Python：从空列表中弹出

Python: pop from empty list

提问by Karvy1

I am using below line in a loop in my code

我在我的代码中循环使用下面的行

importer = exporterslist.pop(0)

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

如果 exporterslist 没有条目或有条目null，则返回error: IndexError: pop from empty list。如何绕过没有条目的exporterslist？

One idea I can think of is if exporterslist is not null then importer = exporterslist.pop(0)else get the next entry in the loop. If the idea is correct, how to code it in python?

我能想到的一个想法是，如果 exporterslist 不为 null，则importer = exporterslist.pop(0)获取循环中的下一个条目。如果想法是正确的，如何在python中对其进行编码？

采纳答案by NightShadeQueen

You're on the right track.

你在正确的轨道上。

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

回答by Utsav T

Use this:

用这个：

if exporterslist:
    importer = exporterslist.pop(0)

回答by Padraic Cunningham

You can also use a try/except

您也可以使用 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 dequea better option as deque.popleft() is 0(1).

如果您总是从前面弹出，您可能会发现deque是更好的选择，因为 deque.popleft() 是 0(1)。

回答by Josh Woods

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

您也可以通过确定列表的长度是否为 1 或更多，仅当列表中有项目时 .pop() ：

if len(exporterslist) > 1:
    importer = exporterslist.pop()

回答by Gergely M

This one..

这个..

exporterslist.pop(0) if exporterslist else False

exporterslist.pop(0) if exporterslist else False

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

.. 与@nightshadequeen 的接受答案有些相同，只是更短：

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

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

或者也许你可以用它来获得任何回报：

exporterslist.pop(0) if exporterslist else None

exporterslist.pop(0) if exporterslist else None

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