Python - converting a string of numbers into a list of int [duplicate]

Answer 1

I guess the dirtiest solution is this:

list(eval('0, 0, 0, 11, 0, 0, 0, 11'))

Answer 2

Try this :

import re
[int(s) for s in re.split('[\s,]+',example_string)]

Answer 3

You can also use list comprehension on splitted string

[ int(x) for x in example_string.split(',') ]

Answer 4

it should work

example_string = '0, 0, 0, 11, 0, 0, 0, 0, 0, 19, 0, 9, 0, 0, 0, 0, 0, 0, 11'
example_list = [int(k) for k in example_string.split(',')]

Answer 5

Split on commas, then map to integers:

map(int, example_string.split(','))

Or use a list comprehension:

[int(s) for s in example_string.split(',')]

The latter works better if you want a list result, or you can wrap the map() call in list().

This works because int() tolerates whitespace:

>>> example_string = '0, 0, 0, 11, 0, 0, 0, 0, 0, 19, 0, 9, 0, 0, 0, 0, 0, 0, 11'
>>> list(map(int, example_string.split(',')))  # Python 3, in Python 2 the list() call is redundant
[0, 0, 0, 11, 0, 0, 0, 0, 0, 19, 0, 9, 0, 0, 0, 0, 0, 0, 11]
>>> [int(s) for s in example_string.split(',')]
[0, 0, 0, 11, 0, 0, 0, 0, 0, 19, 0, 9, 0, 0, 0, 0, 0, 0, 11]

Splitting on just a comma also is more tolerant of variable input; it doesn't matter if 0, 1 or 10 spaces are used between values.

Answer 6

number_string = '0, 0, 0, 11, 0, 0, 0, 0, 0, 19, 0, 9, 0, 0, 0, 0, 0, 0, 11'

number_string = number_string.split(',')

number_string = [int(i) for i in number_string]