Python DFS Shortest Path Search with weighted, undirected graph

Question

I'm having trouble parsing through this graph dictionary:

routes = {'a': [('b', 5.0), ('c', 8.0)], 'c': [('a', 8.0), ('d', 2.0)], 
'b' [('a', 5.0), ('d', 6.0)], 'e': [('d', 12.0), ('g', 3.0)], 
'd':  [('b', 6.0),('c', 2.0), ('e', 12.0), ('f', 2.0)], 'g': [('e', 3.0),
('f', 7.0)],'f': [('d',2.0), ('g', 7.0)]}

How do I separate out the values of each edge while running the dictionary through a DFS Search looking at 2 keys? I'm not very familiar with dict.

So far I have,

def dfs(graph, start, end, path):
    path = path + [start]
    if start == end: 
        paths.append(path)
    for node in graph.childrenOf(start):
        if node not in path:
            dfs(graph, node, end, path)

I need to return the smallest weighted path, so I need the numbers in the values separated out and summed as the program runs.

What part exactly of the dictionary you want? Keys, values or part of the values (which parts)? — nbro
– nbro, Commented Feb 6, 2015 at 21:12
Keys are the colon left side stuff and the right side stuff are, you guessed, the values — nbro
– nbro, Commented Feb 6, 2015 at 21:17
I need to return the smallest weighted path, so I need the numbers in the values separated out and summed as the program runs. — N.B.48
– N.B.48, Commented Feb 6, 2015 at 21:18
defaultdict(dict) is convenient. If you need something faster, there are other data structures which may be slightly less convenient but much faster. — Brian Cain
– Brian Cain, Commented Feb 6, 2015 at 21:21

Anderson Vieira · Accepted Answer · 2015-02-06 21:23:27Z

1

You can build your graph using a dict of dicts:

routes = {'a': {'b': 5.0, 'c': 8.0}, 'c': {'a': 8.0, 'd': 2.0}}

Then routes['a']['b'] will return the weight, which is 5.0 in this case. If you need to get all the children of a node you can do routes['a'].keys() and that will return ['c', 'b'].

edited Feb 6, 2015 at 21:23

answered Feb 6, 2015 at 21:17

Anderson Vieira

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tzaman · Accepted Answer · 2015-02-06 21:20:01Z

0

routes is your graph, in adjacency-list format. Each key is a node, the value being a list of (other_node, edge_weight) tuples. So what you're looking for is how "graph.childrenOf" should work.

Assuming start, end are simple string node names (like 'a') you can just look them up in the dictionary; e.g. graph[a] will give you [('b', 5.0), ('c', 8.0)]. You can iterate directly over both the node and weight using python's nifty built-in tuple unpacking, like so:

# replace your for loop with this:
for node, weight in graph[start]:

answered Feb 6, 2015 at 21:20

tzaman

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