What is the best way to create a new empty list in Python?
l = []
or
l = list()
I am asking this because of two reasons:
- Technical reasons, as to开发者_运维百科 which is faster. (creating a class causes overhead?)
- Code readability - which one is the standard convention.
Here is how you can test which piece of code is faster:
% python -mtimeit "l=[]"
10000000 loops, best of 3: 0.0711 usec per loop
% python -mtimeit "l=list()"
1000000 loops, best of 3: 0.297 usec per loop
However, in practice, this initialization is most likely an extremely small part of your program, so worrying about this is probably wrong-headed.
Readability is very subjective. I prefer []
, but some very knowledgable people, like Alex Martelli, prefer list()
because it is pronounceable.
list()
is inherently slower than []
, because
there is symbol lookup (no way for python to know in advance if you did not just redefine list to be something else!),
there is function invocation,
then it has to check if there was iterable argument passed (so it can create list with elements from it) ps. none in our case but there is "if" check
In most cases the speed difference won't make any practical difference though.
I use []
.
- It's faster because the list notation is a short circuit.
- Creating a list with items should look about the same as creating a list without, why should there be a difference?
I do not really know about it, but it seems to me, by experience, that jpcgt is actually right. Following example: If I use following code
t = [] # implicit instantiation
t = t.append(1)
in the interpreter, then calling t gives me just "t" without any list, and if I append something else, e.g.
t = t.append(2)
I get the error "'NoneType' object has no attribute 'append'". If, however, I create the list by
t = list() # explicit instantiation
then it works fine.
Just to highlight @Darkonaut answer because I think it should be more visible.
new_list = []
or new_list = list()
are both fine (ignoring performance), but append()
returns None
, as result you can't do new_list = new_list.append(something)
.
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