Re: [igraph] Isolated Edges and Vertex Attributes

[Tamas Nepusz]

> I know how to delete nodes that have no edges at all but is there a way to 
> delete isolated
> edges?
Find the connected components of your graph using clusters(), then delete
those vertices that participate in clusters of size 2. clusters() gives you
a result with three components:

cl <- clusters(g)
cl$membership -- is a vector where element i contains the cluster index of
vertex i-1 (unless you are using igraph 0.6 whose R interface uses 1-based
indexing, not 0-based)
cl$csize -- is a vector where element i gives you the size of cluster i
cl$no -- gives you the number of clusters

So, once you have cl, you can use which(cl$csize <= 2) to find the indices
of the clusters you will have to delete, then use cl$membership to find the
indices of the vertices you will have to delete:

small.clusters <- which(cl$csize <= 2)
vertices.to.delete <- which(cl$membership %in% small.clusters) - 1
g <- delete.vertices(g, vertices.to.delete)

> Secondly I have a data frame with 8 columns each representing a node
> attribute (except first column).
Then probably the easiest is to use graph.data.frame to construct your graph
directly from the data frame -- wouldn't that work in your case?

> g <- set.vertex.attribute(g, "x", index=V(g), value = n[,2])
Note that an easier way to achieve the same is V(g)$x <- n[,2]

> Also when I try to print the graph with the attributes using
> 
> print(g, vertex.attributes=igraph.par("print.vertex.attributes"),
> names=TRUE, quote.names=TRUE)
> 
> all I get is edges information and no attributes.
I'm not sure how this works in R (I use igraph from Python mostly), so there
could be an easier way to achieve the same thing, but the following line
definitely works:

sapply(list.vertex.attributes(g), function(x) { get.vertex.attribute(g, x) })

This gives you a table where each column corresponds to a vertex attribute
and columns are named according to the attributes. If you use this
frequently, you might want to save it as a function.

-- 
T.