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I have a parcel layer and a trees layer, I want to label each parcel with the number of trees it contains and the sum of DBH. I wrote a script that calculates this and it works stand-alone, but I can't figure out how to get it to work with a Label Expression. Here is the code I've tried (ArcMap 10.2):

import arcpy
import numpy
def FindLabel():
  mxd = arcpy.mapping.MapDocument("CURRENT")
  parcels = arcpy.mapping.ListLayers(mxd, "Parcel*")[0]
  trees = arcpy.mapping.ListLayers(mxd, "Surveyed Trees*")[0]
  arcpy.SelectLayerByLocation_management(parcels, "CONTAINS", trees, "#", "NEW_SELECTION")
  arcpy.SelectLayerByLocation_management(trees, "WITHIN", parcels, "#", "NEW_SELECTION") 
  with arcpy.da.SearchCursor(parcels, ["OBJECTID"]) as cur:
    for row in cur:
      arcpy.SelectLayerByLocation_management(trees, "WITHIN", parcels, "#", "NEW_SELECTION")
      treeCount = int(arcpy.GetCount_management(trees).getOutput(0))
      field = arcpy.da.TableToNumPyArray(trees, ("DBH"), skip_nulls = True)
      sum = field["DBH"].sum()
      message = "Trees = " + str(treeCount) + "\n" + "DBH = " + str(sum)
      return message

This sends ArcMap into a long loop and then seems to return the last calculated value for each parcel, so everything is labeled "1 Tree 7 DBH".

Maybe I shouldn't be using a SearchCursor?

Solution

Here's the code I ended up using. It's still processor intensive, but this fits our workflow fine.

import arcpy
def FindLabel ( [Loc_Name] ):
  mxd = arcpy.mapping.MapDocument("CURRENT")
  trees = arcpy.mapping.ListLayers(mxd, "ProgramSurveyGround*")[0]
  arcpy.MakeTableView_management(trees, "treesTable")
  parcels = arcpy.mapping.ListLayers(mxd, "ParcelProcessing*")[0]
  Query = ' "Loc_Name"' + " = '" + [Loc_Name] + "'"
  treeSelect = arcpy.SelectLayerByAttribute_management("treesTable", "NEW_SELECTION", Query)
  treeCount = int(arcpy.GetCount_management("treesTable").getOutput(0))
  dbhArray = arcpy.da.TableToNumPyArray("treesTable", ('DBH'), skip_nulls = True)
  dbhCount = dbhArray['DBH'].sum()
  myText = "Trees : " + str(treeCount) + "\n" + "DBH : " + str(dbhCount)
  return myText
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    I would have done a Spatial Join using the polygon as a target and points as the source to create the "count" field and a Sum of the DBH within the parcels?. I'm not sure I see how the labels would work for you. Why not just keep the points and label those? Sep 14, 2015 at 15:14
  • Completely agree with @jbchurchill, your existing approach is very inefficient. Do a spatial join, if you don't want "yet another" dataset, simply copy the count field into your existing polygon featureclass.
    – Hornbydd
    Sep 14, 2015 at 19:16
  • I'm looking for a way of labeling these polygons dynamically because we get data updates all the time. Maybe a script the uses a spatial join is best but I'm trying to avoid that. For now, I don't see why I can run this as a stand alone script and get the values I'm looking for but can't get them to appear as labels. Any other suggestions would be appreciated.
    – KJYDavis
    Sep 14, 2015 at 19:34

1 Answer 1

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Three calls to arcpy.SelectLayerByLocation_management and one call to arcpy.da.TableToNumPyArray!

Suppose you have solved the bug in your code but calling 4 process costly geoprocessing functions in a label expression is not a rational choice and doesn't follow the best practice.

  1. You can either pre-process the data and use spatial join (Merge Rule = Count)
  2. Create relationship classes between the two layer and count the number of related features and show as label. This is both dynamic and process friendly. However you should take care of point feature deletion and addition and auto relate the point feature to the polygon upon different update events.
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  • If someone can point me in the direction of solution 2, it would be appreciated. I've created a relationship class and can see the related features using the info tool, but can't figure out how to get a count of related features without doing a similarly inefficient set of geoprocessing operations. I looked here, but this doesn't look any better than what I had link
    – KJYDavis
    Sep 15, 2015 at 13:28
  • I would recommend to close this question (e.g. by marking as answer) and ask a new question and put your source code in the new question, then we can tell you what is the problem with your code. Sep 15, 2015 at 13:34

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