ArcGIS Pro Python Field Calculating

Question

I am trying to use Python in ArcGIS Pro to cleanse some data via imported Excel sheets. The expression works, but the Code Block if-else statement to return nothing if !Street_Id! is returns an error.

Address

str(!Street_ID!) + ' ' + !Street!.split("/")[0] 

Code Block

def f(x): ' '
if !Street_ID! = 'Null'
return x
else return str(!Street_ID!) + ' ' + !Street!.split("/")[0]

Error

File"<string>",line 2
if!Street_ID! = 'Null'
   ^
SynatxError: invalid syntax

Answer 1

There's a bunch of things wrong with your code:

• Incorrect code invocation
• Incorrect function definition (placing a non-comment to the right of the colon)
• Poor function naming (far better to keep everything you use, so you don't have to debug every time you use it, but that requires descriptive naming)
• Failure to pass in the values to be used in the expression string
• No indentation (required part of the Python language)
• Failure to place a space between the if keyword and logical expression
• Using assignment (=) instead of equivalence (==) for a logical test
• Using expression field delimiters in Python code
• Assuming a NULL value will match the string 'Null' (instead it maps to None)
• Improper use of early exit (after a return you don't need an else)
• Performing string math (using format is best practice)

When the smoke clears, the corrected result looks like this:

---

Expression: usefulName(!Street!,!StreetID!,'nullValue')

---

Expression type: Python

---

Code block:

def usefulName(street_val,street_id,x):
    if (street_id == None):
        return x
    return "{:s} {:s}".format(street_id,street_val.split("/")[0])

---

If you want to avoid early exit (some organizations/programs forbid it as confusing), then the code block would look like:

def usefulName(street_val,street_id,x):
    result = x
    if (street_id != None):
        result = "{:s} {:s}".format(street_id,street_val.split("/")[0])
    return result

And the ternary logic form would look like:

def usefulName(street_val,street_id,x):
    return "{:s} {:s}".format(street_id,street_val.split("/")[0]) if street_id else x

Which actually compresses down to a simple Python expression of:

"{:s} {:s}".format(!StreetID!,!Street!.split("/")[0]) if !StreetID! else 'nullValue'

though that's probably a bit obscure for reuse.

Answer 2

You need a double equal sign, a colon at the ends, and proper indentation for sure.

Consider writing your code in a Python IDE that shows syntax errors during code development.

 def f(x):
    if !Street_ID! == 'Null':
        return x
    else:
        return str(!Street_ID!) + ' ' + !Street!.split("/")[0]