It seems so "dirty" emptying a list in this way:
while len(alist) > 0 : alist.pop()
Does a clear way exist to do that?
You can create an empty list using an empty pair of square brackets [] or the type constructor list() , a built-in function that creates an empty list when no arguments are passed. Square brackets [] are commonly used in Python to create empty lists because it is faster and more concise.
The clear() method removes all items from the list.
Using del() The del() function you can selectively remove items at a given index or you can also remove all the elements, making the list empty. In the below example we take a list, remove the element at index 2.
The clear() method removes all the elements from a list.
This actually removes the contents from the list, but doesn't replace the old label with a new empty list:
del lst[:]
Here's an example:
lst1 = [1, 2, 3] lst2 = lst1 del lst1[:] print(lst2)
For the sake of completeness, the slice assignment has the same effect:
lst[:] = []
It can also be used to shrink a part of the list while replacing a part at the same time (but that is out of the scope of the question).
Note that doing lst = []
does not empty the list, just creates a new object and binds it to the variable lst
, but the old list will still have the same elements, and effect will be apparent if it had other variable bindings.
If you're running Python 3.3 or better, you can use the clear()
method of list
, which is parallel to clear()
of dict
, set
, deque
and other mutable container types:
alist.clear() # removes all items from alist (equivalent to del alist[:])
As per the linked documentation page, the same can also be achieved with alist *= 0
.
To sum up, there are four equivalent ways to clear a list in-place (quite contrary to the Zen of Python!):
alist.clear() # Python 3.3+
del alist[:]
alist[:] = []
alist *= 0
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