Finding all connected vertices with a specific vertex property

I am very new to graph tool and graphs in general and hence this naive question.

I have created a 2D square lattice graph as such:

L=4
g = gt.lattice([L,L])  
idx = g.vertex_index.copy()
x = g.new_vertex_property("double")
y = g.new_vertex_property("double")
x.a = idx.a % L
y.a = idx.a // L
pos = group_vector_property([x,y])
graph_draw(g,pos,vertex_text=g.vertex_index)

And then I add a property called ‘act_stat’ to each vertex:
(act_stat can be 0 or 1):

act_stat = g.new_vertex_property("int")
act_stat.a = np.random.randint(2,size=L**2)
graph_draw(g,pos,vertex_text=act_stat)

How do I randomly find a vertex which has an ‘act_stat = 0’, and find all the vertices with ‘act_stat=0’ connected to it either directly or through a line of ‘act_stat=1’ vertices.

I do not want to delete the ‘act_stat=1’ vertices using filtering because the goal is to find the path of minimum distances between the randomly chosen ‘act_stat=0’ and all other ‘act_stat=0’ vertices connected to it either directly or through a line of ‘act_stat=1’ vertices.

Any tips on which functions to use will also be greatly appreciated.

From what I understand, I need to use Djikstra’s search algorithm to find the shortest paths between a randomly choses source ‘act_stat=0’ vertex and all other ‘act_stat=0’ vertices connected to it directly or via ‘act_stat=1’.

But how do I find all the ‘act_stat=0’ vertices connected to the source ‘act_stat=0’ vertex? Should I first run some kind of depth-first search to find a list of all ‘act_stat=0’ vertices and then for each vertex in that list, run a Djikstra search to find the path of minimum distance to that vertex?

This is not really a graph-tool question, but a general graph algorithm problem.

In any case, you can use graph filtering. For each source and target of type 0, you change them to type 1, and you remove all the remaining nodes of type 0. Then you compute the shortest path as usual. The direct connections of type 00 you consider separately.

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Thank you for this useful suggestion!