How to split a string of space separated numbers into integers?

I have a string "42 0" (for example) and need to get an array of the two integers. Can I do a .split on a space?

Use str.split():

>>> "42 0".split()  # or .split(" ")
['42', '0']

Note that str.split(" ") is identical in this case, but would behave differently if there were more than one space in a row. As well, .split() splits on all whitespace, not just spaces.

Using map usually looks cleaner than using list comprehensions when you want to convert the items of iterables to built-ins like int, float, str, etc. In Python 2:

>>> map(int, "42 0".split())
[42, 0]

In Python 3, map will return a lazy object. You can get it into a list with list():

>>> map(int, "42 0".split())
<map object at 0x7f92e07f8940>
>>> list(map(int, "42 0".split()))
[42, 0]

text = "42 0"
nums = [int(n) for n in text.split()]

l = (int(x) for x in s.split())

If you are sure there are always two integers you could also do:

a,b = (int(x) for x in s.split())

or if you plan on modifying the array after

l = [int(x) for x in s.split()]

This should work:

[ int(x) for x in "40 1".split(" ") ]

Of course you can call split, but it will return strings, not integers. Do

>>> x, y = "42 0".split()
>>> [int(x), int(y)]
[42, 0]

or

[int(x) for x in "42 0".split()]

Other answers already show that you can use split() to get the values into a list. If you were asking about Python's arrays, here is one solution:

import array
s = '42 0'
a = array.array('i')
for n in s.split():
    a.append(int(n))

Edit: A more concise solution:

import array
s = '42 0'
a = array.array('i', (int(t) for t in s.split()))

You can split and ensure the substring is a digit in a single line:

In [1]: [int(i) for i in '1 2 3a'.split() if i.isdigit()]
Out[1]: [1, 2]

Given: text = "42 0"

import re
numlist = re.findall('\d+',text)

print(numlist)

['42', '0']

Use numpy's fromstring:

import numpy as np

np.fromstring("42 0", dtype=int, sep=' ')

>>> array([42,  0])