[FIXED] Create list of single item repeated N times

Issue

I want to create a series of lists, all of varying lengths. Each list will contain the same element e, repeated n times (where n = length of the list).

How do I create the lists, without using a list comprehension [e for number in xrange(n)] for each list?

Solution

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.

Answered By - Mark Byers