text = "42 0"
nums = [int(n) for n in text.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()]
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()]
Given: text = "42 0"
import re
numlist = re.findall('\d+',text)
print(numlist)
['42', '0']
Just use strip() to remove empty spaces and apply explicit int conversion on the variable.
Ex:
a='1 , 2, 4 ,6 '
f=[int(i.strip()) for i in a]
This should work:
[ int(x) for x in "40 1".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()))
Here is my answer for python 3.
some_string = "2 3 8 61 "
list(map(int, some_string.strip().split()))
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]
I suggest checking if the substring is a digit:
In [1]: [int(i) for i in '1 2 3a'.split() if i.isdigit()]
Out[1]: [1, 2]
Source: Stackoverflow.com