Creating a list in Python with multiple copies of a given object in a single line
You can use the *
operator :
L = ["a"] * 10
L = [0] * 10
L = [["x", "y"]] * 10
Be careful this create N copies of the same item, meaning that in the third case you create a list containing N references to the ["x", "y"]
list ; changing L[0][0]
for example will modify all other copies as well:
>>> L = [["x", "y"]] * 3
>>> L
[['x', 'y'], ['x', 'y'], ['x', 'y']]
>>> L[0][0] = "z"
[['z', 'y'], ['z', 'y'], ['z', 'y']]
In this case you might want to use a list comprehension:
L = [["x", "y"] for i in range(10)]
itertools.repeat()
is your friend.
L = list(itertools.repeat("a", 20)) # 20 copies of "a"
L = list(itertools.repeat(10, 20)) # 20 copies of 10
L = list(itertools.repeat(['x','y'], 20)) # 20 copies of ['x','y']
Note that in the third case, since lists are referred to by reference, changing one instance of ['x','y'] in the list will change all of them, since they all refer to the same list.
To avoid referencing the same item, you can use a comprehension instead to create new objects for each list element:
L = [['x','y'] for i in range(20)]
(For Python 2.x, use xrange()
instead of range()
for performance.)
You could do something like
x = <your object>
n = <times to be repeated>
L = [x for i in xrange(n)]
Substitute range(n) for Python 3.