[Bmesh] It has an easy way to know if two vertices belong to an edge?

Coding Python Support
1

I have two vertices. And I have a script that says if they belong or not belong to an edge:

if V2 in [x for y in [a.verts for a in V1.link_edges] for x in y if x != V1]:


But I do not know if this is a good solution.
My question is: have a better solution?

2

It might be more efficient to try to create an edge between the two verts. If there already is one, an exception will be raised. Otherwise, an edge will be created - thus, you need to remove that edge again.

3

Thanks for the tip :slight_smile: @CoDEmanX
I decided to do a performance test:

import bpy, bmesh, timeit

def test_edge1(V1, V2):
    try:
        edge = bm.edges.new([V1, V2])
    except ValueError:
        print('test_edge1: this edge exists')


def test_edge2(V1, V2):
    if V2 in [x for y in [a.verts for a in V1.link_edges] for x in y if x != V1]:
        print('test_edge2: this edge exists')


obj = bpy.context.edit_object
me = obj.data
bm = bmesh.from_edit_mesh(me)


V1 = bm.edges[0].verts[0]
V2 = bm.edges[0].verts[1]


if __name__ == '__main__':
    test1 = timeit.timeit("test_edge1(V1, V2)", setup="from __main__ import test_edge1, V1, V2", number=100)
    test2 = timeit.timeit("test_edge2(V1, V2)", setup="from __main__ import test_edge2, V1, V2", number=100)
    print('resultado:', test1/test2, 'diferenca:', test1-test2)


#bmesh.update_edit_mesh(me, True)

And the result is that really your solution presents a vanishingly performance a little better. It depends on a few factors in mesh.
But I will still continue with the first solution. Because you do not need to remove the edge.