Clearing a list
I find it annoying that I can't clear a list. In this example:
a = []
a.append(1)
a.append(2)
a = []
The second time I initialize a to a blank list, it creates a new instance of a list, which is in a different place in memory, so I can't use it to reference the first, not to mention it's inefficient.
The only way I can see of retaining the same pointer is doing something like the following:
for i in range(len(a)):
a.pop()
This seems pretty long-winded though, is there a better way of solving this?
Asked by: Rafael934 | Posted: 28-01-2022
Answer 1
You are looking for:
del L[:]
Answer 2
I'm not sure why you're worried about the fact that you're referencing a new, empty list in memory instead of the same "pointer".
Your other list is going to be collected sooner or later and one of the big perks about working in a high level, garbage-collected language is that you don't normally need to worry about stuff like this.
Answered by: Hailey602 | Posted: 01-03-2022Answer 3
this can help you::)
L[:] = []
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