topic Re: How do I create a new label class using arcpy? in Python Questions

How do I create a new label class using arcpy?

Per_Åke_MattiasWallin — Fri, 31 Jul 2020 01:01:52 GMT

I am using ArcPro 2.6 and I can modify existing label classes no problem. But what I want to do is to create a new label class using arcpy- I just cannot figure it out, what am I missing? All example’s I come across starts with listLabelClasses which assumes the class already exists.

Thanks.

Re: How do I create a new label class using arcpy?

DanPatterson — Fri, 31 Jul 2020 01:49:22 GMT

From here

Layer—ArcGIS Pro | Documentation

listLabelClasses ({wildcard})

Returns a Python list of LabelClass objects in a layer.

Which means you have to have something to create the labels from, and as you know...

LabelClass—ArcGIS Pro | Documentation
with the warning

Not all layers support labeling, so it is useful to test this ahead of time using the supports method on the Layer object. For example:
if lyr.supports("SHOWLABELS"):

where you can modify them.

But I don't know what you would use as a base except for an existing file

Re: How do I create a new label class using arcpy?

Per_Åke_MattiasWallin — Fri, 31 Jul 2020 23:43:26 GMT

Thank you for your reply Dan.

I was able to create a new label class using this:

# Create a new label class
new_label_class = arcpy.cim.CreateCIMObjectFromClassName('CIMLabelClass','V2')
new_label_class.name = 'Class 2'
cim_industry.labelClasses.append(new_label_class)

However... whenever I try to use the layer that I created the new label class on it crashes ArcPro so something gets corrupted... which they do wanrn about here, Python CIM access—ArcGIS Pro | Documentation

Re: How do I create a new label class using arcpy?

Per_Åke_MattiasWallin — Tue, 22 Jun 2021 08:26:13 GMT

I just noticed that there is a feature request https://community.esri.com/ideas/6533 from 2012 with 9 votes...

Re: How do I create a new label class using arcpy?

Anonymous User — Wed, 11 Nov 2020 02:04:37 GMT

I am also noticing the same thing. I use arcpy.cim.CreateCIMObjectFromClassName('CIMLabelClass','V2') and then append it into the list of label classes. When I go to interact with the new label class in the GUI ArcPro immediately crashes.

Re: How do I create a new label class using arcpy?

Nicholas_G — Tue, 02 Mar 2021 06:33:02 GMT

It may be easier to try the arcpy.cim.CIMLabelClass() to create a new label class

Essentially I think you can get your layer object and export out a version of the cim object like this.

cimObject = LayerObj.getDefinition(‘v2’)

From here I would use arcpy.cim.CIMLabelClass() to create a new label placement object and set all of the properties to your liking.

Then append this new object to the cimObject label class list.

Something like cimObject.labelClasses.append(new label class obj)

Check the cimObject.labelClasses to see if its in there.

Then go ahead and set its definition like this

LayerObj.setDefinition(cimObject)

This should work, I don’t think you will see it in the Pro interface under the labelling tab but the label class should be added to your layer object.

Re: How do I create a new label class using arcpy?

peter_poti_sk — Wed, 28 Jun 2023 06:52:02 GMT

Hi, have you finally solved problem with adding new label class?

Re: How do I create a new label class using arcpy?

DestinyKelley1 — Thu, 27 Jul 2023 20:17:53 GMT

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().

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)

Hope this helps other users!