连接起来（从两个组件中随机选择）（这是我之前一个帖子中的解决方案）。请参考此帖。

graph1 <-  graph_from_data_frame(g, directed = FALSE)
E(graph1)$weight <- g$values
cl <- components(graph1)
graph2 <- with(
  stack(membership(cl)),
  add.edges(
    graph1,
    c(combn(sapply(split(ind, values), sample, size = 1), 2)),
    weight = 0.01))

现在，我想在那些具有最高度数的节点之间添加一个边，例如：ID_00180的度数为14（附图左侧组件），ID_00104的度数为3（附图顶部组件）。在合并这两个组件时，我想在ID_00180和ID_00104之间添加一条边（而不是随机选择）。

如果任何组件具有多个相同数量的最高度数，例如：附图右下角的组件（所有节点的度数均为2），那么我们可以从最高度数的节点中随机选择任何一个。比如说，我们可以在ID_00180和任何节点之间添加一条边。

可重现数据。

g <- structure(list(query = structure(c(1L, 1L, 1L, 2L, 2L, 3L, 4L, 
                                        5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L), .Label = c("ID_00104", 
                                                                                                            "ID_00136", "ID_00169", "ID_00178", "ID_00180"), class = "factor"), 
                    target = structure(c(16L, 19L, 20L, 1L, 9L, 9L, 6L, 11L, 
                                         13L, 15L, 4L, 8L, 10L, 14L, 2L, 3L, 5L, 7L, 12L, 17L, 18L
                    ), .Label = c("ID_00169", "ID_00288", "ID_00324", "ID_00394", 
                                  "ID_00663", "ID_00790", "ID_00846", "ID_00860", "ID_00910", "ID_00959", 
                                  "ID_01013", "ID_01047", "ID_01130", "ID_01222", "ID_01260", "ID_06663", 
                                  "ID_06781", "ID_06786", "ID_06791", "ID_09099"), class = "factor"), 
                    values = c(0.654172560113154, 0.919096895578551, 0.925821596244131, 
                                0.860406091370558, 0.746376811594203, 0.767195767195767, 
                                0.830379746835443, 0.661577608142494, 0.707520891364902, 
                                0.908193484698914, 0.657118786857624, 0.687664041994751, 
                                0.68586387434555, 0.874513618677043, 0.836646499567848, 0.618361836183618, 
                                0.684163701067616, 0.914728682170543, 0.876297577854671, 
                                0.732707087959009, 0.773116438356164)), row.names = c(NA, 
                                                                                      -21L), class = "data.frame")

- 0Knowledge

Answer 1

## Save the layout from graph1
set.seed(2021)
LO = layout_nicely(graph1)
plot(graph1, layout=LO)

MaxNode <- rep("", max(cl$membership))
for(i in 1:max(cl$membership)) {
    MaxNode[i] <- names(which.max(degree(graph1)[cl$membership == i])) 
}
graph2 <-add.edges(graph1, combn(MaxNode, 2), weight = 0.01)
plot(graph2, layout=LO)

- G5W

Answer 2

更新

graph2 <- add.edges(
  graph1,
  combn(
    sapply(
      decompose(graph1),
      function(p) sample(names(V(p))[degree(p) == max(degree(p))], 1)
    ), 2
  ),
  weight = 0.01
)

plot(graph2, layout = layout_nicely(graph1))

提供

---

之前的回答

您可以尝试

out <- combn(
  decompose(graph1),
  2,
  FUN = function(x) {
    add.edges(
      graph1,
      sapply(x, function(p) sample(names(V(p))[degree(p) == max(degree(p))], 1)),
      weight = 0.01
    )
  },
  simplify = FALSE
)

sapply(out,plot)

- ThomasIsCoding

---

非常感谢您的及时回答。但是，我只需要一个包含所有组合的图形（例如：我的问题代码中的graph2对象包含6个组合。我认为您的代码给出了单独图形的最大节点连接！您介意检查一下代码吗？ - 0Knowledge

@0Knowledge 看一下我的更新 - ThomasIsCoding