nav[aria-label="Primary Navigation"] { padding: 0; & ul { list-style: none; width: 100%; display: flex; flex-direction: row; justify-content: start; align-items: start; gap: 30px; padding: 0; & li { margin: 0; } & ul li { list-style: none; } } }

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

Anonymous User

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)

Find more posts tagged with

Accepted answers

Anonymous User

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

D_Hornby_ddh_uni

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?

All comments

DestinyKelley1

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)

anyry

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

Anonymous User

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

D_Hornby_ddh_uni

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?

Join the Conversation

Signed in members can post, follow updates, and more. New here? Register a free account.

Quick Links

Community Help
Find useful guides, FAQs, and documents to help you navigate and make the most of Esri Community.