I know a list comprehension will do this, but I was wondering if there is an even shorter (and more Pythonic?) approach.
I want to create a series of lists, all of varying length. Each list will contain the same element e, repeated n times (where n = length of the list). How do I create the lists, without doing
[e for number in xrange(n)]
for each list?
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.
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(' * 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!!!
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
>>>  * 4 [5, 5, 5, 5]
Be careful when the item being repeated is a list. The list will not be cloned: all the elements will refer to the same list!
>>> x= >>> y=[x] * 4 >>> y [, , , ] >>> y = 6 >>> y [, , , ]