Converting a String to List in Python

Question

Hi, I'm trying to create a list from arguments I receive in a url.

e.g I have:

 user.com/?users=0,1,2

Now when I receive it in the request it comes as a string. I want to make a list out of "0,1,2" [0,1,2]

Answer 1

To go from "0,1,2" to ['0','1','2'] its just "0,1,2".split(",")

So if you have it in a variable users, then users.split(",") will give you the list.

If you need them as ints instead of strings, it would be [int(x) for x in users.split(',')].

Answer 2

To convert the string to a list, use split.

To convert the list of strings to a list of integers, use a list comprehension with int.

So putting it all together, it looks something like this:

s = '0,1,2'
l = [int(x) for x in s.split(',')]

Results:

[1, 2, 3]

Answer 3

Use the split method. Example:

>>> "0,1,2".split(",")
['0', '1', '2']

Or even,

>>> [int(x) for x in "0,1,2".split(",")]
[0, 1, 2]

Answer 4

import ast
x=ast.literal_eval('0,1,2')
print(x)
# (0, 1, 2)

ast.literal_eval is like eval, but completely safe since it restricts the string to literals such as strings, numbers, tuples, lists, dicts, booleans and None.

Another alternative, not yet mentioned, is to use map:

x=map(int,'0,1,2'.split(','))

Answer 5

You can use following code:

s = 'user.com/?users=0,1,2'
s.rpartition('?users=')[2].split(',')

Answer 6

This question was originally tagged Django, so I'll proceed with that in mind.

Inside your view function, the request object has a GET attribute that is an instance of a QueryDict. If you always know that you are going to get a comma separated list of integers for the key "users", you could do something like this in your view function:

users_list = request.GET('users', "").split(',')

That will give you a list of strings, or an empty list if "users" wasn't supplied in GET. If you wanted a list of integers you could process it further with a list comprehension:

users_list = [int(x) for x in users_list]