topic Re: Creating a new label class in Arc Pro using the arcpy CIM in Python Questions

Creating a new label class in Arc Pro using the arcpy CIM

Anonymous User — Sat, 11 Dec 2021 22:24:15 GMT

I am attempting to create a new label class for a layer using the arcpy CIM, code below. The new label class gets created, but ArcGIS Pro immediately crashes when I attempt to interact with it in the GUI. I have tried in the python window and in a script.

aprx = arcpy.mp.ArcGISProject('CURRENT')
m = aprx.listMaps("Map")[0]
lyr = m.listLayers("SOME LAYER")[0]
ldef = lyr.getDefinition('V2')

newlblclass = arcpy.cim.CreateCIMObjectFromClassName('CIMLabelClass','V2')
newlblclass.name = "NEW CLASS"
newlblclass.visibility = True
ldef.labelClasses.append(newlblclass)
lyr.setDefinition(ldef)

Re: Creating a new label class in Arc Pro using the arcpy CIM

DuncanHornby — Thu, 12 Nov 2020 12:06:16 GMT

May be there are required properties that need setting? All you set is the name and visibility, but what about the field that the label is derived from?

Re: Creating a new label class in Arc Pro using the arcpy CIM

Anonymous User — Mon, 16 Nov 2020 18:49:09 GMT

Correct. After researching there are many "nested" CIMobjects that make up a label class cim object. All of these objects need to be created and assigned in order for the label class object to function properly.

Reference: https://pro.arcgis.com/en/pro-app/arcpy/mapping/python-cim-access.htm

Re: Creating a new label class in Arc Pro using the arcpy CIM

anyry — Tue, 11 Jan 2022 15:02:01 GMT

Can you please share your label class creation solution? I have the same problem and I can't solve it. Thanks

Re: Creating a new label class in Arc Pro using the arcpy CIM

DestinyKelley1 — Fri, 28 Jul 2023 13:08:44 GMT

My solution is different, and may be of help to you if you are still experiencing this issue. I have figured out a way to bypass the corruption of the CIM definition when attempting to create a new label class with ArcPy CIM. To test this out, you can use any layer those default class is named "Class 1." Instead of using arcpy.cim.CreateCIMObjectFromClassName('CIMLabelClass','V2') or arcpy.cim.CIMLabelClass() as suggested by other users, I used cimObject.labelClasses.copy().

It's a little awkward, but I hope it helps!

The code to try out is below:

import arcpy p = arcpy.mp.ArcGISProject('CURRENT') m = p.listMaps('Map')[0] layer = m.listLayers("Your Layer Name Here")[0] cimObject = layer.getDefinition('V2') lblClasses = cimObject.labelClasses for lblClass in lblClasses: if lblClass.name == "Class 1": nlc_index = cimObject.labelClasses.index(lblClass) lblClass_copy = cimObject.labelClasses.copy()[nlc_index] cimObject.labelClasses.append(lblClass_copy) layer.setDefinition(cimObject) break cimObject = layer.getDefinition('V2') lblClasses = cimObject.labelClasses for lblClass in lblClasses: if lblClass.name == "Class 1": nlc_index = cimObject.labelClasses.index(lblClass) if nlc_index != 0: lblClass.name = "Dummy" layer.setDefinition(cimObject)

Re: Creating a new label class in Arc Pro using the arcpy CIM

JingchaoZhou — Tue, 28 Nov 2023 18:13:09 GMT

Just wondering if you can change the label class name before you append it in the first for loop

something like this

for lblClass in lblClasses: if lblClass.name == "Class 1": nlc_index = cimObject.labelClasses.index(lblClass) lblClass_copy = cimObject.labelClasses.copy()[nlc_index] lblClass_copy.name = "new_class_name" cimObject.labelClasses.append(lblClass_copy) layer.setDefinition(cimObject) break

I have not tested this, but just wondering if this will optimize the code and avoid the second for loop....