2

I have done a lot of research and cannot seem to get the following expression right. I have three (3) attribute columns - ID, Diameter, and Material, I would like to label sewer lines with the unique id, diameter, and material. E.g. GS1945 12" VCP. Ultimately the script would leave out any information the is NULL.

def FindLabel ([ID],[Diameter],[Material])
     if IsNull([Diameter]) and IsNull([Material]) then
          return [ID] 
     elif IsNull([Diameter]) then
          return ([ID] + " " + [Material])
     elif IsNull([Material]) then
          return ([ID] + " " + [Diameter] + "''" )
     else:
          return ([ID] + " " + [Diameter] + "''" + [Material])
  • As identified by @Bjorn you are mixing scripting languages and have incorrect syntax. His answer pretty much nails it. – Hornbydd Jun 6 '17 at 22:54
6

Your code appears to be a mix of VBScript and Python. Also, use variables as parameters in a function definition instead of attribute identifiers. IsNull is not a Python command; use a boolean test instead. Making those corrections, and assuming the 3 parameters are strings, your code would look like:

def FindLabel(id, diameter, material):
    if not diameter and not material:
        return id 
    elif not diameter:
        return (id + " " + material)
    elif not material:
        return (id + " " + diameter + "''" )
    else:
        return (id + " " + diameter + "''" + material)

By further exploiting variables, concatenation, and string formatting, the code can be simplified to:

def FindLabel(id, diameter, material):
    label = id
    if diameter:
        label += " {}''".format(diameter)
    if material:
        label += " {}".format(material)
    return label
  • This is generally correct. I had to change a few minor things, I assume to work with ArcGIS's Label Expression Python Parser over directly with Python. Do you recommend any reading resources to understand formatting, etc. I will likely answer my own question with the correct format that worked for me, unless there is another appropriate way to handle this. – LandArch Jun 7 '17 at 13:19
  • If your ultimate solution is significantly different, you should post it as an answer, and mark it as the correct one. – Bjorn Jun 15 '17 at 20:19
1

@Bjorn was generally correct above. I needed to tweak things slightly, I assume to pork with ArcGIS's Python Parser over straight Python. The expression that worked for me is as follows:

def FindLabel ([ID], [Diameter], [Material]):
    if not [Diameter] and not [Material]:
        return [ID] 
    elif not [Diameter]:
        return ([ID] + " " + [Material])
    elif not [Material]:
        return ([ID] + " " + [Diameter] + "''")
    else:
        return ([ID] + " " + [Diameter] + "''" + " " + [Material])

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