CODEWITHSUNDEEP
pythongraph-theoryigraphdirected-graph
ahajib
ahajib

Reputation: 13510

get adjacency matrix of a directed graph in python-igraph

My question might sound stupid but how can I get the adjacency matrix of a directed graph?

Let me explain. I have the following code:

g = Graph()
g.add_vertices(5)
g.add_edges([(0,1),(1,2)])
print(g)
print(g.get_adjacency())

The first print returns:

IGRAPH U--- 5 2 --
+ edges:
0--1 1--2

While the second one returns:

[[0, 1, 0, 0, 0]
 [1, 0, 1, 0, 0]
 [0, 1, 0, 0, 0]
 [0, 0, 0, 0, 0]
 [0, 0, 0, 0, 0]]

What if I want my graph to be directed and not undirected?

Thanks for the help

Upvotes: 0

Views: 5751

Answers (1)

ali_m
ali_m

Reputation: 74172

You need to initalize your graph as directed before you start adding edges to it:

gd = Graph(directed=True)
gd.add_vertices(5)
gd.add_edges([(0,1),(1,2)])
print(gd.get_adjacency())
# [[0, 1, 0, 0, 0]
#  [0, 0, 1, 0, 0]
#  [0, 0, 0, 0, 0]
#  [0, 0, 0, 0, 0]
#  [0, 0, 0, 0, 0]]

The .to_directed() method that Matthew mentioned won't do what you want, because it creates edges in both directions for each undirected edge:

print(g.is_directed())    # your original undirected example graph
# False

g.to_directed()

print(g.is_directed())
# True

print(g.get_adjacency())  # adjacency matrix is still symmetric
# [[0, 1, 0, 0, 0]
#  [1, 0, 1, 0, 0]
#  [0, 1, 0, 0, 0]
#  [0, 0, 0, 0, 0]
#  [0, 0, 0, 0, 0]]

You can prevent both directed edges from being created by passing .to_directed(mutual=False), but in this case igraph just picks an arbitrary single direction for each edge. The fundamental problem is that information about the directions of edges is simply not stored if the graph is undirected.

Upvotes: 2

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