Circular loops in R package iGraph

Question

Using iGraph, how can I represent self-reflexive nodes with circle shaped curves? By default, these curves are represented by a pinched or tear drop shaped loop.

Answer 1

As Spacedman said, you would need to do quite some programming to do this. You could plot a graph without self-loops and then add them (graphs are basically a scatterplot and you can use points and similar functions to add lines to them), but this is not trivial (especially since you need to know the edge of nodes, not their center) and will cause the selfloops to be plotted on top of everything else which might not look good.

This weekend I have updated qgraph with how self-loops work. qgraph can be used to plot networks and should play nicely with igraph. e.g.:

# An adjacency matrix:
A <- matrix(1,3,3)

library("igraph")
# igraph graph and layout:
Graph <- graph.adjacency(A)
Layout <- layout.circle(Graph)

# Plot in qgraph:
library("qgraph")
qgraph(get.adjacency(Graph,sparse=FALSE),layout=Layout,diag=TRUE,directed=TRUE)

I am quite content with how these self-loops turned out and they seem to be more to what you describe. So this could be an option. However, my loops are just as hardcoded. For reference, I compute the edge of a node (starting and ending point of the loop) with the inner function qgraph:::Cent2Edge and compute the shape of the loop (spline) with the inner function qgraph:::SelfLoop.

Answer 2

Inside plot.igraph you can see that loops are drawn using a plot.bezier function, and all the control for that is pretty much hard coded there. You'd have to rewrite large chunks of plot.igraph to call a plot.circle function you'd have to write to do this.

Also, I'm guessing you don't want complete circles, but circle segments that start on the edge of the vertex symbol (the default blue circle with the vertex number in it) and end (possibly with an arrowhead) on another part of the edge of the vertex symbol? Or do you want circles that touch the symbol like the bezier teardrop loops do?

Either way, the answer seems to be 'no, not without doing some programming or submitting a feature request to the igraph guys'

I posted an earlier answer saying the layout functions were involved, but that's not true - the layout functions only position the vertices, and it is plot.igraph's job to draw the edges.