Mykhailo

Thank you for your help!

How can I split an ifc file, for example, that consists of six walls into six ifc files? I tried to do it as follows: for i in range(len(assembly_walls)): element_id = assembly_walls[i].id() model = ifcopenshell.file(schema=ifc_object.schema) model.add(ifc_object.by_type('IfcElementAssembly')[i]) model.write(output_file_path). However, I was getting only this part: ISO-10303-21; HEADER; FILE_DESCRIPTION(('ViewDefinition [CoordinationView]'),'2;1'); FILE_NAME('','2024-04-29T19:34:38',(),(),'IfcOpenShell...