将字符串转换为Python中的列表-ConvertingaStringtoListinPython

31  

Use the split method. Example:

使用分割方法。例子:

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

Or even,

甚至,

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

6  

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

这个问题最初被标记为Django，所以我将继续考虑这个问题。

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:

在视图函数中，请求对象有一个GET属性，它是QueryDict类型的实例。如果你总是知道你会为关键的“用户”得到一个逗号分隔的整数列表，你可以在你的视图函数中做这样的事情:

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:

这将给您一个字符串列表，或者一个空列表，如果“用户”没有被提供。如果你想要一个整数列表，你可以通过列表理解进一步处理:

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

3  

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.

literal_eval类似于eval，但完全安全，因为它将字符串限制为字符串、数字、元组、列表、dicts、布尔值和None等常量。

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

另一个尚未提及的替代方法是使用map:

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

2  

To convert the string to a list, use split.

要将字符串转换为列表，请使用split。

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

若要将字符串列表转换为整数列表，请使用带有int的列表理解。

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

把它放在一起，看起来是这样的

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

Results:

结果:

[1, 2, 3]

2  

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

从“0 1 2”(' 0 ',' 1 ',' 2 ']只是“0 1 2”.split(",")

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

因此，如果在一个变量用户中有它，那么用户。split(“，”)将会给出列表。

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

如果您需要它们作为ints而不是string，那么它将是[int(x)在user .split('，')]中的x。

0  

You can use following code:

您可以使用以下代码:

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

-2  

You have eval to convert string to "real code":

您有eval将字符串转换为“真正的代码”:

Example:

例子:

>>> l = u"[('0','None'),('2','Taxable Goods'),('4','Shipping')]"
>>> type(l)


>>> t = eval(l)
>>> type(t)