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?
To repeat the elements of the list n times, we will first copy the existing list into a temporary list. After that, we will add the elements of the temporary list to the original list n times using a for loop, range() method, and the append() method.
To create a list of n placeholder elements, multiply the list of a single placeholder element with n . For example, use [None] * 5 to create a list [None, None, None, None, None] with five elements None . You can then overwrite some elements with index assignments.
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
.
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