DFS on a graph using a python generator

I am using a generator to do a full search on a graph, the real data set is fairly large, here is a portion of the code i wrote on a small data set:

class dfs:
    def __init__(self):
        self.start_nodes = [1,2]  # not used, see below explanation
        self.end_nodes = [5,7] # not used, see below explanation
    _graph={
        1 : [4,5,6,2],
        2 : [1,3,5],
        4 : [1,5],
        3 : [5,2],
        5 : [3,1,4,6,2,7],
        6 : [1,5],
        7 : [5],
    }

    def __iter__(self):
        return self.find_path(self._graph, 2, 7)

    def find_path(self, graph, start, end, path=[]):
        path = path + [start]
        if start == end:
            yield path
        if not graph.has_key(start):
            return
        for node in graph[start]:
            if node not in path:
                for new_path in self.find_path(graph, node, end, path):
                    if new_path:
                        yield new_path


d = dfs()
print list(d)

when run this outputs all the paths from '2' to 开发者_如何学运维'7' as expected:

[[2, 1, 4, 5, 7], [2, 1, 5, 7], [2, 1, 6, 5, 7], [2, 3, 5, 7], [2, 5, 7]]

What I would like to do is modify this generator so that it does the same thing except i get the paths back for a set number of start and end points, ie self.start_nodes and self.end_nodes.

Since my generator is a recursive function it makes it difficult to loop on the different start and end points, been scratching my head over this any ideas?

Perhaps I'm misunderstanding your question, but it seems to me that you want to replace your __iter__ function with something like this:

def __iter__(self):
    for start in self.start_nodes:
        for end in self.end_nodes:
            for path in self.find_path(self._graph, start, end):
                yield path

You can find more information about generators in this question.