Convert comma separated string to list

Question

I want to save a list of integers as a field in my model. Since this field is not set by default in Django, I do this with a CommaSeparatedIntegerField named x. In my opinion, I take this line and create a list of integers from it. When a model instance is created with a parameter n, I want it to xbe set in length n, with each element set to zero.

Here is the model:

class Foo(models.Model):
    id = models.IntegerField(default = 0)
    x = models.CommaSeparatedIntegerField(max_length = 10)

@classmethod
def create(cls, _id, n):
    user = cls(id = _id)
    user.class_num_shown = '0,' * n

Then I create an instance:

f = Foo.create(1, 4)
f.save()

And load it from the database and convert the string to a list:

f = Foo.objects.get(id = 1)
x_string = f.x
x_list = x_string.split(',')
print x_list

But it deduces [u'0,0,0,0,'], and not what I want, it will be [0,0,0,0]. How can I achieve the desired result?

Answer 1

values = map(int, '0,1,2,3,'.rstrip(',').split(','))

Answer 2

split() , . "".

, , .

, , .

>>> data = '0,0,0,0,'
>>> values = [int(x) for x in data.split(',') if x]
>>> values
[0, 0, 0, 0]