I'm looking to visualize the nodes and edges of my graph in the physical layout that corresponds to the spatial location each node corresponds to in a segmented image. I have the spatial information for each segmented epithelial cell and the node it corresponds to in xy and am just wondering if there is a plugin that already exists to do this.
MATLAB: How to visualize graph nodes on top of image with segmented cells they correspond to
graphimage segmentationMATLABnetwork analysisnodes
Related Solutions
The simplest solutions use integer weights, two hidden nodes with sign activation functions, purelin output activation functions and polar binary targets [-1 1 1 -1].
Since there are only 4 training equations for estimating (2+1)*2+(2+1)*1 = 9 weights, solutions are far from being unique without other constraints.
You can play around with the equations and determine your own version of "simplest solutions".
These can be easily transformed into solutions for continuous activation functions by replacing sign with tansig. The transformation of purelin to tansig or logsig is also straightforward.
Good Luck.
Greg
There are some utilities for this on the file exchange, and there is the isocaps function which works with isosurface, but there isn't a builtin utility for adding "skirts" to the surface that the surf function creates.
It's actually relatively easy to do by hand. Here's a simple example:
% Create some interesting sample data
[x,y,z] = peaks;z = z+3;x = x(20:end,20:end);y = y(20:end,20:end);z = z(20:end,20:end);[m,n] = size(z);% Create the original surface
surf(x,y,z);% Create 4 surfaces around the edges
hold onwalls(1) = surf([x( 1,:); x( 1,:)],[y( 1,:); y( 1,:)],[zeros(1,n); z( 1,:)])walls(2) = surf([x(end,:); x(end,:)],[y(end,:); y(end,:)],[zeros(1,n); z(end,:)])walls(3) = surf([x(:, 1), x(:, 1)],[y(:, 1), y(:, 1)],[zeros(m,1), z(:, 1)])walls(4) = surf([x(:,end), x(:,end)],[y(:,end), y(:,end)],[zeros(m,1), z(:,end)])set(walls,'FaceColor','interp')
But this only works if all of your Z values are on one side of the origin. If they cross, you need to do something similar to what I described in this blog post about a similar 2D problem.
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