You can also write:
[e] * n
You should note that if e is for example an empty list you get a list with n references to the same list, not n independent empty lists.
Performance testing
At first glance it seems that repeat is the fastest way to create a list with n identical elements:
>>> timeit.timeit('itertools.repeat(0, 10)', 'import itertools', number = 1000000)
0.37095273281943264
>>> timeit.timeit('[0] * 10', 'import itertools', number = 1000000)
0.5577236771712819
But wait – it’s not a fair test…
>>> itertools.repeat(0, 10)
repeat(0, 10) # Not a list!!!
The function itertools.repeat doesn’t actually create the list, it just creates an object that can be used to create a list if you wish! Let’s try that again, but converting to a list:
>>> timeit.timeit('list(itertools.repeat(0, 10))', 'import itertools', number = 1000000)
1.7508119747063233
So if you want a list, use [e] * n. If you want to generate the elements lazily, use repeat.