Question

What would be a more efficient way to do the following:

directors = get_element_or_none(title_node, 'Director')
producers = get_element_or_none(title_node, 'Producer')
writers = get_element_or_none(title_node, 'Writer')

if directors:
    directors = [director.strip() for director in directors.split(',')]
if producers:
    producers = [producer.strip() for producer in producers.split(',')]
if writers:
    writers = [writer.strip() for writer in writers.split(',')]

Answer 1

To always produce a list (possibly empty):

directors = [director.strip() for director in directors.split(',')] if directors else []
# etc.

or use map(str.strip, ...):

directors = map(str.strip, directors.split(',')) if directors else []

but in Python 3 that requires an explicit call to list():

directors = list(map(str.strip, directors.split(','))) if directors else []

because map() returns an iterator instead.

or use a helper function:

def tolist(commaseparated):
    return [s.strip() for s in commaseparated.split(',')] if commaseparated else []

directors = tolist(directors)
producers = tolist(producers)
writers   = tolist(writers)

or, with the map version:

def tolist(commaseparated):
    return map(str.strip, commaseparated.split(',')) if commaseparated else []

The split-and-strip operation could probably be merged into one function with the get_element_or_none() call, but that depends on what else you might use the tolist() functionality for.