3
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I am trying to find the shortest code that generates a full mesh of the elements fed into it in graphviz format e.g. the full mesh of [0, 1, 2] is:

2 -> 0

2 -> 1

1 -> 0

I have the code below and I think there might be a way to combine the if and for loops, and avoid needing to directly create the empty string

def graph(l):
    if not l:
        return ""
    i = l.pop()
    out = ""
    for z in l:
        out += "  {}->{}\n".format(i, z)
    return out + graph(l)
print graph(range(3))
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  • \$\begingroup\$ Did you hear about meshpy ? Does it help you ? \$\endgroup\$ – Grajdeanu Alex. Aug 18 '16 at 18:11
  • \$\begingroup\$ I'm not trying to generate a mesh in the model sense but a mesh in the network topology sense where every node is connected to every other node, mostly I want to see how mutch smaller I can make the code \$\endgroup\$ – Mark Omo Aug 18 '16 at 18:13
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If I get it correctly, the shortest way that I can think of is by using itertools.combinations. Basically, you want all unique combinations, so:

from itertools import combinations

def mesh(L):
    return ["->".join(map(str, comb)) for comb in combinations(L, 2)]

Which will return:

['1->2', '1->3', '2->3']

For: mesh([1, 2, 3])

If you then want to print them as above, just:

for x in mesh([1, 2, 3]):
    print(x)
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