Concatenating fields with label expressions using Python

Question

I am trying to combine three numeric fields together for a feature's label. I only want to label the feature if it is over 0. The labeled numbers need to be separated a dash ("-"). This is what I attempted to use as the label expression in ArcMap 10.2:

def FindLabel ( [a1], [a2], [a3] ):
 if ([a3]) == 0 and int([a2]) == 0 and ([a1]) == 0:
 label = " "
 elif ([a3]) == 0 and int([a2]) == 0
 label = [a1]
 elif ([a3]) == 0
 label = [a1] + "-" + [a2]
 else:
 label = [a1] + "-" + [a2] + "-" + [a3]
 return label

• papadoo: What errors or issues did you have?
• juturna: I get this when I try to verify the expression or hit OK: "Error 0 on line 0. SyntaxError: invalid syntax(<string>, line 4)."
• recurvata: What are the field types for a1, a2, a3, and are there Null values for any?
• juturna: All three field types are String and there are some null values.
• Michael Stimson: You're missing the colon on the end of the elif expressions, that would be why it's not working. For example change elif ([a3]) == 0 and int([a2]) == 0 to elif ([a3]) == 0 and int([a2]) == 0: for correct syntax. If you have null values the expression is [A3] == None for null.

Answer 1

This function doesn't need to be too complex. Convert the items you want into a list, then join the result.

def FindLabel([A1], [A2], [A3]):
 args = [A1, A2, A3]
 items = [str(x) for x in args if bool(x) and int(x) > 0]
 if any(items):
 return '-'.join(items)
 else:
 return ' '

If this were proper Python, you would normally use *args to accept any number of arguments into a list.

• Michael Stimson: That's not how an ArcGis label expression works MikeT, have a read of this resources.arcgis.com/en/help/main/10.1/index.html#//… that functionallity will work if you supply that as another sub and call it from the find_label sub - that name is special, but if you name it join_label and call it from find_label it works great!
• Mike T: @MichaelMiles-Stimson OK, I see how it works.
• Michael Stimson: Awesome! Great expression for its simplicity and diversity: '_'.join, ' '.join etc.. works just as well. I like the use of any(items), that's not one I've seen before, usually I do if len(items) != 0 but that can mean a list of empty items. Thanks for that, I'll be using it in the future.
• Paul: +1 for a nice pythonic approach that isn't limited to only 3 fields with set names.

Answer 2

There are missing semi-colons and problems with the parentheses. then you should make sure that you always have a string output.

EDIT : sorry, I've given a solution for the field calculator. This solution is updted for the labelling engin.

def FindLabel ( [A1], [A2], [A3] ):
 a1=[A1]
 a2=[A2]
 a3=[A3]
 if ( (int(a1) == 0) and (int(a2) == 0) and (int(a3) == 0)):
 label = " "
 elif ( (int(a2) == 0) and (int(a3) == 0):
 label = str(a1)
 elif (int(a3) == 0):
 label = a1 + "-" + a2
 else:
 label = a1 + "-" + a2 + "-" + a3
 return label

note that the following test are equivalent

 int(a1) == 0 

and

 a1='0'

for testing if a field is null, you can use

 if not a1:

• juturna: Thanks for the response. I'm (very) new to Python. My fields are all Strings. I tried the code you wrote, as well as the following (since my fields are Strings). Neither worked and I get the same error. ` def FindLabel ( a1, a2, a3 ): if ( (int(a3) == 0) and (int(a2) == 0) and (int(a1) == 0)): label = " " elif ( (int(a3) == 0) and (int(a2) == 0): label = str(a1) elif (int(a3) == 0): label = str(a1) + "-" + a2 else: label = str(a1) + "-" + a2 + "-" + str(a3) return label `
• GISHuman: +1 for the edited solution for labelling. If this doesn't work I'd guess that something is wrong with the data
• papadoo: your first elif statement is checking a2 twice, I'm assuming you meant to check a2 and a3.
• radouxju: @papadoo you are right, it's a typo. Corrected by now.