1

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?

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  • Unless there's a typo that seems like a bug. Out of curiosity, did you try !Street_Name!.split(" ")[0:-1] – KJYDavis Dec 2 '16 at 0:29
  • @ KJYDavis Yes. I did. – geoJshaun Dec 2 '16 at 2:32
6

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

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  • +1 Since the first code has an advantage over removing the extra spaces within the phrase and seems like 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
  • @Paul That's odd because I'm making this selection prior to the calculation - Street_Name like ('% _') - so the records should all have at least one space and single final character. if there was a space at the end of a record it wouldn't be added to the selection. I there were no spaces then the select criteria (a space) between % and _ should exclude those records prior to the calculation attempt. I'll give your code a shot. – geoJshaun Dec 2 '16 at 2:25
0

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()

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  • @faith_dur Thanks but still the same result. – geoJshaun Dec 1 '16 at 23:40

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