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   length of the list    How do I create the lists  without using a list comprehension  e for number in xrange n   for each list

User · Answer

Create List of Single Item Repeated n Times in Python

Depending on your use-case, you want to use different techniques with different semantics.

Multiply a list for 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 usually 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)

Multiply the list where we want the same item repeated

Multiplying a list gives us the same elements over and over. The need for this is rare:

[iter(iterable)] * 4

This is sometimes used to map an iterable into a list of lists:

>>> iterable = range(12)
>>> a_list = [iter(iterable)] * 4
>>> [[next(l) for l in a_list] for i in range(3)]
[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11]]

We can see that a_list contains the same range iterator four times:

>>> a_list
[<range_iterator object at 0x7fde73a5da20>, <range_iterator object at 0x7fde73a5da20>, <range_iterator object at 0x7fde73a5da20>, <range_iterator object at 0x7fde73a5da20>]

Mutable items

I've used Python for a long time now, and I have seen very few use-cases where I would do the above with mutable objects.

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.

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'])]

User · Answer

gt  gt  gt   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    gt  gt  gt  x  5   gt  gt  gt  y  x    4  gt  gt  gt  y   5    5    5    5    gt  gt  gt  y 0  0    6  gt  gt  gt  y   6    6    6    6

User · Answer

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

User · Answer

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

User · Answer

e    n   should work

User · Answer

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    gt  gt  gt  timeit timeit  itertools repeat 0  10     import itertools   number   1000000  0 37095273281943264  gt  gt  gt  timeit timeit   0    10    import itertools   number   1000000  0 5577236771712819   But wait - it s not a fair test      gt  gt  gt  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    gt  gt  gt  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

[python] Create list of single item repeated N times

Create List of Single Item Repeated n Times in Python

Multiply a list for Immutable items

Multiply the list where we want the same item repeated

Mutable items

Caveats for using the immutable method with mutable items:

Examples related to python

Examples related to list-comprehension

Examples related to multiplication

Examples related to replicate