How to create an integer array in Python?

Question

It should not be so hard. I mean in C,

int a[10];

is all you need. How to create an array of all zeros for a random size. I know the zeros() function in NumPy but there must be an easy way built-in, not another module.

Answer 1

>>> a = [0] * 10
>>> a
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

Answer 2

a = 10 * [0]

gives you an array of length 10, filled with zeroes.

Answer 3

two ways:

x = [0] * 10
x = [0 for i in xrange(10)]

Edit: replaced range by xrange to avoid creating another list.

Also: as many others have noted including Pi and Ben James, this creates a list, not a Python array. While a list is in many cases sufficient and easy enough, for performance critical uses (e.g. when duplicated in thousands of objects) you could look into python arrays. Look up the array module, as explained in the other answers in this thread.

Answer 4

If you are not satisfied with lists (because they can contain anything and take up too much memory) you can use efficient array of integers:

array.array('i')

See here

If you need to initialize it,

a=array.array('i',(0 for i in range(0,10)))

Answer 5

Use the array module. With it you can store collections of the same type efficiently.

>>> import array
>>> import itertools
>>> a = array_of_signed_ints = array.array("i", itertools.repeat(0, 10))

For more information - e.g. different types, look at the documentation of the array module. For up to 1 million entries this should feel pretty snappy. For 10 million entries my local machine thinks for 1.5 seconds.

The second parameter to array.array is a generator, which constructs the defined sequence as it is read. This way, the array module can consume the zeros one-by-one, but the generator only uses constant memory. This generator does not get bigger (memory-wise) if the sequence gets longer. The array will grow of course, but that should be obvious.

You use it just like a list:

>>> a.append(1)
>>> a.extend([1, 2, 3])
>>> a[-4:]
array('i', [1, 1, 2, 3])
>>> len(a)
14

...or simply convert it to a list:

>>> l = list(a)
>>> len(l)
14

Surprisingly

>>> a = [0] * 10000000

is faster at construction than the array method. Go figure! :)

Answer 6

import random

def random_zeroes(max_size):
  "Create a list of zeros for a random size (up to max_size)."
  a = []
  for i in xrange(random.randrange(max_size)):
    a += [0]

Use range instead if you are using Python 3.x.

Answer 7

If you need to initialize an array fast, you might do it by blocks instead of with a generator initializer, and it's going to be much faster. Creating a list by [0]*count is just as fast, still.

import array

def zerofill(arr, count):
    count *= arr.itemsize
    blocksize = 1024
    blocks, rest = divmod(count, blocksize)
    for _ in xrange(blocks):
        arr.fromstring("\x00"*blocksize)
    arr.fromstring("\x00"*rest)

def test_zerofill(count):
    iarr = array.array('i')
    zerofill(iarr, count)
    assert len(iarr) == count

def test_generator(count):
    iarr = array.array('i', (0 for _ in xrange(count)))
    assert len(iarr) == count

def test_list(count):
    L = [0]*count
    assert len(L) == count

if __name__ == '__main__':
    import timeit
    c = 100000
    n = 10
    print timeit.Timer("test(c)", "from __main__ import c, test_zerofill as test").repeat(number=n)
    print timeit.Timer("test(c)", "from __main__ import c, test_generator as test").repeat(number=n)
    print timeit.Timer("test(c)", "from __main__ import c, test_list as test").repeat(number=n)

Results:

(array in blocks) [0.022809982299804688, 0.014942169189453125, 0.014089107513427734]
(array with generator) [1.1884641647338867, 1.1728270053863525, 1.1622772216796875]
(list) [0.023866891860961914, 0.035660028457641602, 0.023386955261230469]