Is preallocating a list more efficient?

Question

The first one is this:

n = int(input())
a = []
for i in range(n):
    a.append(int(input()))

The second one is this:

n = int(input())
a = [0]*n
for i in range(n):
    a[i] = int(input())

I have always been using first one, but the second one just came up to my mind. I don't know how append works, but I thought the second one could be more efficient than the first one.

Can I get answer of this problem?

If you can help it, get all the inputs in a single line separated by spaces or commas, and then split and convert to integer to create the list you are looking for. — recentadvances, Nov 08 '20 at 02:37
Does this answer your question? [Python - Create a list with initial capacity](https://stackoverflow.com/questions/311775/python-create-a-list-with-initial-capacity) Per the accepted answer, "Conclusion: [preallocating the list] barely matters. Premature optimization is the root of all evil." — Brian61354270, Nov 08 '20 at 02:38

score 1 · Answer 1 · edited Dec 21 '20 at 01:58

I believe there should not be a definitive answer as it will depend on which version of Python, which platform, etc.

But in general, this is what you want to do to test it out real quick:

Create a file, say, testcode.py and make this:

def first():
    n = 100000
    a = []
    for i in range(n):
        a.append(int("42"))

def second():
    n = 100000
    a = [0]*n
    for i in range(n):
        a[i] = int("42")

Then run with timeit:

$ python -m timeit -s 'from testcode import first, second' -- 'first()'
10 loops, best of 5: 27.1 msec per loop
$ python -m timeit -s 'from testcode import first, second' -- 'second()'
10 loops, best of 5: 23.6 msec per loop

Is preallocating a list more efficient?

1 Answers1