Export matplotlib triangle mesh to an asymptote surface

Question

This is a followup of a question of mine at stackoverflow partially reproduced here for clarity.

I'm playing a bit with scikit-image marching cubes algorithm. Here is a simplified version of the example given in the docs.

from matplotlib import pyplot as plt
import numpy as np
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
from skimage import measure


x = np.linspace(-1, 1, 11)

X, Y, Z = np.meshgrid(x, x, x, indexing = 'ij')

def f(x, y, z):
    return x

verts, faces = measure.marching_cubes(f(X, Y, Z), 0.6)

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

mesh = Poly3DCollection(verts[faces])
ax.add_collection3d(mesh)

ax.set_xlim(0, 10)
ax.set_ylim(0, 10)
ax.set_zlim(0, 10)

plt.show()

Here is the resulting surface:

Now I'd like to draw the surface with asymptote. So basically, I have to find a way to export the coordinates of the triangles. Easy: each triangle is an element of verts[faces] represented by 3 lists, the elements being the 3 coordinates of each vertice. So I can parse verts[faces] and write to a file path3 triangle = (x1, y1, z1) -- (x2, y2, z2) -- (x3, y3, z3) -- cycle for each triangle. Then, for each triangle, I draw the surface: draw(surface(triangle)). But is there a better way to define the surface for asymptote?

Note that in real life, the surface is far more complex than a simple plane :)