Python: pop from empty list

I am using below line in a loop in my code

importer = exporterslist.pop(0)

If exporterslist has no entries it returns error: IndexError: pop from empty list. How can I bypass exporterslist with no entries in it?

One idea I can think of is if exporterslist is not None then importer = exporterslist.pop(0)
else get the next entry in the loop.
If the idea is correct, how to code it in python?

Answers:

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Method 1

You’re on the right track.

if exporterslist: #if empty_list will evaluate as false.
    importer = exporterslist.pop(0)
else:
    #Get next entry? Do something else?

Method 2

This one..

exporterslist.pop(0) if exporterslist else False

..is somewhat the same as the accepted answer of @nightshadequeen’s just shorter:

>>> exporterslist = []   
>>> exporterslist.pop(0) if exporterslist else False   
False

or maybe you could use this to get no return at all:

exporterslist.pop(0) if exporterslist else None

>>> exporterslist = [] 
>>> exporterslist.pop(0) if exporterslist else None
>>>

Method 3

You can also use a try/except

try:
    importer = exporterslist.pop(0)
except IndexError as e:
    print(e)

If you are always popping from the front you may find a deque a better option as deque.popleft() is 0(1).

Method 4

Use this:

if exporterslist:
    importer = exporterslist.pop(0)

Method 5

You can also .pop() only if the list has items in it by determining if the length of the list is 1 or more:

if len(exporterslist) >= 1: #if the length of the list is equal or higher than one
    importer = exporterslist.pop()

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0