How can I parse "package://" in a URDF file path?

Question

I have a robot URDF that points to mesh files using "package://".

  <geometry>
    <mesh filename="package://a1_rw/meshes/hip.dae" scale="1 1 1"/>
  </geometry>

I would like to use urdfpy to parse this URDF. However, it is unable to interpret the meaning of "package://".

import os
from urdfpy import URDF

a1_rw = {
    "model": "a1",
    "csvpath": "a1_rw/urdf/a1_rw.csv",
    "urdfpath": "a1_rw/urdf/a1_rw.urdf"
}

model = a1_rw
curdir = os.getcwd()
path_parent = os.path.dirname(curdir)
print("path parent = ", path_parent)
model_path = model["urdfpath"]
robot = URDF.load(os.path.join(path_parent, model_path))

Here is the error message:

$ python3.8 calc_parallax.py
path parent =  /home/ben/Documents/git_workspace/a1_test
Traceback (most recent call last):
  File "calc_parallax.py", line 18, in <module>
    robot = URDF.load(os.path.join(path_parent, model_path))
  File "/home/ben/.local/lib/python3.8/site-packages/urdfpy/urdf.py", line 3729, in load
    return URDF._from_xml(node, path)
  File "/home/ben/.local/lib/python3.8/site-packages/urdfpy/urdf.py", line 3926, in _from_xml
    kwargs = cls._parse(node, path)
  File "/home/ben/.local/lib/python3.8/site-packages/urdfpy/urdf.py", line 161, in _parse
    kwargs.update(cls._parse_simple_elements(node, path))
  File "/home/ben/.local/lib/python3.8/site-packages/urdfpy/urdf.py", line 137, in _parse_simple_elements
    v = [t._from_xml(n, path) for n in vs]
  File "/home/ben/.local/lib/python3.8/site-packages/urdfpy/urdf.py", line 137, in <listcomp>
    v = [t._from_xml(n, path) for n in vs]
  File "/home/ben/.local/lib/python3.8/site-packages/urdfpy/urdf.py", line 181, in _from_xml
    return cls(**cls._parse(node, path))
  File "/home/ben/.local/lib/python3.8/site-packages/urdfpy/urdf.py", line 161, in _parse
    kwargs.update(cls._parse_simple_elements(node, path))
  File "/home/ben/.local/lib/python3.8/site-packages/urdfpy/urdf.py", line 137, in _parse_simple_elements
    v = [t._from_xml(n, path) for n in vs]
  File "/home/ben/.local/lib/python3.8/site-packages/urdfpy/urdf.py", line 137, in <listcomp>
    v = [t._from_xml(n, path) for n in vs]
  File "/home/ben/.local/lib/python3.8/site-packages/urdfpy/urdf.py", line 1146, in _from_xml
    kwargs = cls._parse(node, path)
  File "/home/ben/.local/lib/python3.8/site-packages/urdfpy/urdf.py", line 161, in _parse
    kwargs.update(cls._parse_simple_elements(node, path))
  File "/home/ben/.local/lib/python3.8/site-packages/urdfpy/urdf.py", line 127, in _parse_simple_elements
    v = t._from_xml(v, path)
  File "/home/ben/.local/lib/python3.8/site-packages/urdfpy/urdf.py", line 181, in _from_xml
    return cls(**cls._parse(node, path))
  File "/home/ben/.local/lib/python3.8/site-packages/urdfpy/urdf.py", line 161, in _parse
    kwargs.update(cls._parse_simple_elements(node, path))
  File "/home/ben/.local/lib/python3.8/site-packages/urdfpy/urdf.py", line 127, in _parse_simple_elements
    v = t._from_xml(v, path)
  File "/home/ben/.local/lib/python3.8/site-packages/urdfpy/urdf.py", line 581, in _from_xml
    meshes = load_meshes(fn)
  File "/home/ben/.local/lib/python3.8/site-packages/urdfpy/utils.py", line 225, in load_meshes
    meshes = trimesh.load(filename)
  File "/home/ben/.local/lib/python3.8/site-packages/trimesh/exchange/load.py", line 111, in load
    ) = parse_file_args(file_obj=file_obj,
  File "/home/ben/.local/lib/python3.8/site-packages/trimesh/exchange/load.py", line 623, in parse_file_args
    raise ValueError('string is not a file: {}'.format(file_obj))
ValueError: string is not a file: /home/ben/Documents/git_workspace/a1_test/a1_rw/urdf/package://a1_rw/meshes/trunk.dae

Is there any way to get urdfpy (or another urdf parser) to parse this correctly?