[Mne_analysis] Identifying adjacent sources

A.C.G. Thwaites acgt2 at cam.ac.uk
Sat May 11 06:09:01 EDT 2013
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Hi Steve,

I have a function for this, it accepts a .surf mesh, a center vertex 
number, and a 'distance' d and it returns a list of vertices d vertices 
away from the centre vertex (so distance 1 would return a list of every 
vertex adjacent to the centre vertex, distance 2 would be the vertices that 
form a ring around the adjacent vertices, etc).

I can't remember if it does the full mesh or allows you to search through a 
downsampled mesh (ie just the sources you have estimated).

I'll check and send it to the list on monday.

Andy




On May 11 2013, Stephen Politzer-Ahles wrote:

>Hello,
>
>Is there a straighforward method to figure out, for any cortical source or
>vertex, what the adjacent cortical sources/vertices are?
>
>More specifically, I have extracted .amp timecourse files for each
>participant and condition that look something like this:
>    100779 9.76119e-13 9.8006e-13 1.01097e-12 1.0584e-12
>    102810 5.77402e-13 4.74985e-13 6.13183e-13 8.53066e-13
>    103338 7.69642e-13 1.13359e-12 1.48948e-12 1.78157e-12
>    103374 6.44688e-13 8.72736e-13 1.09166e-12 1.24282e-12
>where the first number in each row is the vertex that timecourse comes from
>(and the following columns are the current estimates at each timepoint).
>I'm interested in doing a permutation-based spatiotemporal clustering test,
>so what I'd like to be able to do is, for each of those vertices, to get a
>list of which sources are adjacent to it. (Not necessarily which vertices
>are adjacent to it, since my understanding is that the cortical sources
>represent just a subset of the vertices present in the original
>tesselation.) I assume that after triangulation there must be some record
>of this, or some way to find it out, but I can't figure out where to find
>it. By the way, my structural images were segmented in Freesurfer, so the
>triangulations are in .surf files rather than .tri files.
>
>Any advice is greatly appreciated! Thank you,
>Steve Politzer-Ahles
>
>



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