Question

我有一个二分图和一个数据框，每一行与图的第一面的每个顶点相关联。图形没有连接，当我找到图形的最大组件时，我必须对数据框进行子集化（这不会给我正确的答案）。另一种可能的选择是将数据框的行设置为图形第一面顶点的属性（我不知道该怎么做）！这是一个玩具示例：

edgelist = matrix(c("A","a","A","b","B","b","C","c","D","c"),ncol=2,byrow=T)
bg <- graph.data.frame(edgelist, directed=F)
V(bg)$type <- V(bg)$name %in% edgelist[,1]
summary(bg)
V(bg)[V(bg)$type==1]
df = data.frame(id=c("A","B","C","D"), x=runif(4,10,50), y=sample(4), z=rnorm(4))

gclust = clusters(bg)
numClust = gclust$no 
numLCC = gclust$csize[1] 
bg2 = induced.subgraph(bg, which(gclust$membership ==which.max(gclust$csize)))
remained_set1 <- V(bg2)[V(bg2)$type==1]

df[as.character(df[,1])%in%remained_set1,] # wrong answer

Answer 1

尝试使用顶点的名称并翻转subset(df, id %in% names(remained_set1))

foreach (string dirPath in Directory.GetDirectories(sPath, "*",
         SearchOption.AllDirectories))
         Directory.CreateDirectory(dirPath.Replace(sPath, dPath));

foreach (string newPath in Directory.GetFiles(sPath, "*.*",
        SearchOption.AllDirectories))
        File.Copy(newPath, newPath.Replace(sPath, dPath), true);

这将返回示例中A和B的行。

设置二分图的一种模式的顶点的矢量属性

1 个答案: