> After proceeding with clique measurement, is there any way to identify/
> select just the �strong� clique like that:
that would for now only be possible, if you use
transformation->links->merge->contrary directed
to merge the <-> links to undirected links and then use
links->selection->visual properties->direction
to select the directed links (i.e. those who do not have a reverse
partner) and delete them
unfortunately for now the clique detection uses only the underlying
undirected graph (that is, each edge is replaced by an undirected edge
by the algorithm)
> b) Clique information is stored as a �integer list� by Visone. So, to
....
> Is there any way to obtain/extract from Visone a list like that (to a
> bigger network, the idea is to obtain a list of actors grouped around
> each clique):
>
> [1] A, B, C
this is not possible directly in visone, but I'm quite sure, you can do
this after sending the graph to Knime (they are more flexible)
should not be to complicated using the "feature extractor" node combined
with the "group by" node
> c) The �clustering coefficient� attaches value 1 to actors that are
> �alone�, without relations with any other actor. Should I consider
> this measure just to connected actors? In other words: to disregard
ok, that one is a bit more complicated:
from the visone part I can say that this is probably due to the
definition used in the implementation of clustering coefficient
the actual question then refers to the interpretation of clustering
coefficient, which is a more delicate question for which a cannot offer
a good answer
I think the interpretation depends on the data at hand, but I'd advise
to consider literature on the topic, since I'm not too familiar with the
qualitative interpretation of clustering coefficient.
However, it seems to me (this is a thought, not a statement) that the
isolates should be handled special in this case (e.g. leave them out and
report the average of the rest+ the number of isolates)
But I cannot stress enough, that this depends on what you are doing and
is probably answered in a better way elsewhere.
hope this helps (and sorry for the delay, your questions are quite
complicated sometimes)
Uwe