Remove first seven characters from field string using Field Calculator

Question

I wrote an expression/codeblock that removes the words "Nearby:" from the field value string if it has it.

Codeblock:

import string
def splitme(s):
    if s[:7] == "Nearby:": 
        aList = string.split(s, ":")
        return aList[1]

    else:
        pass

Expression:

splitme(!Street1!)

It removes "Nearby:" on all fields that contain it in the string but it also deletes the strings in the field records that are fine and do not contain "Nearby:". I thought the if/else check would handle this.

Answer 1

as written, the function returns a string with 'Nearby:' removed only if the 'Nearby:' test is true (if 'nearby:' exists within the field string. If the test is false, the function is set to 'pass', which returns no value (deleting the original value in the process)

to remedy, a value should be passed for both cases - true and false - something like

def splitme(s):
    if s[:7] == "Nearby:": 
        aList = string.split(s, ":")
        return aList[1]
    else:
        return(s) #return the original string if the test fails
        #pass

Answer 2

Assuming "Nearby:" is at the beginning of the line (and never in the middle), you can simplify this logic with a one-liner, without the need of a codeblock:

str(!Street1!).replace('Nearby:', '').lstrip()