Create list of single item repeated N times

Question

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 = len

Answer 1

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.

Answer 2

>>> [5] * 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=[5]
>>> y=[x] * 4
>>> y
[[5], [5], [5], [5]]
>>> y[0][0] = 6
>>> y
[[6], [6], [6], [6]]

Answer 3

Itertools has a function just for that:

import itertools
it = itertools.repeat(e,n)

Of course itertools gives you a iterator instead of a list. [e] * n gives you a list, but, depending on what you will do with those sequences, the itertools variant can be much more efficient.

Answer 4

As others have pointed out, using the * operator for a mutable object duplicates references, so if you change one you change them all. If you want to create independent instances of a mutable object, your xrange syntax is the most Pythonic way to do this. If you are bothered by having a named variable that is never used, you can use the anonymous underscore variable.

[e for _ in xrange(n)]

Answer 5

[e] * n

should work

Answer 6

Create List of Single Item Repeated n Times in Python

Immutable items

For immutable items, like None, bools, ints, floats, strings, tuples, or frozensets, you can do it like this:

[e] * 4

Note that this is best only used with immutable items (strings, tuples, frozensets, ) in the list, because they all point to the same item in the same place in memory. I use this frequently when I have to build a table with a schema of all strings, so that I don't have to give a highly redundant one to one mapping.

schema = ['string'] * len(columns)

Mutable items

I've used Python for a long time now, and I have never seen a use-case where I would do the above with a mutable instance. Instead, to get, say, a mutable empty list, set, or dict, you should do something like this:

list_of_lists = [[] for _ in columns]

The underscore is simply a throwaway variable name in this context.

If you only have the number, that would be:

list_of_lists = [[] for _ in range(4)]

The _ is not really special, but your coding environment style checker will probably complain if you don't intend to use the variable and use any other name.

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Caveats for using the immutable method with mutable items:

Beware doing this with mutable objects, when you change one of them, they all change because they're all the same object:

foo = [[]] * 4
foo[0].append('x')

foo now returns:

[['x'], ['x'], ['x'], ['x']]

But with immutable objects, you can make it work because you change the reference, not the object:

>>> l = [0] * 4
>>> l[0] += 1
>>> l
[1, 0, 0, 0]

>>> l = [frozenset()] * 4
>>> l[0] |= set('abc')
>>> l
[frozenset(['a', 'c', 'b']), frozenset([]), frozenset([]), frozenset([])]

But again, mutable objects are no good for this, because in-place operations change the object, not the reference:

l = [set()] * 4
>>> l[0] |= set('abc')    
>>> l
[set(['a', 'c', 'b']), set(['a', 'c', 'b']), set(['a', 'c', 'b']), set(['a', 'c', 'b'])]