I'm trying to determine if an object is a linked duplicate in a script.

My test scene is as simple as possible, 2 cubes, one a linked duplicate of another.

The following I have tried to no avail:

  • bpy.data.objects["Cube.001"].library (returns None)
  • bpy.data.objects["Cube.001"].original (returns "Cube.001")
  • bpy.data.objects["Cube.001"].is_property_set("linked") (fails to run)

I attempted to grep around in the source code to find a solution but I, unfortunately, couldn't find anything that worked.

  • $\begingroup$ bpy.objects["Foo"].data $\endgroup$
    – batFINGER
    Jul 23 '20 at 19:33


Linked duplicates will have the same linked data, if there is linked data. For an empty ob.data is None

import bpy
from collections import defaultdict

scene = bpy.context.scene

data_links = defaultdict(list)
for ob in scene.objects:
for k, v in data_links.items():
    print(k, v)

Sample output, notice there are 4 objects using the mesh "Cube.001"

<bpy_struct, Mesh("Cube.001")> [bpy.data.objects['Cube'], bpy.data.objects['Cube.001'], bpy.data.objects['Cube.002'], bpy.data.objects['Cube.003']]
<bpy_struct, PointLight("Lamp")> [bpy.data.objects['Lamp']]
<bpy_struct, Camera("Camera")> [bpy.data.objects['Camera']]
None [bpy.data.objects['Empty'], bpy.data.objects['Field']]
  • $\begingroup$ My first brush with defaultdict. Python is very neat and tidy, sometimes.. :) $\endgroup$ Jul 24 '20 at 13:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.