Faster way to build a tree graph using Networkx Python?

Question

Is there any faster, nicer way of building a Networkx tree. Currently, my code is

for numb in range(0,len(previous)):
    nodos = list(chunks(current,3))
        for i in range(0,3):
            G.add_edge(previous[numb],nodos[numb][i])

This works in the following way: 1. The tree has 3 branches (or edges). I have two arrays:

previous = [x,y,z] #These nodes have already been added to the graph
current = [xx,xy,xz, xy,yy,yz, xz,yz,zz] #This nodes need to be added.

Ideally, I should do the following:

1. Start with x in previous:
1.1 Pick the first 3 nodes in current (i.e. xx,xy,xz)
1.1.1 Add the nodes-edges: x->xx, x->xy, x->xz

So far my codes does:

1. Start with x in previous
2. Partition current into chunks of 3 items: [[xx,xy,xz], [xy,yy,yz], [xz,yz,zz]]
3. Loop through all the nodes in these chunks:
4. Add x->xx, loop again, add x->xy... etc.

My implementation is extremely inefficient. How would you do this efficiently? Thank you

Aric · Accepted Answer · 2014-11-14 04:15:09Z

3

You could use the helper function from https://github.com/networkx/networkx/blob/master/networkx/generators/classic.py#L50

def _tree_edges(n,r):
    # helper function for trees
    # yields edges in rooted tree at 0 with n nodes and branching ratio r
    nodes=iter(range(n))
    parents=[next(nodes)] # stack of max length r
    while parents:
        source=parents.pop(0)
        for i in range(r):
            try:
                target=next(nodes)
                parents.append(target)
                yield source,target
            except StopIteration:
                break

print list(_tree_edges(13,3))
#[(0, 1), (0, 2), (0, 3), (1, 4), (1, 5), (1, 6), (2, 7), (2, 8), (2, 9), (3, 10), (3, 11), (3, 12)]
import networkx as nx
G = nx.Graph(_tree_edges(13,3))

If you want nodes other than integers you can either relabel or specify them on input like this

def _tree_edges(nodes,r):
    # helper function for trees
    # yields edges in rooted tree with given nodes and branching ratio r
    nodes=iter(nodes)
    parents=[next(nodes)] # stack of max length r
    while parents:
        source=parents.pop(0)
        for i in range(r):
            try:
                target=next(nodes)
                parents.append(target)
                yield source,target
            except StopIteration:
                break

nodes = list('abcdefghijklm')
print list(_tree_edges(nodes,3))

edited Nov 14, 2014 at 4:15

answered Nov 12, 2014 at 21:10

Aric

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1 Comment

Oniropolo Over a year ago

Thank you! I didnt know this existed. But I still have a problem, this works great if n,r are numbers, but in my case they are letters. so [(x,xx),(x,xy),(x,xz),(y,xy) .. and so on]. But I cant use this function if r is not a number. Any ideas of how to get around this?

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