Shouldn't this return everything but the last word in the street name?
!Street_Name!.split(" ")[:-1]
Python parser is checked. Receiving nulls after calculation. If I try to slice[0], I get the first word. If I try to slice [-1] I get the last word. Why won't [:-1] return anything?
You probably need to join it back afterwards, otherwise you might be trying to calculate a field with a list:
" ".join('101 1st Street'.split(" ")[:-1])
As @faith_dur noted though, this will fail when there are no spaces in the string. A better way to do this would be with rsplit:
for s in ['101 1st Street', '500 main street extension', 'NoSpacesHere']:
print(s.rsplit(' ', 1)[0])
Which prints:
101 1st
500 main street
NoSpacesHere
rsplit is the elegant way of doing this instead of checking if the phrase is just one-word.
– fatih_dur
Dec 2 '16 at 0:43
Most probably there is/are white space(s) at the end of your strings. Try !Street_Name!.strip().split(" ")[:-1].
EDIT
I missed the colon in your slicing query. Please see below:
"xxxx yyyy".split(" ")[-1] returns--> yyyy
"xxxx yyyy".split(" ")[:-1] returns--> ['xxxx']
"xxxxyyyy".split(" ")[:-1] returns--> []
Basically, [:-1] returns anything but the last slice ("xxxx yyyy zzzz".split()[:-1] returns ['xxxx', 'yyyy']) and if you have only one word in your street address, naturally it returns an empty list. I think you are looking for
!Street_Name!.strip().split(" ")[:-1] if len(!Street_Name!.strip().split(" "))>1 else !Street_Name!.strip()
!Street_Name!.split(" ")[0:-1]– KJYDavis Dec 2 '16 at 0:29