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; } } }

Stacked Label Expression in Python

BrynDunbar

I have successfully rendered a stacked label in ArcMap using a java script labeling expression, but I cannot figure out how to do the same thing with Python in the Label Expression box.

This is what I used for java script:

function FindLabel ( [LOT], [SUBDIVISION] )

{

return [LOT] + "\r" + [SUBDIVISION];

}

How do I do this same thing by using python?

I just want to stack the my Lot label on top of my Subdivision label, but using python.

Thank you in advance for any thoughts or suggestions.

Find more posts tagged with

Accepted answers

XanderBakker

You could change the expression to:

def FindLabel([LOT] ,[SUBDIVISION]):
  return "{0}\n{1}".format([LOT], [SUBDIVISION])

It suppresses the error. However, this will return None if there is no data:

The nice part of using the python expression is that you can enhance the result, replacing the None for a different text

def FindLabel([LOT] , [SUBDIVISION]):
  return "{0}\n{1}".format([LOT] if [LOT] != None else "No LOT",  [SUBDIVISION] if [SUBDIVISION] != None else "No SUBDIVISION")

... will result in:

... or adding some formatting to the text:

def FindLabel([LOT] , [SUBDIVISION]):
  return "{0}\n{1}".format([LOT] if [LOT] != None else "<CLR red='255'>No LOT</CLR>",  [SUBDIVISION] if [SUBDIVISION] != None else "<CLR red='255'>No SUBDIVISION</CLR>")

will give:

All comments

DarrenWiens2

Although it sounds like you've already found an answer to the problem, but you could take advantage of try/except to label, or skip, null values:

def FindLabel ( [LOT], [SUBDIVISION] ):
  try:
    return [LOT] +  "\r"  +  [SUBDIVISION]
  except:
    return 'this had a null value'

Another succinct way to write this is:

def FindLabel ( [LOT], [SUBDIVISION] ):
  if [LOT] and [SUBDIVISION]:
      return [LOT] + '\r' + [SUBDIVISION]

BrynDunbar

Thanks Christian! Your response and this thread definitely helped me find the solution that I was looking for. This is what I have so far.

def FindLabel ([LOT], [SUBDIVISION]):

if [LOT]:

if [SUBDIVISION]:

label = [LOT] + '\n' + [SUBDIVISION]

else:

label = " "

return label

It labels Lots and Subdivisions, but when they are Null, you can't see the label because I made it " " say nothing but a space. I am sure there are better ways to write the code, but this seems to be doing what I want for now.

Thank you again.

BrynDunbar

Thank you so much for taking the time to respond with such great ideas and examples. I will definitely spend more time with label expressions, using these examples to help me construct more complex labels. I am still quite new to coding, but it is quite satisfying when I can understand some code and when I can write code that produces something, especially with no errors.

XanderBakker

I wouldn't go for label classes in this case, since in the example above, with a simple line I am tackling 4 classes,,,

XanderBakker

You could change the expression to:

def FindLabel([LOT] ,[SUBDIVISION]):
  return "{0}\n{1}".format([LOT], [SUBDIVISION])

It suppresses the error. However, this will return None if there is no data:

The nice part of using the python expression is that you can enhance the result, replacing the None for a different text

def FindLabel([LOT] , [SUBDIVISION]):
  return "{0}\n{1}".format([LOT] if [LOT] != None else "No LOT",  [SUBDIVISION] if [SUBDIVISION] != None else "No SUBDIVISION")

... will result in:

... or adding some formatting to the text:

def FindLabel([LOT] , [SUBDIVISION]):
  return "{0}\n{1}".format([LOT] if [LOT] != None else "<CLR red='255'>No LOT</CLR>",  [SUBDIVISION] if [SUBDIVISION] != None else "<CLR red='255'>No SUBDIVISION</CLR>")

will give:

deleted-user-rmQdIupSdOqe

The other option is to use the labeling method "Define classes of features and label each class differently" option, setting up a SQL expression first, to filter out data that has NULL values.

sql for null values.JPG

Then your python label expression for that label class (my label class is Default) is:

stacked label python example.JPG

deleted-user-rmQdIupSdOqe

It is the NULL values in the fields that make these python expressions error out. The answer above solves that problem.

I have one that looks like this, where sometimes [ContractName] and [ContractYear] are null, so I nest in the if statements.

def FindLabel ([ContractName], [ContractYear]):

if [ContractName]:

if [ContractYear]:

label = [ContractName] + '\n' + [ContractYear]

else:

label = "No" + '\n' + "Contract"

return label Stacked Python Labels.JPG

ChristianHinderman

I had a similar problem recently. I think one of the label fields must have a null value. If you only have to worry about null values in subdivision (and both fields are strings) this should work:

def FindLabel ( [LOT], [SUBDIVISION] 😞

if [SUBDIVISION]:

label = "Name: " + [LOT] + "\n" + [SUBDIVISION]

else:

label = "Name: " + [LOT]

return label

BrynDunbar

Thank you for the reply.

I typed this into the Expression box with Python selected as the Parser and Advanced checked.

def FindLabel ( [LOT] , [SUBDIVISION] 😞

return "Name: " + [LOT] + '\n' + [SUBDIVISION]

I got an error,

Error 0 on line 0.

Error running expression:

FindLabel(ESRIExpressionArg0,ESRIExpressionArg1)

Tracedback (most recent call last):

File"<expression>", line 1, in <module>

File"<string>", line 2, in FindLabel

TypeError: cannot concatenate 'str' and 'NoneType' objects.

Does this error have to do with 'null' values? Or do I need to cast the variable or do a type conversion?

XanderBakker

It is very similar:

def FindLabel ( [LOT] , [SUBDIVISION]  ):
  return [LOT] +  "\n"  +  [SUBDIVISION]

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.