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

Question

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?

Solution 1

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]

Solution 2

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

Solution 3

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()]

Solution 4

This should work:

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

Solution 5

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()]

View more solutions